JEE Main + AdvancedJEE Main + AdvancedClass XIBasics of PhysicsScoring + conceptual
Units & Measurements
The grammar of physics — units, dimensions, significant figures and the honest reporting of error
🎯 Overview Chapter hero + roadmap
🔬 Interactive 3D · Assemble base-unit powers and watch dimensional homogeneity decide if an equation can be right.
Every result in physics begins with an act of comparison. When we say a length is 287.5 cm or that a pendulum takes 1.62 s to swing, we are quietly asserting that the world can be pinned to a number and a unit at the same time. That pairing is not decoration; it is the whole point. A physical quantity is meaningless as a bare number and equally meaningless as a bare unit, and this chapter is where you learn to respect both halves at once. Measurement is the bedrock of physics because it is the single bridge between an equation on paper and an experiment on a bench. Strip it away and you are left with algebra that no one can ever confirm or refute. This is why 'Units and Measurement' sits at the very front of the syllabus: it is the grammar every later chapter silently obeys. 🔉⇢
The first idea to internalise is that the universe of physical quantities, vast as it looks, is built from a tiny toolkit. We do not need a fresh unit for every quantity because quantities are interrelated. A small set of base quantities is chosen, standardised against carefully guarded reference standards, and everything else is expressed as a combination of those base units. Length, mass and time were historically enough for mechanics, but the modern International System of Units, the SI, rests on seven base quantities: length (metre), mass (kilogram), time (second), electric current (ampere), thermodynamic temperature (kelvin), amount of substance (mole) and luminous intensity (candela). Units built by combining these, such as the newton or the joule, are called derived units, and the complete collection of base and derived units is a system of units. 🔉⇢
There is real history and real engineering behind these choices. Before the SI was adopted, scientists in different countries worked in the CGS, FPS and MKS systems, and a single experiment could be reported three incompatible ways. The SI, developed by the international bureau of weights and measures and most recently revised in 2018, fixed this by tying each base unit to a constant of nature rather than to a fragile artefact. The metre is now defined through the fixed speed of light, the second through the caesium frequency, and the kilogram through the Planck constant. You are not expected to memorise those numerical values, but you should absorb the philosophy: definitions are revised so that measurements can grow ever more precise without the standard itself drifting. 🔉⇢
Because measured quantities span from the diameter of a proton to the size of a galaxy, we lean heavily on scientific notation and prefixes. Writing a number as $a \times 10^{b}$, with $a$ between 1 and 10, lets us compare a hydrogen atom ($\sim 10^{-10}$ m) with the diameter of the Earth ($\sim 10^{7}$ m) and see instantly that they differ by seventeen orders of magnitude. The exponent $b$ is the order of magnitude of the quantity, and estimating it is a skill in itself. Prefixes such as nano, micro, kilo and giga are just a spoken form of these powers of ten, and the decimal structure of the SI means conversions inside the system never involve awkward factors. Learning to move fluently between $2.308$ cm, $0.02308$ m and $23.08$ mm without panicking is one of the quiet superpowers this chapter hands you. 🔉⇢
The deepest of those superpowers is dimensional thinking. The nature of a physical quantity, as distinct from its magnitude, is captured by its dimensions: the powers to which the base quantities are raised. We write these in square brackets, so length is $[L]$, mass is $[M]$ and time is $[T]$. Volume is three lengths, so its dimensional formula is $[M^{0} L^{3} T^{0}]$; force is mass times acceleration, so it is $[M L T^{-2}]$. Notice that velocity, initial velocity, change in velocity and speed all share the dimension $[L T^{-1}]$, because dimensions care about the quality of a quantity, not its particular value. This abstraction, treating dimensions as algebraic symbols you can multiply and cancel, is what makes dimensional analysis so powerful. 🔉⇢
Dimensional analysis earns its keep in two ways. First, it checks equations through the principle of homogeneity: only terms with identical dimensions may be added or subtracted, so every term in a correct equation must share one dimension. Testing $x = x_{0} + v_{0}t + \tfrac{1}{2}at^{2}$ shows each term reduces to $[L]$, so the relation is dimensionally consistent. But heed the sharp warning NCERT repeats: a dimensionally correct equation need not be exactly correct, while a dimensionally wrong equation must be wrong. Dimensions cannot see pure numbers like $\tfrac{1}{2}$, nor distinguish $\tfrac{1}{2}mv^{2}$ from $\tfrac{3}{16}mv^{2}$. Second, the method can deduce relations: assuming a pendulum's period depends on length, mass and gravity as a product gives $T = k\sqrt{l/g}$, though the dimensionless constant $k = 2\pi$ must come from theory or experiment, never from dimensions alone. 🔉⇢
Alongside this abstract power runs a moral thread: honesty about uncertainty. Every measurement carries error, so a result should be reported in a way that reveals its own precision. The reliable digits together with the first uncertain digit are the significant figures, and the number of them signals how good the measurement is. When we write a period as $1.62$ s, we are declaring that the 1 and 6 are certain and only the 2 is doubtful. Reporting extra digits is not more scientific; it is misleading, because it claims a precision the instrument never delivered. The rules for counting significant figures, the special care needed for trailing zeros, and the recommendation to use scientific notation to remove ambiguity are all in service of one virtue: never pretend to know more than you measured. 🔉⇢
This honesty must survive arithmetic. When measured values are combined, the answer cannot be more precise than its ingredients. In multiplication and division the result keeps as many significant figures as the least precise factor, so dividing a mass of 4.237 g by a volume of 2.51 cm cubed gives a density reported as $1.69$ g cm$^{-3}$, not the eleven-decimal fantasy a calculator prints. In addition and subtraction the rule instead concerns decimal places: the sum keeps as many decimals as the term with the fewest, so $436.32 + 227.2 + 0.301$ rounds to $663.8$ g. Rounding itself follows a convention, including the even-odd rule for a trailing 5, and in long calculations you keep one guard digit through the intermediate steps to stop rounding errors from piling up. 🔉⇢
Errors deserve their own vocabulary because taming them is what separates a good experimenter from a careless one. Systematic errors push results consistently in one direction and arise from faulty instruments, zero offsets or flawed technique; they can often be found and corrected. Random errors scatter results unpredictably and are beaten down by repetition, which is precisely why 100 readings of a rod's diameter give a more reliable estimate than 5. From the spread we extract the mean absolute error $\Delta \bar{a}$, the relative error $\Delta a / a$, and the percentage error $(\Delta a / a)\times 100\%$. When quantities combine, errors propagate: for products and quotients the relative errors add, and for a power $a^{n}$ the relative error is multiplied by $|n|$. These propagation rules turn a jumble of raw readings into a defensible final number with an error bar attached. 🔉⇢
The chapter also introduces the beautiful precision instruments that make small lengths measurable. Least count is the smallest length an instrument can resolve, and the vernier callipers achieves a fine least count by sliding a scale whose divisions are deliberately slightly shorter than those of the main scale, so that coincidences between the two reveal fractions of a division. The screw gauge goes finer still, converting the rotation of a calibrated circular scale through a known pitch into a linear advance, so that its least count is the pitch divided by the number of circular divisions. Understanding zero error, the difference between pitch and least count, and how to combine main-scale and vernier or circular readings is exactly the kind of concrete, examinable skill that rewards careful practice. 🔉⇢
Finally, a word on why JEE loves this chapter so much. It is short, self-contained and ruthlessly objective, which makes it ideal for the one or two crisp questions that appear almost every year. Dimensional analysis lends itself to elegant single-answer problems; significant-figure and error-propagation questions test whether you truly understand precision rather than just plugging into formulas; and vernier and screw-gauge numericals reward students who can read an instrument correctly under pressure. Better still, mastering this material pays compound interest, because error analysis reappears in every physics practical and dimensional checks quietly rescue you in mechanics, electromagnetism and modern physics whenever a derived formula looks suspicious. Treat this chapter not as a warm-up to be rushed but as the toolkit you will carry through the entire course. 🔉⇢
What you will master 🔉⇢
After this chapter you can identify the seven SI base quantities and their units, and express any derived unit as a combination of them.
After this chapter you can rewrite quantities in scientific notation and read off their order of magnitude to compare vastly different scales.
After this chapter you can write the dimensional formula of a physical quantity and use the principle of homogeneity to check whether an equation can be right.
After this chapter you can deduce a product-type relation among quantities by the method of dimensions, and state clearly what dimensions cannot tell you.
After this chapter you can count significant figures correctly, apply the multiplication and addition rules, and round off using the standard convention.
After this chapter you can distinguish systematic from random errors and compute absolute, mean absolute, relative and percentage errors from a set of readings.
After this chapter you can propagate errors through sums, products, quotients and powers to attach a defensible uncertainty to a calculated result.
After this chapter you can determine the least count of a vernier callipers or screw gauge, account for zero error, and report a measured length with proper precision.
🧠 Concepts Map of the deep dives
This chapter breaks into the concepts below. Each has its own tab with interactive 3D, a full derivation, and a JEE worked example. The Contents tab has the complete prose.
🔉⇢
What you are looking at
A map of the whole chapter. Each box is one concept with its governing relation, and the arrows are the recommended order to learn them in.
Click any box to jump straight to that concept’s tab.
🗺️ Concept map — click any concept to open its tab. Each box shows the concept and its governing formula; the path is the recommended learning order.
The SI System and the Seven Base Units 🔉⇢
The International System of Units (SI) is the internationally accepted system of units in which seven base quantities are each measured against a defined base unit, while every other physical quantity is expressed as a combination of these base units.
Scientific notation writes any measured value as a number between one and ten multiplied by a power of ten, while decimal prefixes and the order of magnitude give a compact way to express and compare very large and very small physical quantities.
Dimensions & Dimensional Formulae 🔉⇢
The dimensions of a physical quantity are the powers to which the seven base quantities must be raised to represent that quantity, and the expression showing this combination is called its dimensional formula.
Dimensional analysis is the method of treating the dimensions $[M^a L^b T^c]$ of physical quantities as algebraic symbols, so that the homogeneity of an equation can be checked and, within limits, the form of a relation among quantities can be deduced.
Significant Figures 🔉⇢
The significant figures of a measured value are the digits known reliably together with the first uncertain digit, and they indicate the precision of the measurement fixed by the least count of the instrument.
Rounding & Arithmetic with Significant Figures 🔉⇢
When approximate measured values are combined, the calculated result must be rounded so that it reflects the precision of the least precise input, using the least-significant-figures rule for multiplication and division and the least-decimal-places rule for addition and subtraction.
Measuring Length: Parallax & Indirect Methods 🔉⇢
Lengths are measured directly with graduated scales for ordinary objects, and by indirect methods such as parallax and triangulation for distances too large or too small to measure directly, so that a single family of methods spans from atomic to cosmic ranges.
Measuring Mass and Time 🔉⇢
Mass is measured against the kilogram over a range spanning many orders of magnitude, using the unified atomic mass unit on the atomic scale, while time is measured against the second realised through the unperturbed hyperfine transition of the caesium atom.
Errors in Measurement: Systematic vs Random 🔉⇢
Errors are the unavoidable differences between a measured value and the true value of a physical quantity; systematic errors shift the observations consistently in one direction while random errors scatter them irregularly about the true value.
Absolute, Relative & Percentage Error 🔉⇢
The absolute error of a single observation is the magnitude of its deviation from the mean, the mean absolute error is the average of these deviations, the relative error is that mean divided by the measured value, and the percentage error is the relative error expressed as a percentage.
Propagation of Errors: Combining Uncertainties 🔉⇢
Propagation of errors is the set of rules that tells how the uncertainty in a computed result follows from the uncertainties in the measured quantities: for a sum or difference the absolute errors add, for a product or quotient the relative errors add, and for a power the relative error is multiplied by the magnitude of the exponent.
Least Count & the Vernier Callipers 🔉⇢
The least count of an instrument is the smallest measurement it can read reliably; for a vernier callipers it equals one main scale division less one vernier scale division, $\text{L.C.} = 1\,\text{MSD} - 1\,\text{VSD}$, and a reading is the main scale reading plus the coinciding vernier division multiplied by the least count.
Least Count & the Screw Gauge 🔉⇢
In a screw gauge the least count is the pitch over the number of divisions on the circular scale, $\text{L.C.} = \frac{\text{pitch}}{N}$, and a reading is the linear scale reading plus the coinciding circular scale division multiplied by the least count, correct for zero error and backlash.
📖 Contents Full chapter — read straight through
The complete chapter in reading order. Each section is narratable; the concept tabs add interactive 3D and worked examples on top of this text.
🔉⇢
What you are looking at
The chapter banner — the physics of this chapter in a single moving picture, with live symbols and values.
Every diagram below it is interactive: drag the controls and the numbers move with the drawing.
The SI System and the Seven Base Units 🔉⇢
🎯 Every derived unit is just powers of the base units. Pick a quantity and watch its mass (M), length (L) and time (T) exponents appear — that trio IS its unit.
🔉⇢
[Q] = Ma Lb Tc = —
What this shows
Every derived unit is just powers of the base units. Pick a quantity and watch its mass (M), length (L) and time (T) exponents appear — that trio IS its unit.
Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.
Definition: The International System of Units (SI) is the internationally accepted system of units in which seven base quantities are each measured against a defined base unit, while every other physical quantity is expressed as a combination of these base units. 🔉⇢
Every measurement of a physical quantity is a comparison with an arbitrarily chosen, internationally accepted reference standard called a unit, so the result is always a number accompanied by a unit. Although the number of physical quantities appears very large, they are interrelated with one another, so we actually need only a limited set of units. A few carefully selected quantities are treated as the fundamental or base quantities, and the units assigned to them are the base units. The units of all other physical quantities can be expressed as combinations of these base units, and such units obtained for the derived quantities are the derived units. Together the base units and the derived units form a complete system of units. This division matters because it keeps measurement economical: once the base units are standardised, the entire structure of mechanics, electricity and thermodynamics follows from combinations of them, without inventing a fresh independent unit for every new quantity a physicist chooses to define. 🔉⇢
In earlier times scientists of different countries used different systems for measurement, and three such systems were in extensive use until recently. The CGS system used the centimetre, gram and second; the FPS or British system used the foot, pound and second; and the MKS system used the metre, kilogram and second. The system now internationally accepted for measurement is the Systeme International d'Unites, abbreviated SI. Its standard scheme of symbols, units and abbreviations was developed by the International Bureau of Weights and Measures (BIPM) and was recently revised by the General Conference on Weights and Measures in November 2018. Because SI uses the decimal system, conversions within the system are simple and convenient, which is one reason it is now used throughout scientific, technical, industrial and commercial work across the world. A shared system also removes ambiguity when experimental results measured in one laboratory are compared with those obtained in another. 🔉⇢
SI has seven base units. Length is measured in the metre (m), mass in the kilogram (kg), time in the second (s), electric current in the ampere (A), thermodynamic temperature in the kelvin (K), amount of substance in the mole (mol) and luminous intensity in the candela (cd). Each base quantity is defined against a properly standardised reference so that laboratories everywhere realise the same unit to the same precision. From these seven base units the derived quantities such as velocity, force and mass density are built: their units are combinations of the base units treated as ordinary algebraic symbols, so that identical units cancel between numerator and denominator. Some derived units are given special names such as the newton, joule and watt, yet each of these can still be traced back to a combination of the seven base units, which keeps the whole measurement scheme internally consistent. 🔉⇢
The 2018 revision made SI more fundamental by fixing the numerical values of seven defining constants rather than relying on a physical artefact such as a metal cylinder. The second is defined by taking a fixed numerical value of the caesium frequency $\Delta\nu_{Cs}$; the metre by fixing the speed of light in vacuum $c$; the kilogram by fixing the Planck constant $h$; the ampere by fixing the elementary charge $e$; the kelvin by fixing the Boltzmann constant $k$; the mole by fixing the Avogadro constant $N_A$; and the candela by fixing the luminous efficacy $K_{cd}$ of a specified monochromatic radiation. Because every definition now fixes the numerical value of a constant of nature, the base units no longer depend on any single object that could be damaged or drift, and they can be realised to ever greater accuracy as measuring techniques get improved. The values themselves need not be remembered; they only indicate the extent of accuracy to which the constants are known. 🔉⇢
Besides the seven base units, two more units are defined to describe angle. The plane angle $d\theta$ is defined as the ratio of the length of arc $ds$ to the radius $r$, and its unit is the radian (rad). The solid angle $d\Omega$ is defined as the ratio of the intercepted area $dA$ of a spherical surface, described about the apex as centre, to the square of its radius $r$, and its unit is the steradian (sr). Because both are ratios of like quantities, the radian and the steradian are dimensionless quantities. A complete circle subtends $2\pi$ rad at its centre and a complete sphere subtends $4\pi$ sr. These angular measures are essential when connecting rotational quantities to translational quantities, yet they add no new dimension to the measurement system, which is why they are treated separately from the seven base quantities rather than as an eighth base quantity in their own right. 🔉⇢
It is worth remembering that a base unit and a derived unit differ only in role, not in status: the base units are the seven independently chosen standards, while every derived unit is a combination of them treated as algebraic symbols. When magnitudes of physical quantities are multiplied or divided, their units are handled in the same manner as ordinary algebraic symbols, so identical units in the numerator and denominator cancel, exactly as identical dimensions do. This is why the complete system of units, both the base units and the derived units, stays internally consistent: any measurement, however elaborate, can be reduced to the seven base units, and a change to the definition of one base unit propagates in a controlled way to every derived unit built upon it, without disturbing the numerical value of any measurement already expressed in base units. 🔉⇢
Derivation 🔉⇢
Start from the derived quantity force, defined by Newton's second law as the product of mass and acceleration: $F = m\,a$.
Write mass in its base unit kilogram and acceleration as length per time squared: $a$ has the unit $\mathrm{m\,s^{-2}}$.
Combine the base units as algebraic symbols: the unit of force is $\mathrm{kg}\times\mathrm{m\,s^{-2}} = \mathrm{kg\,m\,s^{-2}}$.
This combination of base units is given the special name newton (N), so $1\ \mathrm{N} = 1\ \mathrm{kg\,m\,s^{-2}}$.
Because the derived unit reduces entirely to base units, no new independent standard is needed, illustrating why only seven base units are required.
⚠️ JEE trap: Students often think the kilogram is still defined by a metal cylinder kept in France; since 2018 it is defined by fixing the numerical value of the Planck constant $h$, and no base unit now depends on a physical artefact. 🔉⇢
🎯 Any measured value is a mantissa m between 1 and 10 times a power of ten; the exponent n is what an SI prefix (nano, kilo, giga…) is a shorthand for. Slide n in steps of three and watch the prefix change.
🔉⇢
x = m × 10n (SI prefix names 10n)
What this shows
Any measured value is a mantissa m between 1 and 10 times a power of ten; the exponent n is what an SI prefix (nano, kilo, giga…) is a shorthand for. Slide n in steps of three and watch the prefix change.
Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.
Definition: Scientific notation writes any measured value as a number between one and ten multiplied by a power of ten, while decimal prefixes and the order of magnitude give a compact way to express and compare very large and very small physical quantities. 🔉⇢
Physical measurements routinely span an enormous range, from the diameter of a hydrogen atom to the diameter of the earth, so writing every value as an ordinary decimal is clumsy and error-prone. To handle this range, every number is expressed in scientific notation as $a\times 10^{b}$, where $a$ is a number between one and ten and $b$ is any positive or negative exponent of ten. This notation is ideal for reporting measurement because it makes the precision explicit: all the zeroes appearing in the base number $a$ are significant, so no confusion arises about trailing zeroes. Reporting a length as $4.700\times10^{2}\ \mathrm{cm}$ instantly conveys four significant figures, whereas the plain form can be misread. Scientific notation also makes numerical computation easier, because multiplying and dividing powers of ten reduces to adding and subtracting their exponents. 🔉⇢
The order of magnitude of a physical quantity is obtained from its scientific notation. To get an approximate idea of the number we round the base number $a$ to one when $a \le 5$ and to ten when $5 \lt a \le 10$. The quantity can then be expressed approximately as $10^{b}$, and the exponent $b$ is called the order of magnitude of the physical quantity. When only an estimate is required, we say the quantity is of the order of $10^{b}$. For example the diameter of the earth, about $1.28\times10^{7}\ \mathrm{m}$, is of the order of $10^{7}\ \mathrm{m}$ with order of magnitude seven, while the diameter of a hydrogen atom, about $1.06\times10^{-10}\ \mathrm{m}$, is of the order of $10^{-10}\ \mathrm{m}$ with order of magnitude negative ten. Thus the diameter of the earth is seventeen orders of magnitude larger than that of the hydrogen atom, a comparison that a single subtraction of exponents makes obvious. 🔉⇢
Prefixes are standardised names attached to a unit to denote decimal multiples and sub-multiples, and they work hand in hand with scientific notation. Common multiplying prefixes include kilo ($10^{3}$), mega ($10^{6}$), giga ($10^{9}$) and tera ($10^{12}$), while sub-multiple prefixes include milli ($10^{-3}$), micro ($10^{-6}$), nano ($10^{-9}$) and pico ($10^{-12}$). A prefix simply replaces the corresponding power of ten, so that $1\ \mathrm{nm} = 10^{-9}\ \mathrm{m}$ and $1\ \mathrm{GHz} = 10^{9}\ \mathrm{Hz}$. Because SI is built on the decimal system, conversions between prefixed units are always powers of ten and never awkward fractions, which is a major convenience of the system. General guidelines for using symbols for SI units and prefixes are laid down so that the same quantity is written the same way everywhere, avoiding ambiguity in scientific and technical work. 🔉⇢
Order-of-magnitude reasoning is a powerful estimation tool in its own right. By replacing every quantity in a problem with its order of magnitude and combining the exponents, a physicist can quickly judge whether a proposed answer is plausible before doing a detailed calculation. For instance, if a result is expected to be of the order of $10^{3}$ and a calculation yields something of the order of $10^{6}$, the discrepancy signals a probable error in units or arithmetic. This kind of quick check is valued because it does not commit us to a particular choice of units and needs no worry about conversions among multiples and sub-multiples. It complements, rather than replaces, an exact computation: the order of magnitude sets the scale and the significant figures then fix the precision within that scale. 🔉⇢
Care is still needed when moving between the compact and the expanded forms. It is customary to write the decimal after the first digit, so $4.700\ \mathrm{m}$ becomes $4.700\times10^{2}\ \mathrm{cm}$ and $4.700\times10^{-3}\ \mathrm{km}$; the power of ten is irrelevant to the determination of significant figures. When adding or subtracting numbers written in scientific notation, the exponents must first be made equal so that the decimal places line up. When multiplying or dividing, the base numbers combine while the exponents add or subtract. Keeping quantities in scientific notation throughout a multi-step computation reduces the risk of dropping a factor of ten, one of the most common and costly mistakes in measurement, and it keeps the indication of precision intact from the first line of working to the final reported result. 🔉⇢
In practice the prefixes, the scientific notation and the order of magnitude are used together to simplify measurement notation and numerical computation, giving an indication of the precision of the numbers. A physical measurement of a very small or very large quantity is first written in scientific notation, an appropriate prefix is then chosen so that the numerical value is convenient, and the order of magnitude is quoted whenever only an estimate of the size is required. Because SI uses the decimal system, moving between a prefix and its power of ten never introduces an awkward factor, so conversions within the system remain simple and the significant figures of the measurement are preserved throughout the calculation. 🔉⇢
Derivation 🔉⇢
Take a measured length written in ordinary form: the diameter of the earth is about $12800000\ \mathrm{m}$.
Shift the decimal point so that one non-zero digit remains to its left, counting the shifts: moving seven places gives the base number $1.28$.
The number of places shifted becomes the exponent of ten: $12800000\ \mathrm{m} = 1.28\times10^{7}\ \mathrm{m}$.
Round the base number for an estimate: since $1.28 \le 5$, replace it by one, giving $10^{7}\ \mathrm{m}$.
Read off the order of magnitude as the exponent, seven, so the earth's diameter is of the order of $10^{7}\ \mathrm{m}$.
⚠️ JEE trap: Many students count the order of magnitude as the leftmost digit of the number itself; it is actually the exponent of ten after the value is written in scientific notation, with $a$ rounded so that $a \le 5$ maps to one and $5 \lt a \le 10$ maps to ten. 🔉⇢
Dimensions & Dimensional Formulae 🔉⇢
🎯 The dimensional formula of a quantity is fixed the moment you write its definition. Pick one and read off its M, L, T exponents — the same formula in every system of units.
🔉⇢
defining relation:— → [Q] = Ma Lb Tc
What this shows
The dimensional formula of a quantity is fixed the moment you write its definition. Pick one and read off its M, L, T exponents — the same formula in every system of units.
Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.
Definition: The dimensions of a physical quantity are the powers to which the seven base quantities must be raised to represent that quantity, and the expression showing this combination is called its dimensional formula. 🔉⇢
The nature of a physical quantity is described by its dimensions. All the physical quantities represented by derived units can be expressed in terms of some combination of the seven fundamental or base quantities, and we call these base quantities the seven dimensions of the physical world. They are denoted with square brackets: length has the dimension [L], mass [M], time [T], electric current [A], thermodynamic temperature [K], luminous intensity [cd] and amount of substance [mol]. The dimensions of a physical quantity are the powers, or exponents, to which the base quantities are raised to represent that quantity. Using square brackets around a quantity means that we are dealing with the dimensions of that quantity, not its magnitude. This idea is powerful precisely because it strips away the numerical value and keeps only the algebraic character of the quantity, letting us compare quantities that look completely different in an equation. 🔉⇢
In mechanics all the physical quantities can be written in terms of the dimensions [L], [M] and [T]. Consider volume, which is expressed as the product of length, breadth and height, or three lengths. Hence the dimensions of volume are $[\mathrm{L}]\times[\mathrm{L}]\times[\mathrm{L}] = [\mathrm{L}^{3}]$. Because volume is independent of mass and time, it has zero dimension in mass and zero dimension in time, and is said to possess three dimensions in length. Force, as the product of mass and acceleration, is mass times length divided by time squared, so its dimensions are $[\mathrm{M}][\mathrm{L}][\mathrm{T}^{-2}] = [\mathrm{M\,L\,T^{-2}}]$: one dimension in mass, one in length and negative two in time, with zero dimension in every other base quantity. In this representation the magnitudes are not considered; it is only the quality of the physical quantity that enters. 🔉⇢
Because dimensions capture only the type of quantity, several distinct physical quantities can share the same dimensions. A change in velocity, an initial velocity, an average velocity, a final velocity and a speed are all equivalent in this context, since each is a length divided by a time and so has the dimensions $[\mathrm{L\,T^{-1}}]$. This is why dimensional analysis cannot, by itself, distinguish quantities that happen to have the same dimensions, and it is also why a pure number, such as an angle formed as the ratio length over length or a refractive index formed as the ratio of two speeds, has no dimensions at all. Recognising which combinations of base quantities recur across mechanics, and which quantities collapse to the same dimensional formula, is a key skill that makes later checking and deducing of relations straightforward. 🔉⇢
The expression which shows how and which of the base quantities represent the dimensions of a physical quantity is called the dimensional formula of that quantity. For example the dimensional formula of volume is $[\mathrm{M^{0}\,L^{3}\,T^{0}}]$, that of speed or velocity is $[\mathrm{M^{0}\,L\,T^{-1}}]$, that of acceleration is $[\mathrm{M^{0}\,L\,T^{-2}}]$ and that of mass density is $[\mathrm{M\,L^{-3}\,T^{0}}]$. An equation obtained by equating a physical quantity with its dimensional formula is called the dimensional equation of that quantity, so that we write $[V] = [\mathrm{M^{0}\,L^{3}\,T^{0}}]$, $[v] = [\mathrm{M^{0}\,L\,T^{-1}}]$, $[F] = [\mathrm{M\,L\,T^{-2}}]$ and $[\rho] = [\mathrm{M\,L^{-3}\,T^{0}}]$. The dimensional equation is obtained from the equation representing the relation between the physical quantities, so any defining relation immediately yields a dimensional formula. 🔉⇢
To write the dimensional formula of any quantity, start from its defining equation, replace each quantity on the right by its own dimensional formula, and combine the base quantities as algebraic symbols, treating identical dimensions in numerator and denominator as cancelling. Work quantity by quantity: acceleration is velocity divided by time, so $[\mathrm{L\,T^{-1}}]/[\mathrm{T}] = [\mathrm{L\,T^{-2}}]$; force is mass times that, giving $[\mathrm{M\,L\,T^{-2}}]$; work is force times length, giving $[\mathrm{M\,L^{2}\,T^{-2}}]$; and power is work divided by time, giving $[\mathrm{M\,L^{2}\,T^{-3}}]$. Keeping the base quantities in a fixed order such as [M][L][T][A][K][mol][cd] and writing an exponent of zero where a base quantity is absent makes the formula unambiguous. Extensive tables of dimensional formulae are compiled for ready reference, but the ability to build any one of them from a defining relation is far more valuable than memorising the list. 🔉⇢
Because dimensions record only the powers of the base quantities and never their magnitudes, they give a compact description of the nature of a physical quantity that is independent of the system of units chosen. This is what makes the dimensional formula so useful as a preliminary description: a large and wide variety of physical quantities, derived from the equations representing the relationships among other quantities, can be reduced to a single expression in the base quantities and collected in tables for ready reference. Writing each formula in a fixed order of the base quantities, and inserting a zero exponent wherever a base quantity is absent, keeps the representation unambiguous and prepares the ground for checking the consistency of any equation in which the quantity appears. 🔉⇢
Derivation 🔉⇢
Begin from the defining relation for mass density: density is mass divided by volume, $\rho = m/V$.
Write the dimensional formula of each quantity on the right: mass is $[\mathrm{M}]$ and volume is $[\mathrm{L}^{3}]$.
Substitute and combine the base quantities as algebraic symbols: $[\rho] = [\mathrm{M}]/[\mathrm{L}^{3}] = [\mathrm{M\,L^{-3}}]$.
Insert an exponent of zero for every absent base quantity to make the formula explicit: $[\rho] = [\mathrm{M\,L^{-3}\,T^{0}}]$.
Read off the powers: density has one dimension in mass, negative three in length and zero in time, matching its known dimensional formula.
⚠️ JEE trap: A common error is to treat the numerical factor in a formula, such as the one-half in kinetic energy, as if it affects the dimensions; dimensional formulae record only the powers of the base quantities and are completely unaffected by pure numbers. 🔉⇢
🎯 Only one combination of L and g has the dimension of time. Try each trial form and watch its M, L, T exponents — the pendulum period must land on M⁰ L⁰ T¹.
🔉⇢
[T] should be M0 L0 T1 = —
What this shows
Only one combination of L and g has the dimension of time. Try each trial form and watch its M, L, T exponents — the pendulum period must land on M⁰ L⁰ T¹.
Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.
Definition: Dimensional analysis is the method of treating the dimensions $[M^a L^b T^c]$ of physical quantities as algebraic symbols, so that the homogeneity of an equation can be checked and, within limits, the form of a relation among quantities can be deduced. 🔉⇢
Every measurement is a number together with a unit, and every unit rests on a dimension. The nature of a physical quantity is described by its dimensions: the powers to which the base quantities length, mass and time are raised to represent that quantity. When we place square brackets round a quantity we mean the dimensions of that quantity, and inside those brackets the symbols [M], [L] and [T] are treated as algebraic symbols. Dimensional analysis is the method of carrying those symbols through a calculation. Because a change of units cannot change the nature of a physical quantity, its dimensional formula is fixed: velocity is [L T^-1], acceleration is [L T^-2], and force, as mass times acceleration, is [M L T^-2]. This fixed nature is what makes dimensions a fast preliminary test of the consistency of an equation. 🔉⇢
🎯 Significant figures are the digits a measurement actually justifies — all the certain ones plus one estimated. Choose how many the instrument supports and watch the reported length round.
Significant figures are the digits a measurement actually justifies — all the certain ones plus one estimated. Choose how many the instrument supports and watch the reported length round.
Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.
Definition: The significant figures of a measured value are the digits known reliably together with the first uncertain digit, and they indicate the precision of the measurement fixed by the least count of the instrument. 🔉⇢
Every measurement involves errors, so the result of a measurement should be reported in a way that indicates the precision of measurement. Normally the reported result is a number that includes all the digits known reliably plus the first digit that is uncertain. The reliable digits plus the first uncertain digit are known as the significant digits or significant figures. If the period of oscillation of a simple pendulum is reported as $1.62\ \mathrm{s}$, the digits one and six are reliable and certain while the digit two is uncertain, so the measured value has three significant figures. A length reported as $287.5\ \mathrm{cm}$ has four significant figures. Reporting more digits than the significant ones is superfluous and misleading, because it would give a wrong idea about the precision of the measurement, which is always limited by the least count of the measuring instrument that was actually used. 🔉⇢
The rules for counting significant figures follow from a single principle: a change of units does not change the number of significant figures. The length $2.308\ \mathrm{cm}$ has four significant figures, and the same value written as $0.02308\ \mathrm{m}$, $23.08\ \mathrm{mm}$ or $23080\ \mathrm{\mu m}$ still has the digits two, three, zero and eight as significant, namely four. This shows that the location of the decimal point is of no consequence in determining the number of significant figures. From this the working rules emerge, and they are best remembered as a small set of statements about non-zero digits and the several kinds of zero, since it is the treatment of zeroes that causes almost all of the confusion in practice. 🔉⇢
First, all the non-zero digits are significant. Second, all the zeroes between two non-zero digits are significant, no matter where the decimal point is, if at all. Third, if the number is less than one, the zeroes on the right of the decimal point but to the left of the first non-zero digit are not significant, so that in $0.002308$ the leading zeroes merely fix the decimal position. Fourth, the digit zero conventionally placed to the left of a decimal for a number less than one, as in $0.1250$, is never significant. These four rules already settle most cases, and each of them is consistent with the principle that a mere change of units cannot create or destroy a significant figure in a genuine measurement. 🔉⇢
The trailing zeroes need special care because their meaning depends on whether a decimal point is present. In a number without a decimal point the terminal or trailing zeroes are not significant, so $123\ \mathrm{m} = 12300\ \mathrm{cm} = 123000\ \mathrm{mm}$ still has only three significant figures. In a number with a decimal point the trailing zeroes are significant, so $3.500$ and $0.06900$ each have four significant figures. A length reported as $4.700\ \mathrm{m}$ therefore has four significant figures, because the zeroes are written deliberately to convey the precision of measurement; if they were not significant it would be superfluous to write them at all. Since $4.700\ \mathrm{m} = 4700\ \mathrm{mm}$, the rule about a decimal-free number would wrongly suggest two significant figures, and this ambiguity is exactly why scientific notation is preferred for reporting a measurement. 🔉⇢
Two further points complete the picture. Reporting a measurement in scientific notation as $a\times10^{b}$ removes every ambiguity about trailing zeroes, because the power of ten is irrelevant to the count and all zeroes appearing in the base number $a$ are significant; thus $4.700\times10^{-3}\ \mathrm{km}$ unambiguously carries four significant figures. Finally, the multiplying or dividing factors which are neither rounded numbers nor measured values are exact and have an infinite number of significant figures. In a relation such as the circumference $s = 2\pi r$ the factor two is exact and may be written as $2$, $2.0$ or $2.0000$ as required, and a mathematical constant such as $\pi$ is known to as many significant figures as we wish. Distinguishing an exact number from a measured value is essential, because an exact number never limits the precision of a calculated result. 🔉⇢
Finally, the number of significant figures indicates the precision of measurement, which depends on the least count of the measuring instrument, and a choice of change of different units does not change the number of significant figures in a measurement. This important remark keeps the whole scheme consistent: whether a length is expressed in metre, in centimetre or in millimetre, the reliable digits and the first uncertain digit stay the same, and only the position of the decimal point moves. Reporting a measurement with more digits than the significant figures is superfluous and misleading, because it would convey a wrong idea about the precision that the instrument can actually deliver, whereas reporting too few discards reliable information that the measurement genuinely contains. The significant figures thus record, in a single compact form, both the measured value and the precision of that measurement, which is why they must be preserved carefully through every conversion of units and every step of a calculation. 🔉⇢
Derivation 🔉⇢
Write the measured length in a fixed unit: $2.308\ \mathrm{cm}$.
Apply the non-zero rule and the enclosed-zero rule: the digits two, three, zero and eight are all significant, giving four significant figures.
Change the unit to test the count: $2.308\ \mathrm{cm} = 0.02308\ \mathrm{m}$, where the two leading zeroes only fix the decimal position and are not significant.
Confirm the count is unchanged: the measured value still has the four significant digits two, three, zero and eight.
Conclude the principle: a change of units cannot change the number of significant figures, so precision resides in the digits, not in the position of the decimal point.
⚠️ JEE trap: Learners frequently assume every zero is significant or that trailing zeroes never count; the truth is conditional, trailing zeroes are significant only when a decimal point is present, which is why measurements are safest reported in scientific notation. 🔉⇢
Rounding & Arithmetic with Significant Figures 🔉⇢
🎯 When you ADD measurements the answer cannot be more precise than the coarsest input: it keeps the least number of decimal places. (Multiplication keeps the least significant figures instead.)
🔉⇢
sum keeps the LEAST number of decimal places
What this shows
When you ADD measurements the answer cannot be more precise than the coarsest input: it keeps the least number of decimal places. (Multiplication keeps the least significant figures instead.)
Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.
Definition: When approximate measured values are combined, the calculated result must be rounded so that it reflects the precision of the least precise input, using the least-significant-figures rule for multiplication and division and the least-decimal-places rule for addition and subtraction. 🔉⇢
The result of a calculation involving approximate measured values, that is values with a limited number of significant figures, must reflect the uncertainties in the original measured values. It cannot be more accurate than the original measured values on which it is based, so in general the final result should not have more significant figures than the original data from which it was obtained. Suppose the mass of an object is measured as $4.237\ \mathrm{g}$ with four significant figures and its volume as $2.51\ \mathrm{cm^{3}}$ with three; mere arithmetic division gives a density of $1.68804780876\ \mathrm{g\,cm^{-3}}$ to eleven decimal places. It would be absurd and irrelevant to record the density to such a precision when the measurements it rests on have far less. The rules for arithmetic with significant figures exist precisely to show the result with a precision consistent with the precision of the input measured values, neither more nor less. 🔉⇢
The first rule governs multiplication and division: the final result should retain as many significant figures as are there in the original number with the least significant figures. In the density example above, the volume has only three significant figures, so the density should be reported to three significant figures as $1.69\ \mathrm{g\,cm^{-3}}$. As a second illustration, if the speed of light is given as $3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ with three significant figures and one year is $3.1557\times10^{7}\ \mathrm{s}$ with five, then the light year works out to $9.47\times10^{15}\ \mathrm{m}$ with three significant figures, because the less precise factor controls the outcome. The rule is intuitive: a product or quotient can be no sharper than its bluntest factor, so the smallest number of significant figures among the inputs sets the ceiling on the result. 🔉⇢
The second rule governs addition and subtraction, and here the count is made in decimal places rather than in significant figures: the final result should retain as many decimal places as are there in the number with the least decimal places. For example the sum of $436.32\ \mathrm{g}$, $227.2\ \mathrm{g}$ and $0.301\ \mathrm{g}$ is, by mere arithmetic, $663.821\ \mathrm{g}$; but the least precise measurement, $227.2\ \mathrm{g}$, is correct to only one decimal place, so the result must be rounded to $663.8\ \mathrm{g}$. Similarly a difference in length is expressed as $0.307\ \mathrm{m} - 0.304\ \mathrm{m} = 0.003\ \mathrm{m} = 3\times10^{-3}\ \mathrm{m}$. It would be wrong to apply the multiplication rule to an addition, because addition and subtraction combine uncertainties through decimal places, and doing otherwise fails to convey the precision of measurement properly. 🔉⇢
The result of a computation with approximate numbers that contains more than one uncertain digit should be rounded off, and the convention is clear in most cases. The preceding digit is raised by one if the insignificant digit to be dropped is more than five, and is left unchanged if it is less than five, so $2.746$ rounds to $2.75$ and $1.743$ rounds to $1.74$. The delicate case is a dropped digit of exactly five with nothing beyond it, and the convention here is round-half-even: if the preceding digit is even the five is simply dropped, and if it is odd the preceding digit is raised by one. Thus $2.745$ rounds to $2.74$ because the preceding digit four is even, while $2.735$ rounds to $2.74$ because the preceding digit three is odd. This even convention avoids a systematic upward bias that would build up if every half were rounded up. 🔉⇢
Two safeguards keep rounding from corrupting a long calculation. First, in any involved or complex multi-step calculation you should retain one digit more than the significant digits in the intermediate steps and round off to the proper significant figures only at the end, otherwise rounding errors can build up; the reciprocal example, where $1/9.58$ carried to an extra digit as $0.1044$ correctly returns $9.58$ on inversion whereas the prematurely rounded $0.104$ returns $9.62$, shows the danger vividly. Second, remember that exact numbers appearing in formulae, such as the factor $2\pi$ in the pendulum period $T = 2\pi\sqrt{L/g}$, carry an infinite number of significant figures and never limit the result, and a constant known to many figures such as the speed of light $2.99792458\times10^{8}\ \mathrm{m\,s^{-1}}$ may be rounded to $3\times10^{8}\ \mathrm{m\,s^{-1}}$ or the value of $\pi$ taken as $3.142$ or $3.14$ as the required precision demands. 🔉⇢
These rules together ensure that the final result of a calculation is shown with the precision that is consistent with the precision of the input measured values, neither claiming more accuracy than the data allow nor discarding precision that the data genuinely carry. In a multi-step computation the intermediate results should be calculated to one more significant figure than the least precise measurement, and only the final answer rounded off to the proper significant figures; otherwise rounding errors can build up and corrupt a result that the original measured values would have supported. Recognising whether an operation is a multiplication, a division, an addition or a subtraction, and applying the matching rule, is therefore the habit that keeps a reported result honest and consistent with the precision of measurement. 🔉⇢
Derivation 🔉⇢
Identify the operation and the inputs: to find density divide mass $4.237\ \mathrm{g}$ (four significant figures) by volume $2.51\ \mathrm{cm^{3}}$ (three significant figures).
Perform the raw arithmetic: $4.237 / 2.51 = 1.68804780876\ \mathrm{g\,cm^{-3}}$, far more digits than the data justify.
Apply the multiplication-division rule: keep as many significant figures as the least precise input, namely three from the volume.
Round the raw result to three significant figures using the standard convention: $1.68804\ldots \to 1.69\ \mathrm{g\,cm^{-3}}$.
Report the value with its unit as $1.69\ \mathrm{g\,cm^{-3}}$, a precision consistent with the least precise measured value.
⚠️ JEE trap: A widespread mistake is to apply the least-significant-figures rule to addition and subtraction; sums and differences must instead be rounded to the least number of decimal places, so $12.9\ \mathrm{g} - 7.06\ \mathrm{g}$ is $5.8\ \mathrm{g}$, not $5.84\ \mathrm{g}$. 🔉⇢
Measuring Length: Parallax & Indirect Methods 🔉⇢
🎯 You cannot lay a ruler to a star. Parallax does it indirectly: a known baseline b and the tiny angle θ it subtends fix the distance D = b/θ. Widen the angle and the star is nearer.
🔉⇢
D = b / θ (1 parsec = 1 AU / 1″)
What this shows
You cannot lay a ruler to a star. Parallax does it indirectly: a known baseline b and the tiny angle θ it subtends fix the distance D = b/θ. Widen the angle and the star is nearer.
Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.
Definition: Lengths are measured directly with graduated scales for ordinary objects, and by indirect methods such as parallax and triangulation for distances too large or too small to measure directly, so that a single family of methods spans from atomic to cosmic ranges. 🔉⇢
A physical quantity such as length is measured by comparison with an internationally accepted reference standard, the metre, so that the result of the measurement is a number accompanied by a unit. For ordinary objects a graduated metre scale is enough, but the range of lengths a physicist must measure is vast, running from the diameter of a hydrogen atom, about $1.06\times10^{-10}\ \mathrm{m}$, to the diameter of the earth, about $1.28\times10^{7}\ \mathrm{m}$, and far beyond to the distances between stars. No single instrument can span such a range, so length measurement is organised as a family of methods, each suited to a particular scale, yet all resting on the same idea of comparison with a standard and all reported with proper significant figures that reflect the precision of the measurement. 🔉⇢
For large distances the parallax method is the workhorse. Parallax is the apparent shift in the position of a distant object against a still more distant background when the object is viewed from two different points separated by a known distance called the basis. The physics is the familiar everyday observation that when you look out of the window of a fast moving train, the nearby trees and houses seem to move rapidly in a direction opposite to the train's motion while distant objects such as hill tops, the moon and the stars seem to be stationary. If the basis is $b$ and the object subtends a small parallax angle $\theta$ at the object, then because the angle equals the ratio of arc to radius, the distance is $D = b/\theta$, with $\theta$ measured in radian. The same principle, using the diameter of the earth's orbit as the basis, lets astronomers reach the distance of nearby stars. 🔉⇢
Triangulation extends the idea when only angles, not a full basis at the target, are convenient to measure. From the two ends of a carefully measured baseline the direction to an inaccessible point, such as a mountain peak across a valley, is sighted, and the two base angles together with the known baseline length fix the triangle completely. Solving the triangle then yields the distance and the height of the peak without ever travelling to it. This indirect approach embodies a recurring strategy in measurement: convert a length that cannot be reached into a combination of a length that can be measured and one or more angles, both of which are then handled with ordinary geometry and reported to the precision that the least precise input allows. 🔉⇢
For very small lengths the same spirit of indirect measurement is used in reverse, magnifying the small so it can be compared with a scale. The unit of length convenient on the atomic scale is the angstrom, with $1\ \mathrm{\mathring{A}} = 10^{-10}\ \mathrm{m}$, and the size of a hydrogen atom is about half an angstrom. To estimate such sizes one may spread a known amount of a substance into a single molecular layer and divide the volume by the area to obtain the thickness, or view a fine object through a microscope of known magnification. A student who measures the width of a human hair as $3.5\ \mathrm{mm}$ in the field of view of a microscope of magnification one hundred concludes that the true thickness is that reading divided by the magnification, a clean example of an indirect length measurement built from a direct reading and a known factor. 🔉⇢
Across all these methods the reported length must carry the correct number of significant figures, because a length is meaningless as an estimate of magnitude unless a standard for comparison is specified. To call a length large or small is meaningless on its own; an atom is small only compared with everyday objects, and an inter-stellar distance is large only compared with terrestrial ones. Order-of-magnitude language ties these scales together, letting us say that the earth's diameter is of the order of $10^{7}\ \mathrm{m}$ while an atom is of the order of $10^{-10}\ \mathrm{m}$, seventeen orders of magnitude apart. Choosing the right method for the scale, expressing the answer in scientific notation, and keeping only the significant digits justified by the least precise input are the three habits that make length measurement reliable from the nuclear to the cosmic range. 🔉⇢
Whatever the method, a length is finally reported by comparison with the standard metre and expressed in scientific notation with the significant figures that the least precise input allows. The same principle of comparison with a reference standard underlies the graduated scale, the parallax observation and the microscope estimate alike, and only the range of the quantity decides which method is convenient. Expressing the result as an order of magnitude, from the atomic scale near the angstrom to the astronomical distance of the stars, lets lengths that differ by many orders of magnitude be compared on a single footing, which is why order-of-magnitude language is inseparable from the measurement of length. 🔉⇢
Derivation 🔉⇢
Set up the parallax geometry: view a distant object from two points separated by a known basis $b$, and measure the small parallax angle $\theta$ it subtends.
Use the definition of plane angle as the ratio of arc length to radius: for a small angle, $\theta = b/D$ where $D$ is the distance to the object.
Rearrange to make the distance the subject: $D = b/\theta$, with $\theta$ expressed in radian.
Note the requirement that $\theta$ be small so the arc approximates the chord, keeping the ratio accurate.
Insert measured values of $b$ and $\theta$ with their significant figures, and report $D$ rounded to the least precise input.
⚠️ JEE trap: Students often try to substitute a parallax angle expressed in degrees directly into $D = b/\theta$; the relation follows from angle defined as arc over radius, so $\theta$ must first be converted to radian, the dimensionless measure of plane angle. 🔉⇢
Measuring Mass and Time 🔉⇢
🎯 A single swing is hard to time; your reaction time swamps it. Time N swings and divide: the per-swing error shrinks by 1/N. Watch the pendulum and read T = t/N.
🔉⇢
T = ttotal / N (divide out the reaction time)
What this shows
A single swing is hard to time; your reaction time swamps it. Time N swings and divide: the per-swing error shrinks by 1/N. Watch the pendulum and read T = t/N.
Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.
Definition: Mass is measured against the kilogram over a range spanning many orders of magnitude, using the unified atomic mass unit on the atomic scale, while time is measured against the second realised through the unperturbed hyperfine transition of the caesium atom. 🔉⇢
Mass is a base quantity whose SI unit is the kilogram, and like length it must be measured over a very large range of magnitudes. The mass of an electron is a very small negative power of ten in kilogram, that of a proton larger, an ordinary object a few kilogram, and astronomical bodies enormously more massive, so the masses met in physics span many orders of magnitude. On the ordinary scale a mass is measured by comparison with a standard mass using a balance, in which an object of unknown mass is compared against known reference masses. A common measurement, such as adding two pieces to a box on a grocer's balance, must then be reported with attention to significant figures, since each observation on the balance carries only a limited precision that the result cannot exceed. The measured value is written as a number accompanied by its unit, exactly as for any other physical quantity. 🔉⇢
On the scale of atoms and molecules the kilogram is far too large to be convenient, so masses are expressed in the unified atomic mass unit. One unified atomic mass unit is a definite small part of the mass of a chosen reference atom, and it is a very small number of kilogram. This atomic unit is tied directly to the amount of substance: one mole contains a fixed number of elementary entities, the Avogadro constant, and an entity may be an atom, a molecule, an ion, an electron or any specified particle. The mass of one mole of a substance, expressed in gram, is numerically its relative mass in the atomic mass unit, so measuring an atomic mass reduces to comparison of the substance with a standard and expressing the result in the atomic unit rather than in the awkwardly large kilogram. Because the mole fixes the number of elementary entities, a measurement of amount of substance and a measurement of mass are closely related. 🔉⇢
Time is the base quantity measured to the highest precision of all, and its SI unit is the second. The second is defined by taking a fixed numerical value of the caesium frequency, the unperturbed ground state hyperfine transition frequency of the caesium atom, so that one second is a definite number of periods of the radiation of that transition. A device that realises the second by counting the periods of this atomic transition is an atomic clock. Because the transition frequency is an unperturbed property of the caesium atom and does not change, rather than a property of a manufactured object, such a clock gives a reference standard of time of extraordinary constancy, against which every other measurement of time is compared. The definition fixes the numerical value of a constant of nature, so the second is realised the same way in every laboratory. 🔉⇢
The masses met in physics range from the very small mass of the electron and the proton to the enormous mass of astronomical bodies, and the intervals of time range from the age of the world down to the very short life of an elementary particle, so both quantities span many orders of magnitude and are handled by the same framework: a defined base unit, a comparison with a standard, and a result expressed in scientific notation with proper significant figures. Many determinations of mass and of time are related, because timing a periodic motion such as the oscillation of a simple pendulum, whose period depends on its length and the acceleration due to gravity, yields a measurement of gravity, and comparison of that against a known mass refines the standard. In every such measurement the reported value must reflect the precision of the least precise observation and be expressed with proper significant figures. 🔉⇢
Because the caesium frequency is an unperturbed property of the atom, the second is now defined with an accuracy and a reliability far greater than any earlier standard based on the motion of the earth, which does change slightly. As measuring techniques are improved, the definitions of the kilogram and the second are revised so that each base unit is tied to a fixed numerical value of a constant rather than to a single object that could change; the values of these constants need not be remembered, since they only indicate the extent of accuracy to which the constants are known. This keeps every measurement of mass and of time consistent, from the atomic scale to the astronomical scale, and lets the result of one measurement be compared reliably with the result of another obtained elsewhere. The progress of technology steadily improves the precision of measurement, and the definitions of the base units are revised to keep up with this progress. Because the kilogram and the second are each now defined by fixing the numerical value of a constant, no measurement of mass or of time depends any longer on a single physical object that could change, drift or be damaged, and the measured value of a quantity stays the same however finely the standard is later realised. 🔉⇢
Derivation 🔉⇢
State the definition of the second: it fixes the numerical value of the caesium frequency, the unperturbed ground state hyperfine transition frequency of the caesium atom.
Interpret the fixed frequency as a fixed count: the second is a definite number of periods of the radiation of the transition.
Build a device that counts these periods of the atomic transition, realising the second directly from a property of the atom.
Because the transition frequency is a property of caesium and not of any manufactured object, every such atomic clock realises the same second.
Conclude that the caesium standard gives a reference standard of time of extraordinary constancy, against which other measurements of time are compared.
⚠️ JEE trap: It is tempting to think an atomic clock keeps time by radioactive decay; in fact it counts the periods of the fixed unperturbed hyperfine transition frequency of the caesium atom, a definite property of the atom, not a random decay. 🔉⇢
Errors in Measurement: Systematic vs Random 🔉⇢
🎯 Systematic and random errors are different beasts. The systematic knob slides the whole cloud off the true value (teal); the random knob only spreads the dots about their mean (brown). Averaging beats random error but never touches systematic.
🔉⇢
systematic shifts the mean; random widens the scatter
What this shows
Systematic and random errors are different beasts. The systematic knob slides the whole cloud off the true value (teal); the random knob only spreads the dots about their mean (brown). Averaging beats random error but never touches systematic.
Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.
Definition: Errors are the unavoidable differences between a measured value and the true value of a physical quantity; systematic errors shift the observations consistently in one direction while random errors scatter them irregularly about the true value. 🔉⇢
Every measurement involves errors, so the result of a measurement should be reported in a way that indicates the precision of measurement. An error is the difference between the measured value of a physical quantity and its true value, and because no instrument and no observer is perfect, some uncertainty is always present. The reported result therefore carries only the reliable digits together with the first uncertain digit, the significant figures, and this precision is limited by the least count of the measuring instrument. Errors are grouped, by their behaviour, into two kinds: systematic errors, which shift every observation in one direction, and random errors, which scatter the observations irregularly about the true value. Deciding which kind is present in a particular measurement is the first step, because the two kinds of error are reduced in completely different ways and demand different treatment from the experimenter. 🔉⇢
Systematic errors tend to occur in one direction, either always positive or always negative, so they bias every observation the same way and are not removed by simply making the measurement again. Their sources are of a few kinds. An instrument error arises from imperfect standardisation, such as a scale whose zero is displaced, so that every observation is greater or smaller than the true value by a fixed amount. A technique error arises when the measuring conditions do not match those intended, and a personal error arises from an observer's individual bias. Because a systematic error acts in a fixed direction, it can be reduced by finding its cause, correcting the zero of the instrument, improving the measuring technique, and comparing the result against a better reference standard. Removing a systematic error improves the accuracy of the measurement, that is, how close the measured value lies to the true value of the quantity. 🔉⇢
Random errors occur irregularly and are therefore positive as often as negative, arising from small changes in the conditions of the measurement and in the observer's judgement from one observation to the next. Unlike a systematic error, a random error cannot be removed by correcting the instrument; but because positive and negative deviations are equally likely, it is reduced by taking a large number of observations and adopting their average, since the deviations tend to cancel. This is exactly why a set of one hundred observations of a diameter is expected to yield a more reliable estimate than a set of only five: the average of many observations reduces the random scatter and improves the precision of the result, that is, how closely repeated observations agree with one another. 🔉⇢
The two kinds of error map onto the paired ideas of accuracy and precision, which a careful measurement always distinguishes. Accuracy is how close a measured value is to the true value of the quantity, and it is degraded mainly by systematic errors that push every observation away from the truth. Precision is how closely repeated observations agree with one another, and it is limited by random errors and by the least count of the instrument. A measurement can be precise yet not accurate, when the observations agree closely but cluster about a wrong value because of an uncorrected zero, or accurate on average yet not precise, when the observations scatter widely but centre on the true value. A vernier callipers or a screw gauge with a fine least count improves the precision of a length measurement, but only correcting a zero error improves its accuracy. 🔉⇢
In practice the experimenter reduces the two kinds of error by different means. To reduce a systematic error, the zero of the instrument is checked, its standardisation is confirmed against a reference standard, the measuring technique is refined so the conditions match those intended, and where possible the quantity is measured again by a different method so that a hidden bias shows itself as a difference. To reduce a random error, the same quantity is measured many times and the average is adopted as the best estimate, while the spread of the observations measures the uncertainty. The final value is then reported with significant figures set by the least count and by the remaining uncertainty, so that the number quoted carries only the digits the measurement can justify. Deciding at the outset whether the error present is mainly systematic or mainly random is thus the first decision in any careful measurement. 🔉⇢
A concrete example makes the distinction clear. Suppose the thickness of a thin brass rod is measured with vernier callipers whose jaws, when closed, already read a small value different from zero: every observation is then greater than the true value by that fixed amount, a systematic zero error corrected by subtracting it from each observation. Now suppose the temperature of the room, or the exact instant at which the observer judges a coincidence of the scale divisions, changes slightly from one observation to the next: the values scatter about the average, a random error reduced only by taking many observations. The progress of technology steadily improves the measuring instrument and the technique, so that both kinds of error are reduced and the precision of the measurement increases; the definitions of the standard units are revised to keep up with this progress, tying every measurement to an ever more reliable reference standard. 🔉⇢
Derivation 🔉⇢
Make repeated observations of a fixed quantity and note that they do not all coincide, revealing that error is present in the measurement.
Separate the deviations by behaviour: those that shift every observation the same way are systematic, those that scatter irregularly are random.
For the systematic part, trace the cause to the instrument zero, the technique or the observer, and correct or restandardise to remove the fixed shift, improving the accuracy.
For the random part, take a large number of observations so that equally likely positive and negative deviations tend to cancel in the average, improving the precision.
Report the average as the best estimate, with significant figures set by the least count and the remaining uncertainty.
⚠️ JEE trap: A frequent confusion is treating accuracy and precision as the same thing; a set of observations that agree closely can still be inaccurate if a systematic zero error shifts them all, so high precision does not by itself guarantee closeness to the true value. 🔉⇢
Absolute, Relative & Percentage Error 🔉⇢
🎯 The same absolute error Δa means very different things on a small quantity and a large one. Fix Δa and grow ā: the band stays the same width but the percentage error shrinks.
The same absolute error Δa means very different things on a small quantity and a large one. Fix Δa and grow ā: the band stays the same width but the percentage error shrinks.
Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.
Definition: The absolute error of a single observation is the magnitude of its deviation from the mean, the mean absolute error is the average of these deviations, the relative error is that mean divided by the measured value, and the percentage error is the relative error expressed as a percentage. 🔉⇢
When a physical quantity is measured several times, the observations differ because of random error, so the best estimate of the true value is taken to be the arithmetic mean of the observations. Suppose a quantity $a$ is measured $n$ times giving values $a_1, a_2, \ldots, a_n$; the mean $\bar{a}$ is their sum divided by $n$, and this mean is adopted as the true value in the absence of any other information. The whole scheme of absolute, relative and percentage error is built on this mean, and it exists to answer a single practical question: having taken a set of scattered observations, by how much may the reported value be trusted, and how should that uncertainty be quoted alongside the number so that the precision of the measurement is conveyed honestly. 🔉⇢
The absolute error in an individual measurement is the magnitude of the difference between that observation and the mean value, written $\Delta a_i = |\,\bar{a} - a_i\,|$. It is expressed in the same unit as the quantity itself and is always taken as a positive magnitude, because we care about the size of the deviation and not its sign. Each of the $n$ observations therefore has its own absolute error, and these individual absolute errors measure how far each particular observation strayed from the best estimate. The absolute error is the most direct expression of uncertainty, since it states plainly, in the unit of the quantity, how much an observation might be off; but on its own it does not tell us whether that amount is a serious or a trivial part of the quantity being measured. 🔉⇢
To summarise the scatter of the whole set in one number, the mean absolute error is defined as the arithmetic mean of the individual absolute errors, $\Delta\bar{a} = \dfrac{1}{n}\sum_{i=1}^{n} \Delta a_i$. The final result of the measurement is then reported as $a = \bar{a} \pm \Delta\bar{a}$, which states that the true value is expected to lie between $\bar{a} - \Delta\bar{a}$ and $\bar{a} + \Delta\bar{a}$. This form makes the precision of measurement explicit, since a small mean absolute error signals closely agreeing observations while a large one warns that the value is loosely determined. Increasing the number of observations $n$ generally lowers the mean absolute error, which is the quantitative reason that the average of many observations yields a more reliable estimate than the average of only a few. 🔉⇢
The absolute error alone cannot say whether an uncertainty is large or small, because that judgement is meaningless without a standard for comparison; an error of one millimetre is negligible in measuring a road but ruinous in measuring the thickness of a wire. The relative error supplies the comparison by dividing the mean absolute error by the mean value, $\dfrac{\Delta\bar{a}}{\bar{a}}$, giving a pure dimensionless ratio. The percentage error is simply the relative error expressed as a percentage, $\left(\dfrac{\Delta\bar{a}}{\bar{a}}\right)\times100\%$. Because it is dimensionless, the relative error lets us compare the quality of measurements of completely different quantities on an equal footing, and it is the natural language for stating how good a measurement is, independent of the unit or the magnitude of the quantity involved. 🔉⇢
A worked comparison shows why the relative measure is indispensable. The accuracy in the measurement of a mass $1.02\ \mathrm{g}$ is $\pm 0.01\ \mathrm{g}$, and another measurement $9.89\ \mathrm{g}$ is also accurate to $\pm 0.01\ \mathrm{g}$; the two have the same absolute error, yet the relative error in the first is $(\pm 0.01/1.02)\times100\% \approx \pm 1\%$ while in the second it is $(\pm 0.01/9.89)\times100\% \approx \pm 0.1\%$. The larger quantity is measured to ten times better relative precision even though the absolute error is identical, because the same absolute uncertainty is a smaller part of a larger value. The relative error of a value specified to a given number of significant figures therefore depends not only on the number of figures but also on the number itself, which is exactly why relative and percentage errors, not absolute errors, are quoted when the quality of a measurement is to be judged. 🔉⇢
In reporting a measured quantity, then, the mean is quoted as the best estimate, the mean absolute error fixes the number of significant figures that may honestly be written, and the relative or percentage error states, in a form independent of the unit, how good the measurement is. Because the relative error of a value depends not only on the number of significant figures but also on the number itself, two measurements with the same absolute error can differ greatly in quality, and only the relative error exposes the difference. This is why a careful experimenter always converts an absolute error into a relative or percentage error before comparing the precision of measurements of different physical quantities, and why the percentage error is the figure most often quoted for the quality of a result. 🔉⇢
Derivation 🔉⇢
Take $n$ repeated observations $a_1, a_2, \ldots, a_n$ of the quantity and compute the mean $\bar{a} = \dfrac{1}{n}\sum a_i$ as the best estimate of the true value.
Form the absolute error of each observation as the magnitude of its deviation from the mean: $\Delta a_i = |\,\bar{a} - a_i\,|$.
Average these absolute errors to obtain the mean absolute error: $\Delta\bar{a} = \dfrac{1}{n}\sum \Delta a_i$, and report $a = \bar{a} \pm \Delta\bar{a}$.
Divide the mean absolute error by the mean to obtain the dimensionless relative error: $\dfrac{\Delta\bar{a}}{\bar{a}}$.
Multiply the relative error by one hundred to express it as the percentage error: $\left(\dfrac{\Delta\bar{a}}{\bar{a}}\right)\times100\%$.
⚠️ JEE trap: Students often assume a small absolute error automatically means a precise measurement; the quality of a measurement is judged by the relative or percentage error, so the same $\pm 0.01\ \mathrm{g}$ is a one percent error on $1.02\ \mathrm{g}$ but only a tenth of a percent on $9.89\ \mathrm{g}$. 🔉⇢
Propagation of Errors: Combining Uncertainties 🔉⇢
🎯 In a product or quotient the percentage errors add, each scaled by its power. Change the combination and watch ΔZ build from ΔA and ΔB.
🔉⇢
ΔZ/Z = |p|·ΔA/A + |q|·ΔB/B = —
What this shows
In a product or quotient the percentage errors add, each scaled by its power. Change the combination and watch ΔZ build from ΔA and ΔB.
Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.
Definition: Propagation of errors is the set of rules that tells how the uncertainty in a computed result follows from the uncertainties in the measured quantities: for a sum or difference the absolute errors add, for a product or quotient the relative errors add, and for a power the relative error is multiplied by the magnitude of the exponent. 🔉⇢
Every measurement involves errors, so the result of any calculation built from measured quantities carries an error of its own. A computed quantity cannot be more accurate than the measured values on which it is based. The rules for the combination of errors answer one question: given the error in each measured quantity, what is the error in a quantity obtained from them by addition, subtraction, multiplication, division or a power? Getting these rules right is the difference between an honest result, reported with a sensible uncertainty, and a result written with more significant figures than the measurement can justify. This is why the estimation of the error in a result is a necessary part of reporting any measured physical quantity, and not an extra step added as an afterthought. 🔉⇢
🎯 The least count of a vernier is one main-scale division shared over N vernier divisions. Add divisions and watch the least count shrink.
🔉⇢
L.C. = 1 MSD ÷ N, 1 MSD = 1 mm = —
What this shows
The least count of a vernier is one main-scale division shared over N vernier divisions. Add divisions and watch the least count shrink.
Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.
Definition: The least count of an instrument is the smallest measurement it can read reliably; for a vernier callipers it equals one main scale division less one vernier scale division, $\text{L.C.} = 1\,\text{MSD} - 1\,\text{VSD}$, and a reading is the main scale reading plus the coinciding vernier division multiplied by the least count. 🔉⇢
The precision of any measurement depends on the least count of the measuring instrument, and the significant figures of a measured value are, in the end, a statement about that least count. The least count is the smallest change in the quantity that the instrument can measure reliably. A plain metre scale has divisions one mm apart, so its least count is one mm, and it cannot honestly report a length to any finer precision; the last, uncertain digit of a metre scale measurement is a guess at a part of the smallest division. The vernier callipers is the instrument that gets past this limit for length, adding one more reliable digit by a neat idea rather than by ruling ever finer divisions on the main scale. 🔉⇢
🎯 A screw gauge's least count is its pitch divided over the circular-scale divisions. Trade pitch against divisions and watch the resolution change.
🔉⇢
L.C. = pitch ÷ n = —
What this shows
A screw gauge's least count is its pitch divided over the circular-scale divisions. Trade pitch against divisions and watch the resolution change.
Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.
Definition: In a screw gauge the least count is the pitch over the number of divisions on the circular scale, $\text{L.C.} = \frac{\text{pitch}}{N}$, and a reading is the linear scale reading plus the coinciding circular scale division multiplied by the least count, correct for zero error and backlash. 🔉⇢
The screw gauge, is the instrument that takes a length measurement to a precision finer than even the vernier callipers, and it does so by using the form of a fine screw. Its precision rests on a simple fact: when a screw turns through one complete turn, it moves along its own axis by a fixed, small distance. By turning a large, easily read turn of a circular scale into a tiny, otherwise tiny move along the axis, the screw gauge makes a very small least count. This is the same idea of trading a coarse motion against a fine one that lies behind the vernier, but here the fine division is made by a turn rather than by a sliding second scale. 🔉⇢
Definition: Dimensional analysis is the method of treating the dimensions $[M^a L^b T^c]$ of physical quantities as algebraic symbols, so that the homogeneity of an equation can be checked and, within limits, the form of a relation among quantities can be deduced. 🔉⇢
🔬 Interactive 3D · Dimensional analysis — interactive 3D. interactive 3D scene
Every measurement is a number together with a unit, and every unit rests on a dimension. The nature of a physical quantity is described by its dimensions: the powers to which the base quantities length, mass and time are raised to represent that quantity. When we place square brackets round a quantity we mean the dimensions of that quantity, and inside those brackets the symbols [M], [L] and [T] are treated as algebraic symbols. Dimensional analysis is the method of carrying those symbols through a calculation. Because a change of units cannot change the nature of a physical quantity, its dimensional formula is fixed: velocity is [L T^-1], acceleration is [L T^-2], and force, as mass times acceleration, is [M L T^-2]. This fixed nature is what makes dimensions a fast preliminary test of the consistency of an equation. 🔉⇢
The one principle on which the whole method rests is the principle of homogeneity of dimensions. The magnitudes of physical quantities may be added together or subtracted from one another only if they have the same dimensions. We cannot add a velocity to a force, nor a length to a mass; the operation is meaningless. It follows that in any equation representing a relationship among physical quantities, every term set apart by a plus or minus sign, and the two sides of the equation, must have one and the same dimensional formula. If the dimensions of all the terms are not the same, the equation is wrong. This is the rule; dimensional analysis is the careful use of it. 🔉⇢
Why does the rule hold? A physical equation must stay true whatever standard we choose for our units. If an equation were dimensionally inconsistent, then a change of units, say from metre to centimetre or from second to minute, would multiply different terms by different numerical factors and the equation would break. Only when every term has the same dimensional formula do all terms scale by exactly the same factor under a change of units, so the equation is kept. Thus dimensional homogeneity is not an arbitrary convention; it follows from the demand that physics not depend on our arbitrary choice of reference standard. A test of consistency of dimensions tells us no more and no less than a test of consistency of units, but it has the advantage that we need not commit ourselves to any particular choice of units, nor worry about conversions among multiples and sub-multiples. 🔉⇢
The first and most common use is checking the correctness of an equation. Suppose a relation is written for the distance travelled by a body that starts from an initial position with an initial velocity under uniform acceleration. To test it, we write the dimensional formula of each term. The position term is a length [L]; the term formed from initial velocity times time is [L T^-1][T] = [L]; the term formed from acceleration times the square of time is [L T^-2][T^2] = [L]. Every term reduces to [L], the dimension of the left side, so the equation passes the homogeneity test. Note that the numerical factor of one half in front of the acceleration term is a pure number and has no dimensions at all; dimensional analysis does not see such factors. 🔉⇢
This is the first and most important limit of the method, and JEE examiners test it often. Dimensional consistency does not guarantee a correct equation. The method is uncertain to the extent of dimensionless quantities and dimensionless functions. A relation may be perfectly homogeneous and still be physically wrong: a pure number may be misplaced, a dimensionless factor such as an angle or a ratio may be left out, or two different physical quantities may happen to have the same dimensional formula. Work and torque both have the dimensions [M L^2 T^-2], yet they are different physical quantities; energy and the moment of a force cannot be told apart by dimensions alone. The correct statement is one sided: if an equation fails the consistency test it is proved wrong, but if it passes it is not thereby proved right. 🔉⇢
A second rule of great use concerns the arguments of special functions. The arguments of trigonometric, logarithmic and exponential functions must be dimensionless. One cannot take the sine of a length or the logarithm of a mass; the quantity fed into such a function must be a pure number, most often the ratio of two similar quantities. An angle, as the ratio of an arc length to a radius, is a ratio of length to length and is therefore dimensionless; the refractive index, as the ratio of the speed of light in vacuum to the speed of light in a medium, is a ratio of speed to speed and is likewise dimensionless. Whenever an expression contains an exponential or an oscillation term, at once check that the combination of quantities inside the bracket reduces to a pure number. If it does not, the expression is wrong, and this one observation settles a large fraction of JEE Main dimensional questions. 🔉⇢
The second great use is deducing a relation among physical quantities. The method of dimensions can sometimes be used to deduce the form of a relation when we already know, from physical reasoning, which quantities the result depends on, provided the number of independent quantities is small, usually not more than three. We assume the quantity depends on the others as a product of powers, with unknown exponents and one dimensionless constant. We then write the dimensional formula of both sides, use homogeneity, and equate the powers of [M], [L] and [T] one by one. This gives a small set of linear equations for the exponents, which we solve. The method is mechanical once the dependence is assumed, and it recovers the correct physics remarkably often. 🔉⇢
The standard example, and the one every JEE student should be able to reproduce, is the period of oscillation of a simple pendulum. We suppose the period depends on the length of the pendulum, the mass of the bob, and the acceleration due to gravity, as a product of powers. Writing the dimensions of both sides and equating the powers of length, mass and time, we find that the exponent of mass must be zero, so the period cannot depend on the mass of the bob at all, a real physical prediction that follows purely from dimensions. We also find that length enters to the power one half and gravity to the power minus one half, so the period is proportional to the square root of length divided by gravity. The dimensionless constant, which is $2\pi$, cannot be obtained by this method; here it does not matter that some number multiplies the right side, because that does not affect its dimensions. 🔉⇢
It is worth being clear about what dimensional deduction can and cannot give, because this is exactly the limit JEE Advanced problems probe. The method gives the exponents of each physical quantity and hence the form of the relation up to a pure number that multiplies it. It cannot give that pure number, it cannot give a dimensionless combination that the result might also depend on, and it fails when the result depends on more independent quantities than there are base dimensions to fix them. If a quantity depends on four independent mechanical quantities, three equations in the exponents, from [M], [L] and [T], cannot fix four unknowns, and the simple product of powers breaks down; the honest answer is that dimensions alone are not enough and more physical input is needed. 🔉⇢
A further point arises when the assumed dependence is not a simple product. Dimensional analysis assumes a product of powers; if the true relation is a sum of terms of different form, or involves a dimensionless factor in a non-simple way, the single product of powers cannot capture it. This is why the method works for the pendulum, for the speed of a wave on a stretched string, and for the range of a projectile in terms of speed and gravity, but stalls for quantities that depend on several independent lengths, where only a dimensionless ratio of those lengths can appear and dimensions cannot fix the function of that ratio. 🔉⇢
In the examination, the working is disciplined and fast. First, identify the dimensional formula of every quantity in the expression, writing each in the standard bracket notation. Second, if you are checking an equation, reduce every term to base dimensions and confirm that they all agree with each other and with the target dimension; a single term that does not match proves the equation wrong. Third, if you are deducing, list the quantities the result depends on, write the product of powers, and form one linear equation per base dimension by equating the exponents. Fourth, solve for the exponents and write the relation, leaving one dimensionless constant in front. Fifth, check the physics: does the relation make sense in limiting cases? A relation that gives an infinite period as gravity goes to zero, for example, is behaving sensibly. 🔉⇢
Dimensional reasoning also guides the conversion of a physical quantity from one system of units to another, a common exercise. Because the magnitude of a quantity times the size of its unit is fixed, a quantity with dimensional formula [M^a L^b T^c] has its numerical value multiplied by definite powers of the ratios of the base units when we change systems. If the units of mass, length and time each change, the new numerical value is the old numerical value scaled by the proper powers of those ratios. This is exactly the reasoning behind showing that a calorie has a particular magnitude in a new system in which the units of mass, length and time are chosen arbitrarily; you express energy as [M L^2 T^-2] and follow how each base unit is scaled. 🔉⇢
There is a deeper reason the method is reliable: dimensions describe the quality of a physical quantity, not its magnitude. In this representation the magnitudes are not considered; it is the type of the physical quantity that enters. A change in velocity, an initial velocity, an average velocity and a final velocity are all equivalent in this context, because all are lengths divided by times and all have the dimensional formula [L T^-1]. This is why dimensional analysis cannot distinguish quantities of the same kind, and why it can never replace a full derivation from the actual definitions and laws. It is a preliminary test and a guide, not a substitute for physics. 🔉⇢
Students often ask why we use a method that cannot even give the factor of $2\pi$. The answer is that dimensional analysis is a fast and reliable filter. In a multi-step calculation it lets you catch an algebraic mistake at once: if your final expression for a speed does not reduce to [L T^-1], you have made a mistake somewhere, and you know it before you ever put in numbers. It also lets you rebuild a half known formula under examination pressure, up to a constant, and it lets you decide which of several given formulae could possibly be correct. In the standard multiple choice setting, several given formulae for the kinetic energy can be ruled out at once because their dimensions do not reduce to [M L^2 T^-2], while dimensional arguments cannot decide between two that share the correct dimensions but differ only by a pure number. 🔉⇢
The recognition of the concept of dimensions, which guides the description of physical behaviour, is of basic importance, because only those physical quantities that have the same dimensions can be added or subtracted. A thorough understanding of dimensional analysis helps in deducing certain relations among different physical quantities and in checking the derivation, accuracy and dimensional consistency of various mathematical expressions. When magnitudes of two or more physical quantities are multiplied, their units and dimensions are treated exactly as ordinary algebraic symbols; identical units and dimensions in the numerator and denominator cancel. This algebra of dimensions runs quietly behind almost every quantitative step in physics, and learning it early pays off across mechanics and the rest of the subject alike. 🔉⇢
To consolidate: dimensional analysis is homogeneity used with algebra. It checks equations by demanding that every term have one dimensional formula; it deduces relations by assuming a product of powers and equating the exponents of the base quantities; and it converts quantities between systems by following how the base units are scaled. Its limits are equally definite: it does not see dimensionless constants and dimensionless functions, it cannot separate quantities of identical dimensions, and it fails when the number of relevant independent quantities is more than the number of base dimensions. Hold both the power and the limits in mind together, and you will neither over trust a merely consistent equation nor under use a method that can save a derivation in seconds. 🔉⇢
A few more checks make the method concrete in practice. Take the relation for the time period written as $T = 2\pi\sqrt{l/g}$: the quantity inside the square root is a length divided by an acceleration, that is [L] divided by [L T^-2], which reduces to [T^2], so the square root has the dimension [T]; the pure number $2\pi$ leaves the dimension unchanged, so the right side is [T] and matches the left side. Consider also the given formulae for the kinetic energy of a body: any formula whose dimensions do not reduce to [M L^2 T^-2] is ruled out at once, and only the formulae with the correct dimensions remain, though dimensions cannot choose the correct pure number among those that remain. Each such check takes only a few seconds and guards against a wrong formula before any numbers are used, which is exactly why dimensional checking is so useful under time pressure. 🔉⇢
It also helps to remember that in computing any physical quantity, the units for the derived quantities in the relationship are treated as though they were algebraic quantities until the desired units are obtained. We may cancel identical units in the numerator and denominator exactly as we cancel identical dimensions. Tracking units and tracking dimensions are the same operation, and both guard the final result. Because dimensions do not depend on the choice of units, a dimensionally correct relation stays correct in the CGS, the FPS and the MKS systems alike, which is why a check by dimensions is often used in place of a check by units when we do not wish to commit ourselves to a particular system of units. 🔉⇢
Finally, treat dimensional homogeneity as a habit, not a last resort. Before you accept any result, look at its dimensions; before you learn a formula, understand which quantities it must contain from dimensional need; and whenever you meet a special function, verify that its argument is dimensionless. This habit turns the principle of homogeneity into an instinct, and that instinct is exactly what marks a fast, accurate JEE performance from a slow one that is full of error. A number of exercises at the end of the chapter are designed to develop exactly this skill in dimensional analysis, and they reward the disciplined working described above. 🔉⇢
Derivation from first principles 🔉⇢
Goal: deduce how the period $T$ of a simple pendulum depends on its length $l$, the mass $m$ of the bob, and the acceleration due to gravity $g$, using only the method of dimensions.
Assume a product-of-powers dependence with a dimensionless constant $k$: $T = k\, l^{x} g^{y} m^{z}$, where $x$, $y$, $z$ are the unknown exponents to be found.
Write the dimensional formula of each quantity: $[T]=[\mathrm{T}]$, $[l]=[\mathrm{L}]$, $[g]=[\mathrm{L\,T^{-2}}]$, $[m]=[\mathrm{M}]$. Substituting gives $[\mathrm{M^0 L^0 T^1}] = [\mathrm{L}]^{x}[\mathrm{L\,T^{-2}}]^{y}[\mathrm{M}]^{z}$.
Collect powers on the right-hand side: $[\mathrm{M^0 L^0 T^1}] = [\mathrm{M}]^{z}\,[\mathrm{L}]^{x+y}\,[\mathrm{T}]^{-2y}$.
Invoke homogeneity: the power of each base quantity must match on both sides. Equate exponents of mass, length and time separately: $z = 0$; $\;x + y = 0$; $\;-2y = 1$.
Solve the linear system: from $-2y = 1$ we get $y = -\dfrac{1}{2}$; then $x = -y = \dfrac{1}{2}$; and $z = 0$, so the period cannot depend on the mass of the bob.
Reassemble: $T = k\, l^{1/2} g^{-1/2} = k\,\sqrt{\dfrac{l}{g}}$. The dimensionless constant $k$ cannot be found by dimensions; experiment and the full derivation give $k = 2\pi$, hence $T = 2\pi\sqrt{\dfrac{l}{g}}$.
Interpretation: dimensions fixed the functional form and the surprising mass-independence, but were silent on the pure number $2\pi$ — a clean illustration of both the power and the limit of the method.
⚠️ JEE trap: A very common JEE trap is to believe that a dimensionally correct equation must be physically correct. In truth the test is one-sided: a dimensionally wrong equation is certainly wrong, but a dimensionally consistent one may still be wrong because dimensions ignore pure numbers, dimensionless factors, and the distinction between quantities such as work and torque that share the formula $[\mathrm{M\,L^2\,T^{-2}}]$. 🔉⇢
Worked example · JEE Advanced 🔉⇢
SITUATION A transverse pulse travels along a stretched string. On physical grounds the wave speed $v$ can depend only on the tension $F$ in the string (a force) and on the linear mass density $\mu$ (mass per unit length).
TARGET Deduce the form of $v$ as a function of $F$ and $\mu$ using dimensional analysis, and state clearly what the method cannot determine.
STRATEGY Assume a product of powers $v = k\,F^{a}\mu^{b}$ with a dimensionless constant $k$, write the dimensional formula of each quantity, apply the principle of homogeneity, and equate the powers of $[\mathrm{M}]$, $[\mathrm{L}]$ and $[\mathrm{T}]$ to solve for $a$ and $b$.
EXECUTE Dimensions: $[v]=[\mathrm{L\,T^{-1}}]$, $[F]=[\mathrm{M\,L\,T^{-2}}]$, $[\mu]=[\mathrm{M\,L^{-1}}]$. Then $[\mathrm{L\,T^{-1}}] = [\mathrm{M\,L\,T^{-2}}]^{a}\,[\mathrm{M\,L^{-1}}]^{b} = [\mathrm{M}]^{a+b}[\mathrm{L}]^{a-b}[\mathrm{T}]^{-2a}$. Equating exponents: for mass $a+b=0$; for time $-2a=-1\Rightarrow a=\dfrac{1}{2}$; hence $b=-\dfrac{1}{2}$; the length equation $a-b = \dfrac{1}{2}-\left(-\dfrac{1}{2}\right)=1$ is satisfied, a useful consistency check. Therefore $v = k\,F^{1/2}\mu^{-1/2} = k\,\sqrt{\dfrac{F}{\mu}}$.
REFLECT Dimensions fix the form $v\propto\sqrt{F/\mu}$ and even provide a self-check through the redundant length equation, but they cannot supply the pure number $k$ (which a full derivation shows to be $1$). This is the method working exactly to its stated limit.
Source: NCERT-derived
Propagation of Errors: Combining Uncertainties 🔉⇢deep concept
Definition: Propagation of errors is the set of rules that tells how the uncertainty in a computed result follows from the uncertainties in the measured quantities: for a sum or difference the absolute errors add, for a product or quotient the relative errors add, and for a power the relative error is multiplied by the magnitude of the exponent. 🔉⇢
🔬 Interactive 3D · Propagation of errors — interactive 3D. interactive 3D scene
Every measurement involves errors, so the result of any calculation built from measured quantities carries an error of its own. A computed quantity cannot be more accurate than the measured values on which it is based. The rules for the combination of errors answer one question: given the error in each measured quantity, what is the error in a quantity obtained from them by addition, subtraction, multiplication, division or a power? Getting these rules right is the difference between an honest result, reported with a sensible uncertainty, and a result written with more significant figures than the measurement can justify. This is why the estimation of the error in a result is a necessary part of reporting any measured physical quantity, and not an extra step added as an afterthought. 🔉⇢
Recall the vocabulary first, since the rules are stated in its terms. If a quantity is measured with an error $\Delta a$, its absolute error, then the relative error is the ratio $\Delta a / a$, a pure number that gives the error as a fraction of the quantity itself, and the percentage error is that ratio times one hundred. The relative error is dimensionless and is the natural measure for products and quotients, while the absolute error has the same units as the quantity and is the natural measure for sums and differences. Much of the confusion among students clears the moment they see that the choice between adding absolute errors and adding relative errors is decided entirely by whether the quantities are added or multiplied. 🔉⇢
Consider first a sum. Suppose a quantity $Z$ is the sum of two measured quantities, $Z = A + B$, each with its own error. Each measured value may lie anywhere within its error, so the extreme values of the sum occur when both quantities err in the same direction. The largest value of $Z$ is obtained by adding both largest values, and the smallest by adding both smallest values. The result is clean and worth stating plainly: when two quantities are added, the absolute errors add. The maximum error in the sum is the sum of the errors in the individual quantities. Nothing here depends on the units, since both quantities and both errors have the same units. 🔉⇢
The subtraction case is the one that traps the unwary, and JEE examiners use it often. When a quantity is the difference of two measured quantities, $Z = A - B$, one is tempted to imagine the errors subtracting. They do not. To find the worst case we again let the quantities err so as to push the result as far from the mean as possible; the extreme difference occurs when one quantity takes its largest value while the other takes its smallest. The absolute errors therefore still add, exactly as for a sum. This has an important result: when we subtract two nearly equal quantities, the error stays fixed while the result itself becomes small, so the relative error can become very large. Subtraction of nearly equal measured numbers ruins precision, and a well planned experiment avoids it where possible. 🔉⇢
Now turn to products and quotients, where relative errors take over from absolute errors. Suppose a result is the product of two measured quantities, $Z = A B$. The cleanest way to see the rule is to write the fractional change of the product. If each factor is uncertain by a small fraction, then to first order in small quantities the fractional change of the product is the sum of the fractional changes of the factors; the product of the two small errors is a very small quantity of second order that we neglect. The rule that follows is the central result of the whole topic: when two quantities are multiplied, the relative errors add. The maximum relative error in a product is the sum of the relative errors in the individual factors. 🔉⇢
The quotient behaves in exactly the same way. If a result is the ratio of two measured quantities, the relative error in the ratio is again the sum of the relative errors of the numerator and the denominator, not their difference. The reasoning follows the subtraction case: to find the worst case we let the numerator err upward and the denominator err downward, or the reverse, and in either extreme the relative errors add rather than cancel. So division, like multiplication, adds relative errors. This likeness between multiplication and division, both adding relative errors, is the single most useful fact to carry into an examination, since most physical formulae are products and quotients of measured quantities. 🔉⇢
Powers are repeated multiplication, and the rule follows at once. If a quantity is raised to some exponent, then because multiplying a quantity by itself adds its relative error to itself that many times, the relative error is multiplied by the magnitude of the exponent. A quantity that is squared contributes twice its relative error to the result; a quantity under a square root, which is the power one half, contributes half its relative error. The magnitude of the exponent is what matters, so a quantity in the denominator raised to some power still contributes a positive relative error equal to that power times its own relative error; the sign of the exponent never makes errors subtract. This is why, in a formula where one quantity enters to a high power, the precision of that one measurement dominates the overall uncertainty and deserves the most careful instrument. 🔉⇢
The general recipe for any formula that is a product of powers of several measured quantities now writes itself. Take the quantity of interest, express it as a product of the measured quantities each raised to its own exponent, and then the maximum relative error in the result is the sum, over all the measured quantities, of the magnitude of each exponent times the relative error of that quantity. Change to a percentage at the very end if a percentage is wanted. This one sentence covers every special case: sums and differences are handled by their own rule on absolute errors, while every product, quotient and power reduces to the exponent weighted sum of relative errors. 🔉⇢
A worked step makes the recipe concrete. Imagine a quantity given by one measured length in the numerator and the square of a measured time in the denominator, times some exact constants. The constants, being exact numbers, contribute no error at all; only the measured length and the measured time carry error. The relative error in the result is therefore the relative error in the length plus twice the relative error in the time, the factor of two arising because the time is squared. If the length is known to one part in two hundred and the time to one part in one hundred, the result is uncertain by one part in two hundred plus two parts in one hundred, and the time measurement, entering as a square, dominates. The lesson for the design of an experiment is immediate: measure most carefully the quantity that enters with the largest exponent. 🔉⇢
It is important to understand why we add errors in the worst case rather than allowing them to partly cancel. The rules described here give the maximum possible error, obtained by assuming that every individual error acts in the least favourable direction at once. This is a safe estimate; it never gives too small an uncertainty. In a careful statistical treatment, where errors are random and independent, the errors are combined by adding their squares and taking the square root, which usually gives a smaller and more realistic estimate. For the standard syllabus and most JEE problems, however, the maximum error rules just given are the expected answer, and you should apply them unless a problem clearly asks for the statistical combination. 🔉⇢
The rules join with the treatment of significant figures, and the two must agree. If a set of experimental data is specified to a certain number of significant figures, a result obtained by combining the data will also be valid to about that many significant figures; the error in the result tells you which digit becomes uncertain, and you round the result so that its last reported digit sits at the position of that uncertainty. Reporting more digits than the error justifies is superfluous and misleading, exactly as reporting a density to eleven decimal places would be absurd when the measured values are known to only three significant figures. The combination of errors and the significant figures are two views of one idea, that the reported result must reflect the precision of the measurements it came from. 🔉⇢
A point worth stating concerns intermediate rounding. When a calculation has several steps, you should retain in the intermediate steps one digit more than the significant figures justified by the least precise measurement, and round to the proper number of significant figures only at the very end. Rounding too hard at each stage lets rounding errors build up and can shift the final uncertain digit. The reciprocal example in the chapter shows this: rounding a reciprocal to three figures and then taking the reciprocal again fails to return the original value, while keeping one extra digit throughout recovers it. The combination of errors assumes you have not already spoiled your numbers by rounding too early. 🔉⇢
Let us collect the four rules into a compact list you can use under time pressure. For a sum or a difference, add the absolute errors, and never let subtraction fool you into subtracting them. For a product or a quotient, add the relative errors, treating division exactly like multiplication. For a power, multiply the relative error by the magnitude of the exponent. And for any formula that mixes these, express it as a product of powers, add the exponent weighted relative errors, handle any additive parts by the rule on absolute errors, and change to a percentage last. With this list the combination of errors becomes almost mechanical, and the only remaining skill is the arithmetic itself. 🔉⇢
The physical meaning behind the arithmetic deserves a final word. An uncertainty is not a mistake; it is an honest statement of how precisely a quantity is known. Combining errors is how we carry that honesty through a calculation so that the final result shows its own precision. A result written as a central value together with an absolute error, or with a percentage, tells the reader far more than a bare number: it says how much of the calculated figure is reliable and how much is only the effect of an imprecise measurement. This is why the estimation of the error in a result is not an optional step but a real part of reporting any measured physical quantity. 🔉⇢
In experimental JEE questions the combination of errors is usually the whole point of the problem. A typical question gives the least count of each instrument, and hence the error in each measured value, asks you to compute a derived quantity through a known formula, and then asks for the percentage error in that quantity. The disciplined path is always the same: change each least count into a relative error, weight each relative error by the exponent with which its quantity appears in the formula, add them, and present the result as a percentage. If any quantity appears in a sum or a difference, switch to absolute errors for that step. Following this path turns a formidable problem with many quantities into a short and reliable sum. 🔉⇢
It is also worth seeing the sum and difference rule and the product and quotient rule as one idea in two forms. In both, we take the worst case by letting each error act in the direction that pushes the result furthest from its mean value, and then we add the contributions. For a sum or a difference the contributions are the absolute errors themselves, because the result has the same units as the quantities; for a product or a quotient the contributions are the relative errors, because the result is a product and the fractional changes add. Once this common origin is clear, the four rules stop being four separate facts to memorise and become one rule applied to two kinds of combination. 🔉⇢
A short numerical illustration ties the ideas together. Suppose the length of a thin rectangular sheet is measured as $16.2$ cm with an error of $0.1$ cm, and its breadth as $10.1$ cm with an error of $0.1$ cm. The relative error in the length is about $0.6\%$ and in the breadth about $1\%$. Since the area is the product of length and breadth, the relative errors add, giving about $1.6\%$ in the area. The area itself is about $163.6$ cm$^2$, so the error in the area is about $2.6$ cm$^2$, and we report the area as roughly $164 \pm 3$ cm$^2$. Every step uses only the product rule and the meaning of relative error, and the reported result carries a sensible uncertainty. 🔉⇢
The relative error of a measured value depends not only on the number of significant figures but also on the number itself. For example, a mass measured as $1.02$ g accurate to $\pm 0.01$ g has a relative error of about $1\%$, while a mass measured as $9.89$ g, also accurate to $\pm 0.01$ g, has a relative error of only about $0.1\%$. The same absolute error therefore gives a much larger relative error for the smaller quantity. This is a useful check when reporting a result: a small measured quantity with a fixed absolute error is known far less precisely, in relative terms, than a large one, and its larger relative error will dominate the combination of errors. 🔉⇢
In addition or subtraction the result should keep as many decimal places as are there in the value with the least decimal places, while in multiplication or division the result should keep as many significant figures as are there in the value with the least significant figures. These arithmetic rules for significant figures and the combination of errors always agree, because both express the same limit set by the precision of the least precise measured value. When you round the final result, place its last reported digit at the position fixed by the error you have just computed, and no further, so that the reported precision matches the precision of the measurement. 🔉⇢
To summarise, the combination of errors rests on whether quantities combine by a sum or difference or by a product, quotient or power. A sum or a difference adds the absolute errors and can ruin the precision of a difference of nearly equal numbers. A product, quotient or power adds the relative errors, with a power weighting the relative error by its exponent. The rules give a safe maximum error, they must agree with the significant figures of the data, and they need care with intermediate rounding. Master these and you can report any derived quantity with a defensible uncertainty, which is exactly what a quantitative science based on measurement requires of every result. 🔉⇢
Derivation from first principles 🔉⇢
Setup: let a derived quantity be a product of powers of measured quantities, $Z = k\,A^{p} B^{q}$, where $k$ is an exact constant and $A$, $B$ carry small absolute errors $\Delta A$, $\Delta B$. We seek the maximum relative error $\dfrac{\Delta Z}{Z}$.
Take natural logarithms of both sides to turn products and powers into sums: $\ln Z = \ln k + p\ln A + q\ln B$.
Differentiate: $\dfrac{dZ}{Z} = p\,\dfrac{dA}{A} + q\,\dfrac{dB}{B}$, since the exact constant $k$ contributes nothing (its differential is zero).
Replace the infinitesimals by finite small errors: $\dfrac{\Delta Z}{Z} = p\,\dfrac{\Delta A}{A} + q\,\dfrac{\Delta B}{B}$ (valid to first order because products of small errors are negligible).
To obtain the maximum possible error, the individual errors are allowed to reinforce rather than cancel, so we take the magnitude of each term and add: $\left(\dfrac{\Delta Z}{Z}\right)_{\max} = |p|\,\dfrac{\Delta A}{A} + |q|\,\dfrac{\Delta B}{B}$.
Read off the special cases: for a product ($p=q=1$) or a quotient ($p=1,\,q=-1$) the relative errors add, $\dfrac{\Delta Z}{Z} = \dfrac{\Delta A}{A} + \dfrac{\Delta B}{B}$; for a power $Z=kA^{p}$ the relative error scales with the magnitude of the exponent, $\dfrac{\Delta Z}{Z} = |p|\,\dfrac{\Delta A}{A}$.
Sum and difference are handled separately from first principles: for $Z = A \pm B$ the extreme value occurs when both quantities err in the reinforcing sense, giving $\Delta Z = \Delta A + \Delta B$ in both cases — the absolute errors add even for a difference.
Convert to percentage at the end by multiplying the relative error by one hundred, and round the central value so its last quoted digit matches the position of the uncertainty.
⚠️ JEE trap: Students often think that when two quantities are subtracted their errors also subtract, and that when quantities are divided the relative errors subtract. Both are wrong: for a difference the absolute errors add (so subtracting nearly equal numbers can give a huge relative error), and for a quotient the relative errors add exactly as for a product. 🔉⇢
Worked example · JEE Advanced 🔉⇢
SITUATION The acceleration due to gravity is found with a simple pendulum from $g = \dfrac{4\pi^{2} L}{T^{2}}$. The length is measured as $L = 1.00\ \text{m}$ with absolute error $\Delta L = 0.01\ \text{m}$, and the period as $T = 2.00\ \text{s}$ with absolute error $\Delta T = 0.02\ \text{s}$.
TARGET Find the maximum percentage error in the computed value of $g$.
STRATEGY Only $L$ and $T$ are measured; $4\pi^{2}$ is an exact constant contributing no error. Apply the product-of-powers rule: $L$ enters to the first power and $T$ to the power $-2$, so the relative error in $g$ is the relative error in $L$ plus twice the relative error in $T$.
EXECUTE Relative errors: $\dfrac{\Delta L}{L} = \dfrac{0.01}{1.00} = 0.01$ and $\dfrac{\Delta T}{T} = \dfrac{0.02}{2.00} = 0.01$. Combine with the correct weights: $\dfrac{\Delta g}{g} = \dfrac{\Delta L}{L} + 2\,\dfrac{\Delta T}{T} = 0.01 + 2(0.01) = 0.03$. As a percentage, $\dfrac{\Delta g}{g}\times 100\% = 3\%$. The period, entering as a square, contributes $2\%$ of the total $3\%$ — twice the share of the length.
REFLECT Even though $L$ and $T$ are each known to $1\%$, $g$ is uncertain to $3\%$ because the period appears squared and its relative error is doubled. This is why, in a pendulum experiment, the timing must be made most precise, typically by measuring many oscillations at once to shrink $\Delta T / T$.
Source: NCERT-derived
Least Count & the Vernier Callipers 🔉⇢deep concept
Definition: The least count of an instrument is the smallest measurement it can read reliably; for a vernier callipers it equals one main scale division less one vernier scale division, $\text{L.C.} = 1\,\text{MSD} - 1\,\text{VSD}$, and a reading is the main scale reading plus the coinciding vernier division multiplied by the least count. 🔉⇢
🔬 Interactive 3D · Vernier callipers — interactive 3D. interactive 3D scene
The precision of any measurement depends on the least count of the measuring instrument, and the significant figures of a measured value are, in the end, a statement about that least count. The least count is the smallest change in the quantity that the instrument can measure reliably. A plain metre scale has divisions one mm apart, so its least count is one mm, and it cannot honestly report a length to any finer precision; the last, uncertain digit of a metre scale measurement is a guess at a part of the smallest division. The vernier callipers is the instrument that gets past this limit for length, adding one more reliable digit by a neat idea rather than by ruling ever finer divisions on the main scale. 🔉⇢
The device has two scales. The main scale is a fixed scale marked in mm, like an ordinary scale, along which the measured object is placed. Beside it slides a small second scale, the vernier scale, fixed to the moving part of the instrument. The measured object is held between the fixed part and the moving part, and the position of the moving part is read where the vernier scale sits against the main scale. The full value of the design lies in the way the vernier scale is divided, since it is this division that gives the extra digit of precision that the main scale alone cannot give. 🔉⇢
Here is the key design. The vernier scale has a fixed number of divisions, and these divisions are ruled so that the full set of them covers a length equal to one division less on the main scale. In the common instrument the sliding scale has a number of divisions covering the same length as one less main scale division; for example a vernier with ten divisions covers the same length as nine main scale divisions. The result is that one vernier scale division is a little smaller than one main scale division. It is this small, planned difference between one main scale division and one vernier scale division that the instrument uses, and that difference is exactly the least count. 🔉⇢
This gives the defining relation, which you should be able to rebuild rather than only recall: the least count is one main scale division less one vernier scale division. Since the full vernier covers one less main scale division, the least count equals the length of one main scale division over the number of divisions on the vernier scale. With a main scale division of one mm and ten vernier divisions, the least count is one mm over ten, that is one tenth of a mm; a vernier with twenty divisions on the sliding scale gives one twentieth of a mm, and a finer vernier gives finer yet. The more divisions on the sliding scale, the smaller the least count and the more precise the instrument, which is exactly why one vernier callipers can be more precise than another. 🔉⇢
Reading the instrument is a two part act, and understanding why it works is worth more than memorising the steps. First, note the main scale value: the main scale mark just before the zero of the vernier scale gives the full part of the measurement, read to the nearest main scale division. This fixes the length to within one main scale division, but leaves the part, the position of the vernier zero inside that last division, yet to be found. That part is exactly what the vernier scale measures, and it does so by a coincidence rather than by a guess. 🔉⇢
The second part is to find which vernier division coincides most closely with a mark on the main scale. Since a vernier division is smaller than a main scale division by exactly the least count, each vernier mark falls a little further behind each main scale mark; the one vernier division that best coincides with a main scale mark tells you, by its number, how many least counts the vernier zero has gone into the last main scale division. So if the fifth vernier division coincides, the vernier zero sits five least counts beyond the main scale mark before it. The vernier scale part of the measurement is therefore this coinciding division number multiplied by the least count. Adding it to the main scale value gives the total: measured value equals main scale value plus the coinciding vernier division multiplied by the least count. 🔉⇢
It is worth pausing on why this coincidence method beats a simple guess by eye. The human eye cannot reliably split one mm into ten equal parts by sight, but it can reliably judge which of many close ruled marks best coincides with a neighbouring mark. The vernier turns the hard task of guessing a part into the easy task of spotting which marks agree, and in doing so it turns an uncertain guessed digit into a reliable read digit. This is the full reason the instrument exists: it makes the least count of a length measurement ten or twenty times smaller than the metre scale, and it does so without asking for better eyesight. 🔉⇢
No real instrument is perfect, and the most important defect is the zero error. Ideally, when the two ends are brought gently together with nothing between them, the zero of the vernier scale should meet exactly with the zero of the main scale, so that the instrument gives zero for zero separation. Often it does not. If the vernier zero lies to the right of the main scale zero when the ends touch, the instrument gives a small positive amount for nothing, and this is a positive zero error; if the vernier zero lies to the left, the instrument gives small and the zero error is negative. The zero error is a fixed error of the same amount in the same direction in every measurement, and unlike a change that varies at random it cannot be reduced by repeating the measurement. 🔉⇢
Correcting for the zero error follows one rule that students constantly reverse: the correct value equals the measured value less the zero error, keeping the sign of the zero error. A positive zero error must be subtracted, which lowers the measured value; a negative zero error, being a negative quantity, is also subtracted, and subtracting a negative amount raises the measured value. To find the zero error itself you read the instrument with the ends closed exactly as you would any measurement, note which vernier division coincides, and multiply by the least count, with the correct sign. Since the zero error is fixed and repeats, once found it can be applied to every later measurement, restoring accuracy. 🔉⇢
A small numerical walk through fixes the method. Suppose the main scale division is one mm and the vernier has ten divisions, so the least count is one tenth of a mm. Suppose the object pushes the vernier zero just past the ten mm mark on the main scale, so the main scale value is ten mm, and suppose the fourth vernier division coincides best with a main scale mark. The vernier part is four multiplied by the least count, that is four tenth of a mm. The measured value is ten plus four tenth, or ten point four mm. If the instrument had a positive zero error of two tenth of a mm, the true measurement would be ten point four less two tenth, giving ten point two mm. Each step is small; the discipline is in never skipping the zero error. 🔉⇢
The significant figures of a vernier measurement are set by its least count, and reporting more digits than the least count supports is meaningless. An instrument with a least count of one tenth of a mm justifies reporting a length to one tenth of a mm and no finer; more digits would give a false idea of the precision of the measurement. This is the same idea that governs significant figures everywhere in the subject: the reliable digits, plus the first uncertain one, are exactly the digits the least count of the instrument allows. A vernier callipers, by making the least count ten or twenty times smaller than the metre scale, increases the number of significant figures we may honestly report. 🔉⇢
The vernier callipers is the instrument of choice for a fixed range of length measurements: the inside and outside diameter of tubes and rods, the depth of a hole, and the length of small solid objects, from a few mm up to a good part of the length of the main scale. Its two pairs of jaws measure outside and inside sizes, and a thin depth rod, moving with the sliding scale, measures depth. Within this range it is a happy middle choice: much more precise than a metre scale, and more convenient and wider in range than the screw gauge, which gives even finer least counts but only over a very small length. Choosing the right instrument for the precision needed is itself part of good measurement. 🔉⇢
Errors that vary at random yet affect the vernier callipers even after the fixed zero error is removed, and these are handled by taking many measurements. Small differences in how firmly the jaws are closed, small errors in judging the coincidence, and real irregularities in the object all make repeat measurements scatter about a central value. The remedy is to take many measurements and average them, since a set of many measurements gives a more reliable estimate than a set of only a few; the average of the amounts by which the measurements differ from their mean then gives the size of the remaining uncertainty. Thus a complete vernier measurement joins a fixed correction for the zero error with an average over many measurements to handle the scatter. 🔉⇢
It helps to see the vernier idea as one case of a general way to beat a coarse least count: place two scales whose divisions differ by a small, known amount, and read the fine part from which of their marks meet rather than from a direct guess. The same idea appears again, turned through a right angle, in the screw gauge, where a linear pitch is over a circular scale; and it appears again in more advanced optical instruments. Recognising the shared idea means you never have to memorise the vernier as a separate recipe: you can rebuild the least count relation, the reading rule and the zero error correction from the one idea that the two scales differ by exactly one least count per division. 🔉⇢
To help the working method for the examination: identify one main scale division and count the vernier divisions, then find the least count as one main scale division over the number of vernier divisions. Check for a zero error by closing the ends and recording the coinciding division with its sign. Take the actual measurement by reading the main scale just before the vernier zero, adding the coinciding vernier division multiplied by the least count, and then subtracting the zero error with its sign. Report the result to the number of significant figures the least count allows, and where scatter matters, repeat and average. Done in this order, a vernier measurement is fast, orderly and sound, which is exactly what a precise result needs. 🔉⇢
It is useful to state the range of the least count in numbers. With a main scale in mm, a vernier of ten divisions gives a least count of $0.1$ mm, a vernier of twenty divisions gives $0.05$ mm, and a vernier of fifty divisions gives $0.02$ mm. In every case the least count is one main scale division over the number of vernier divisions, so a larger number of divisions on the sliding scale gives a smaller least count and a more precise instrument. This is why, when a case asks which of two vernier callipers is the more precise device, the answer is simply the one with the larger number of divisions on the sliding scale, since that instrument has the smaller least count. 🔉⇢
Before any set of measurements it is good practice to test the instrument for its zero error, because a zero error left uncorrected shifts every value by the same amount and cannot be removed by averaging. Close the jaws gently on nothing, read the value exactly as for a real measurement, and record it with its sign as the zero error. Then, for each measured object, take the measured value from the main scale and the vernier scale in the usual way and subtract the zero error with its sign. Because the zero error is the same for every measurement, this one test at the start protects the accuracy of the full set of measurements, and it costs only a moment. 🔉⇢
A final word on why this instrument earns its place in the syllabus. The vernier callipers is the first instrument in which a student sees, in metal, the ideas of least count, fixed error and precision made concrete. Every idea from the theory of measurement appears in it: a reliable value plus one uncertain digit, a fixed zero error that must be correct rather than averaged away, scatter that must be averaged, and a precision set by the design of the instrument. Understanding the vernier well is therefore not only about one device; it is about taking in how careful measurement is actually done, which is the base of a quantitative science built upon measurement. 🔉⇢
A few words on how the vernier callipers fits with measurement will help here. Every measurement has an uncertainty, and the precision of an instrument is set by its least count: the smaller the least count, the smaller the uncertainty in a one observation. The vernier callipers is more precise than a plain metre scale by a factor of $10$ or $20$, so the magnitude of a length such as the diameter of a small cylinder or the thickness of a metal sheet can be reported to more significant figures. Accuracy, which measures how close the result is to the accepted value, yet depends on removing the zero error; precision and accuracy are different, and a careful experimental method looks to both. The estimation of a length is a numerical result with a fixed number of significant figures, and the least count tells us exactly how many digits are reliable. 🔉⇢
It is worth putting the vernier callipers beside the other instruments in a small comparison. A plain metre scale, a vernier callipers, and a screw gauge form a set of increasing precision, and the exercises of the chapter ask the student to pick a suitable instrument for a given length. For a small rod a few centimetre long the vernier callipers is the proper choice; for the thickness of a thin sheet the screw gauge is better; for a large distance a metre scale is enough. The physical quantity being measured, its magnitude, and the precision required together set the instrument. This is the useful skill the chapter shows: to fit the instrument to the measurement so that the result has the right number of significant figures and a small uncertainty. 🔉⇢
A vernier callipers is used with an orderly record of observations. The student takes the main scale value and the vernier division that lines up, writes the least count at the top of the table, and gives each result with its proper unit. Many observations are taken and their average is used as the reliable estimate of the length, since one observation may be in error. The zero error, found before the measurement and used on every value, keeps the result accurate; the least count keeps the result to the correct number of significant figures. In this way the callipers turns a physical length into a numerical value that can be given with a clear magnitude and a small uncertainty, exactly the standard the chapter sets for careful measurement. 🔉⇢
To sum up, the vernier callipers is a measuring instrument whose least count fixes its precision, whose reading is the main scale value plus the vernier reading, and whose zero error must be found and removed for an accurate result. The magnitude of the measured length is a physical quantity reported to the significant figures the least count allows, with a small uncertainty. Treated with a careful method and an orderly record of observations, the instrument gives a reliable estimate of a length, and it stands between the metre scale and the screw gauge as a matter of precision. 🔉⇢
Derivation from first principles 🔉⇢
Design premise: let one main scale division have length $1\,\text{MSD}$. The vernier scale is ruled so that its $n$ divisions together span the same length as $(n-1)$ main scale divisions.
Write this equality of total lengths: $n \times (1\,\text{VSD}) = (n-1) \times (1\,\text{MSD})$, where $1\,\text{VSD}$ is the length of one vernier division.
Solve for one vernier division: $1\,\text{VSD} = \frac{(n-1)}{n}\,(1\,\text{MSD})$, which shows each vernier division is slightly smaller than a main scale division.
Set the least count as the difference between one main scale division and one vernier division: $\text{L.C.} = 1\,\text{MSD} - 1\,\text{VSD}$.
Substitute the expression for $1\,\text{VSD}$: $\text{L.C.} = 1\,\text{MSD} - \frac{(n-1)}{n}\,(1\,\text{MSD}) = \frac{1\,\text{MSD}}{n}$.
Interpret: the least count equals one main scale division over the number of vernier divisions. For $1\,\text{MSD}=1\,\text{mm}$ and $n=10$, $\text{L.C.}=0.1\,\text{mm}$; for $n=20$, $\text{L.C.}=0.05\,\text{mm}$ — more divisions give a smaller least count and higher precision.
Reading rule: total reading $=$ main scale reading (the mark just before the vernier zero) $+$ (coinciding vernier division number) $\times \text{L.C.}$
Correction rule: true reading $=$ observed reading $-$ zero error, retaining the sign of the zero error (positive zero error subtracts, negative zero error adds).
⚠️ JEE trap: A frequent JEE error is to add the zero error instead of subtracting it, or to mishandle its sign. The rule is fixed: corrected reading $=$ observed reading $-$ (zero error with its own sign), so a positive zero error lowers the result while a negative zero error, being subtracted, raises it. 🔉⇢
Worked example · JEE Advanced 🔉⇢
SITUATION A vernier callipers has a main-scale division of $1\,\text{mm}$ and $20$ divisions on the sliding scale that span exactly $19$ main-scale divisions. When the jaws are closed on nothing, the $4$th vernier division coincides with a main-scale mark (a positive zero error). Measuring a rod, the main-scale reading is $12\,\text{mm}$ and the $12$th vernier division coincides.
TARGET Determine the least count, the zero error, and the true diameter of the rod.
STRATEGY Compute the least count as one main-scale division divided by the number of vernier divisions. Convert the closed-jaw coincidence into a zero error using the least count, with a positive sign. Form the observed reading as main-scale reading plus coinciding division times least count, then subtract the zero error with its sign.
REFLECT The result $12.40\,\text{mm}$ is quoted to $0.01\,\text{mm}$, consistent with a least count of $0.05\,\text{mm}$; reporting finer digits would misstate the precision. Forgetting the positive zero-error correction would have overstated the diameter by $0.20\,\text{mm}$, a systematic error no amount of averaging could remove.
Source: NCERT-derived
Least Count & the Screw Gauge 🔉⇢deep concept
Definition: In a screw gauge the least count is the pitch over the number of divisions on the circular scale, $\text{L.C.} = \frac{\text{pitch}}{N}$, and a reading is the linear scale reading plus the coinciding circular scale division multiplied by the least count, correct for zero error and backlash. 🔉⇢
🔬 Interactive 3D · Screw gauge — interactive 3D. interactive 3D scene
The screw gauge, is the instrument that takes a length measurement to a precision finer than even the vernier callipers, and it does so by using the form of a fine screw. Its precision rests on a simple fact: when a screw turns through one complete turn, it moves along its own axis by a fixed, small distance. By turning a large, easily read turn of a circular scale into a tiny, otherwise tiny move along the axis, the screw gauge makes a very small least count. This is the same idea of trading a coarse motion against a fine one that lies behind the vernier, but here the fine division is made by a turn rather than by a sliding second scale. 🔉⇢
Two quantities set the instrument, and the full theory follows from them. The first is the pitch, the distance the screw moves along its axis in one complete turn; on a common instrument the pitch is one mm, meaning one full turn moves the screw one mm along the linear scale. The second is the number of divisions marked round the circular scale, often one hundred. Since one complete turn, which moves the screw by one pitch, turns the circular scale through all of its divisions, each one division of the circular scale is a move along the axis of the pitch over the number of circular divisions. That move per circular division is the least count. 🔉⇢
The defining relation is therefore small and should be rebuilt from the form rather than memorised blindly: the least count equals the pitch over the number of divisions on the circular scale. With a pitch of one mm and one hundred circular divisions, the least count is one mm over one hundred, that is $0.01$ mm, ten times finer than a common vernier. This is why the screw gauge is the instrument of choice for the smallest lengths, such as the diameter of a thin wire or the thickness of a sheet. A larger number of circular divisions, or a smaller pitch, makes the least count smaller and the instrument more precise, though as the exercises note there is a useful limit to how far this can be pushed. 🔉⇢
The instrument shows two scales at right angles. A linear scale runs along the axis, marked in mm and sometimes in half mm, and it gives how many full pitch lengths the screw has gone. A circular scale, round the moving part, gives the part of a turn beyond the last full pitch. The object to be measured is held gently between a fixed face and the moving face of the screw, the screw being turned until the object is just lightly held. The linear scale then gives the coarse part of the length and the circular scale gives the fine part, exactly as the main scale and the vernier scale divide the work in the callipers. 🔉⇢
Reading the screw gauge joins the two scales by addition. First, read the linear scale up to the edge of the circular scale to get the full number of pitch lengths, that is, the coarse value in mm. Second, read the division of the circular scale that lies against the reference line of the linear scale; this coinciding circular division, multiplied by the least count, gives the fine part of a pitch beyond the coarse value. The total measurement is the sum: the linear scale value plus the circular scale division multiplied by the least count. As with the vernier, the fine digit is got reliably from a scale rather than from an uncertain guess, which is the source of the precision. 🔉⇢
The main defect, once again, is the zero error, and it is handled by the same logic as in the callipers. When the two ends are brought gently together with nothing between them, the zero of the circular scale should line up with the reference line of the linear scale. If the circular scale zero has not yet met the reference line, the instrument gives a small positive amount for nothing and the zero error is positive; if the circular scale zero has gone past the reference line, the zero error is negative. The zero error is a fixed error that shifts every measurement by the same amount. The correct value is the measured value less the zero error, taken with its own sign, exactly the rule that governs the vernier. 🔉⇢
The screw gauge gives a second fixed defect not present in the sliding callipers: backlash error, sometimes called back lash. Since a screw and its thread can never be made to fit with no room to spare, there is always a small looseness in the thread. When the direction of the turn of the screw is turned back, the screw does not start to move at once; the circular scale turns through a small angle before the slack is taken up and the face starts to move. If the student turns the screw back while nearing a measurement, this lost motion gives in an error. The standard cure is a matter of method rather than of correction: always make the final turn to the object in one and the same direction, so the thread stay pressed on the same side and the lost motion never enters the measurement. 🔉⇢
The ratchet at the end of the screw is there to control another source of error, the different pressure with which different users close the ends on the object. If the object is pressed too hard the value is too small and the object may bend; if too lightly, the value is too large. The ratchet slips with a soft click once a light, standard pressure is reached, so that every measurement is made at the same light hold whoever takes it. This standard hold turns what would otherwise be an error that depends on the user into one that repeats, and using the ratchet for the final turn is a necessary part of correct method. 🔉⇢
A numerical walk through makes the reading method concrete. Take a screw gauge of pitch one mm with one hundred divisions on the circular scale, so the least count is $0.01$ mm. Suppose the linear scale gives three mm up to the edge of the circular scale, and the forty fifth division of the circular scale lies against the reference line. The circular part is forty five multiplied by $0.01$ mm, that is $0.45$ mm. The measured value is three mm plus forty five hundredths, namely three point four five mm. If the instrument had a negative zero error of $0.03$ mm, the true value would be three point four five less that negative amount, which raises the value to three point four eight mm. Every step is simple; the discipline is in the sign of the zero error and in having turned from one direction to avoid backlash. 🔉⇢
As always, the significant figures we may honestly report are set by the least count. A screw gauge with a least count of $0.01$ mm justifies reporting a thickness to $0.01$ mm, and no finer; more digits than the least count would misstate the precision of the measurement. The screw gauge thus adds one more reliable significant figure over the vernier callipers for the small lengths it suits, which is exactly why it is chosen for measuring the diameter of a fine wire or the thickness of a thin sheet, where that extra digit is real and worth the trouble of the more delicate instrument. 🔉⇢
It is useful to compare the three common length instruments as a ladder of least count. The metre scale gives one mm; the vernier callipers gives $0.1$ or $0.05$ mm by placing a sliding scale beside the main scale; the screw gauge gives $0.01$ mm by dividing a linear pitch with a circular scale. Each instrument is the right tool over a particular range and precision, and choosing correctly is part of good measurement: one would not measure a fine wire with a metre scale, nor a long rod with a screw gauge whose length is only a few mm. The optical instruments named in the exercises, which can measure to within a wavelength of light, extend the ladder yet further for the most demanding measurements. 🔉⇢
There is a natural question, raised directly by the chapter exercises, of whether the accuracy of a screw gauge can be increased without limit simply by putting more divisions on the circular scale. In principle a larger number of circular divisions gives a smaller least count. In practice the divisions become so close together that they can no longer be told apart by eye, and defects such as irregular thread and remaining backlash set a floor below which more division buys no real precision. So there is a sensible limit: beyond a point, more circular divisions give a smaller least count in name but no real gain in the reliability of the measurement, and the honest precision of the instrument stops improving. 🔉⇢
Errors that vary at random remain after the fixed zero error and backlash are dealt with, and they call for the same treatment used everywhere in measurement. Small differences in the hold, slight lack of round shape in the object, and the student's judgement of which circular division coincides all make repeat measurements scatter. Since a set of many measurements gives a more reliable estimate than a set of only a few, the standard practice is to measure the wire or sheet at many places and directions and to take the mean; the average of the amounts by which the measurements differ from their mean then gives the remaining uncertainty. A wire, for example, is measured at many places along its length and in two ways at right angles at each point to show any lack of round shape, and the many measurements are averaged. 🔉⇢
Putting the full method together for use in the examination: find the pitch by turning the screw a known number of complete turns and reading the move along the axis on the linear scale, then divide the pitch by the number of circular divisions to get the least count. Check the zero error by closing the ends with the ratchet and recording the circular scale value with its sign. Take the measurement by reading the linear scale for the full pitch lengths, adding the coinciding circular division multiplied by the least count, always making the final turn in one direction to defeat backlash and using the ratchet for a standard hold. Finally subtract the zero error with its sign, report to the significant figures the least count allows, and average many measurements to handle the scatter. 🔉⇢
Finding the pitch itself deserves a note, since it is a small worked idea in its own right and a common examination step. Rather than trusting a one turn, one turns the screw through many complete turns, gives the total distance gone along the axis on the linear scale, and divides that distance by the number of turns. If the screw moves a certain full number of mm in that many turns, the pitch is that distance over the number of turns. Dividing the pitch in turn by the number of circular divisions gives the least count. This two step method, distance per turn giving the pitch, and pitch per circular division giving the least count, is the logical spine of the instrument and is worth being able to reproduce cleanly. 🔉⇢
It is worth stating the least count of the screw gauge in numbers for the common cases. With a pitch of one mm and one hundred circular divisions the least count is $0.01$ mm; with a pitch of one mm and two hundred circular divisions it is $0.005$ mm; and with a pitch of half a mm and one hundred circular divisions it is again $0.005$ mm. In every case the least count is the pitch over the number of circular divisions, so either a smaller pitch or a larger number of circular divisions makes the instrument more precise. This is exactly why, when a case compares a screw gauge with a vernier callipers, the screw gauge is usually the more precise device for the small lengths it is built to measure. 🔉⇢
Before a set of measurements, test the screw gauge for its zero error just as you would the vernier callipers, since a zero error left uncorrected shifts every value by the same amount and cannot be removed by averaging. Close the ends gently with the ratchet, read the circular scale against the reference line, and record the value with its sign as the zero error. Then, for each object, take the linear scale value and the circular scale value in the usual way and subtract the zero error with its sign. Because the zero error is the same for every measurement, this one test at the start protects the accuracy of the full set of measurements, and like the pitch it should be found before any real measuring starts. 🔉⇢
In summary, the screw gauge turns a coarse, readable turn into a fine move along the axis, so its least count, the pitch over the number of circular divisions, gives $0.01$ mm. A measured value is the linear scale value plus the coinciding circular division multiplied by the least count, correct for a zero error taken with its sign. It carries two fixed defects to manage, the zero error and the backlash of imperfect thread, plus the pressure that depends on the user, which the ratchet makes standard, and scatter that averaging reduces. Master its form and its zero error correction and you hold the most precise of the everyday length instruments, together with a clear picture of how precision itself is built into a measuring device. 🔉⇢
It helps to place the screw gauge within measurement as an idea. The precision of the instrument is set by its least count, and with a least count as small as $0.01$ mm the uncertainty in a one observation is very small, so the magnitude of a small length can be reported to many significant figures. Accuracy, meaning how close the result is to the accepted value, yet needs the zero error removed; precision and accuracy are different, and a careful experimental method looks to both. The result is a numerical value with a fixed number of significant figures, and the least count fixes how many digits are reliable. This is why the screw gauge is the proper instrument for the diameter of a thin wire or the thickness of a metal sheet, where a plain scale would give too few significant figures. 🔉⇢
The screw gauge also needs a careful experimental method. Because a one observation has a small uncertainty, the standard practice is to record many observations of the same length and to report their average, which is more reliable than any one value. The exercises of the chapter treat the screw gauge as the instrument of choice for small lengths, and they ask the student to find the least count from the pitch and the number of circular divisions, to remove the zero error, and to report the result to the proper number of significant figures. Handled in this way, the screw gauge gives a measurement whose precision and accuracy are both understood, and whose magnitude is expressed together with its uncertainty. 🔉⇢
A screw gauge is used with the same orderly record as the callipers. The pitch and the number of circular divisions are noted, the least count is found and written at the top of the table, and the zero error is found and given its sign before any length is taken. For each object the student gives the linear scale value and the circular scale value, finds the result, and writes it with its proper unit. Many observations of the same wire or sheet are taken and their average is used as the reliable estimate, since one observation has a small uncertainty. The thread of the screw and its pitch are the centre of the instrument, and a clear record of pitch, least count, zero error, and observations is the standard the chapter sets for careful measurement. 🔉⇢
Derivation from first principles 🔉⇢
Screw principle: one complete turn of the screw moves it along its axis by a fixed distance called the pitch, so pitch $=$ (axial distance advanced) $\div$ (number of complete turns).
Determine the pitch in practice: turn the screw through $r$ complete turns, read the axial move $d$ on the linear scale, and set $\text{pitch} = \frac{d}{r}$ (for example $d = 5\,\text{mm}$ in $r = 5$ turns gives $\text{pitch} = 1\,\text{mm}$).
The circular scale carries $N$ equal divisions round its circumference, and turning it through all $N$ divisions is exactly one complete turn, moving the screw by one pitch.
Hence one division of the circular scale corresponds to an axial move of one $N$th of a pitch. Set the least count as this smallest move: $\text{L.C.} = \frac{\text{pitch}}{N}$.
Evaluate: for $\text{pitch} = 1\,\text{mm}$ and $N = 100$, $\text{L.C.} = \frac{1\,\text{mm}}{100} = 0.01\,\text{mm}$ — ten times finer than a common vernier callipers.
Reading rule: total reading $=$ linear scale reading $+$ (coinciding circular scale division) $\times \text{L.C.}$
Zero error correction: true reading $=$ observed reading $-$ zero error, retaining the sign (positive zero error subtracts, negative zero error adds).
Backlash rule: always make the final approach to the object by turning the screw in one and the same direction, so the thread clearance (backlash) is taken up on one side and never enters the reading.
⚠️ JEE trap: Students assume a screw gauge can be made arbitrarily precise just by adding circular divisions, and they often ignore backlash. In reality thread clearance and the eye's limit on distinguishing crowded divisions set a floor on real precision, and reversing the turning direction near a reading injects a backlash error that a smaller nominal least count cannot cure. 🔉⇢
Worked example · JEE Advanced 🔉⇢
SITUATION A screw gauge advances $5\,\text{mm}$ along the linear scale in $5$ complete rotations and has $100$ divisions on the circular scale. With the faces closed by the ratchet, the circular scale reads $3$ divisions above the reference line, giving a negative zero error. Measuring a wire, the linear scale reads $2\,\text{mm}$ and the $47$th circular division coincides with the reference line.
TARGET Find the pitch, the least count, the zero error, and the true diameter of the wire.
STRATEGY Get the pitch as axial advance divided by number of rotations, then the least count as pitch divided by the number of circular divisions. Convert the closed-face circular reading into a negative zero error. Form the observed reading as linear reading plus coinciding division times least count, then subtract the zero error with its sign.
EXECUTE Pitch $= \dfrac{5\,\text{mm}}{5} = 1\,\text{mm}$. Least count $= \dfrac{1\,\text{mm}}{100} = 0.01\,\text{mm}$. The circular zero sits $3$ divisions above the line, so zero error $= -3 \times 0.01\,\text{mm} = -0.03\,\text{mm}$. Observed reading $= 2\,\text{mm} + 47 \times 0.01\,\text{mm} = 2\,\text{mm} + 0.47\,\text{mm} = 2.47\,\text{mm}$. True diameter $=$ observed $-$ zero error $= 2.47\,\text{mm} - (-0.03\,\text{mm}) = 2.50\,\text{mm}$.
REFLECT Because the zero error is negative, subtracting it raises the reading from $2.47\,\text{mm}$ to $2.50\,\text{mm}$ — a common sign trap. The value is quoted to $0.01\,\text{mm}$, matching the least count, and the final approach should have been made in one direction so that backlash contributes nothing to the $47$-division circular reading.
Source: NCERT-derived
✍️ Worked Examples Polya 5-move · JEE tier
WE1 · The SI System and the Seven Base Units · JEE Main 🔉⇢
SITUATION The SI unit of pressure, the pascal (Pa), is defined as one newton per square metre.
TARGET Express the pascal purely in terms of the SI base units.
STRATEGY Replace the derived unit newton by its base-unit combination, then divide by the base unit of area, treating units as algebraic symbols.
REFLECT The pascal involves only the base units kilogram, metre and second, confirming that derived units are combinations of base units and that no new base quantity is introduced by pressure.
SITUATION A hydrogen atom has diameter about $1.06\times10^{-10}\ \mathrm{m}$ and the earth has diameter about $1.28\times10^{7}\ \mathrm{m}$.
TARGET Find, by order of magnitude alone, how many times larger the earth's diameter is than the atom's.
STRATEGY Reduce each length to its order of magnitude and subtract the exponents, since dividing powers of ten subtracts their exponents.
EXECUTE Earth is of order $10^{7}\ \mathrm{m}$ and the atom of order $10^{-10}\ \mathrm{m}$; the ratio is of order $10^{7-(-10)} = 10^{17}$.
REFLECT The earth's diameter is seventeen orders of magnitude larger than the atom's, a comparison reached without any lengthy division, showing the estimation power of scientific notation.
Source: NCERT-derived
WE3 · Dimensions & Dimensional Formulae · JEE Main 🔉⇢
SITUATION Kinetic energy is defined by the relation $K = \tfrac{1}{2} m v^{2}$, where $m$ is mass and $v$ is speed.
TARGET Obtain the dimensional formula of energy.
STRATEGY Replace each quantity by its dimensional formula, ignore the pure numerical factor, and combine the base quantities as algebraic symbols.
REFLECT Energy has dimensions $[\mathrm{M\,L^{2}\,T^{-2}}]$, the same as work; this shared dimensional formula is exactly why dimensions alone cannot distinguish the two quantities.
Source: NCERT-derived
WE4 · Significant Figures · JEE Main 🔉⇢
SITUATION A measured mass is recorded as $0.06900\ \mathrm{kg}$ and separately as $2.64\times10^{24}\ \mathrm{kg}$.
TARGET State the number of significant figures in each value.
STRATEGY Apply the zero rules: leading zeroes before the first non-zero digit are not significant, trailing zeroes after a decimal are significant, and in scientific notation all base-number digits count.
EXECUTE In $0.06900$ the leading zeroes are not significant while the two trailing zeroes are, leaving digits six, nine, zero, zero for four significant figures; in $2.64\times10^{24}$ only the base number matters, giving three significant figures.
REFLECT The power of ten never affects the count, and the contrast between leading and trailing zeroes shows why the decimal point's presence must always be checked before counting.
SITUATION Three masses measured on the same balance are $436.32\ \mathrm{g}$, $227.2\ \mathrm{g}$ and $0.301\ \mathrm{g}$, and separately a rod of length $16.2\ \mathrm{cm}$ and breadth $10.1\ \mathrm{cm}$ is measured.
TARGET Report the total mass, and the area of the rectangle, to the correct precision.
STRATEGY Use the least-decimal-places rule for the sum and the least-significant-figures rule for the product, rounding only at the end with the round-half-even convention.
EXECUTE Sum $= 663.821\ \mathrm{g}$, and the least precise term $227.2\ \mathrm{g}$ has one decimal place, so the mass is $663.8\ \mathrm{g}$; area $= 16.2\times10.1 = 163.62\ \mathrm{cm^{2}}$, and with three significant figures in each factor the area is $164\ \mathrm{cm^{2}}$.
REFLECT The same data are governed by two different rules depending on the operation, which is why the operation must be identified before any rounding is applied.
SITUATION A student views a human hair through a microscope of magnification one hundred and finds its average width in the field of view to be $3.5\ \mathrm{mm}$ over twenty observations.
TARGET Estimate the actual thickness of the hair.
STRATEGY Treat this as an indirect measurement: the true length equals the magnified reading divided by the known magnification, keeping significant figures consistent with the reading.
REFLECT Averaging twenty observations reduces the random error, and the result, of the order of $10^{-5}\ \mathrm{m}$, is reported to two significant figures matching the precision of the reading.
Source: NCERT-derived
WE7 · Measuring Mass and Time · JEE Advanced 🔉⇢
SITUATION A box of mass $2.30\ \mathrm{kg}$ measured on a grocer's balance receives two gold pieces of masses $20.15\ \mathrm{g}$ and $20.17\ \mathrm{g}$.
TARGET Find the total mass of the box to the correct significant figures.
STRATEGY Convert all values to a common unit, add, then round to the least number of decimal places among the measurements, since this is an addition.
EXECUTE $2.30\ \mathrm{kg} = 2300\ \mathrm{g}$; total $= 2300 + 20.15 + 20.17 = 2340.32\ \mathrm{g}$, but the box mass is known only to the units place, so the total is $2340\ \mathrm{g} = 2.34\ \mathrm{kg}$.
REFLECT The precise gold masses cannot improve the coarse balance measurement of the box, illustrating that a sum can be no more precise in decimal places than its least precise term.
Source: NCERT-derived
WE8 · Errors in Measurement: Systematic vs Random · JEE Advanced 🔉⇢
SITUATION A vernier callipers reads $0.02\ \mathrm{cm}$ when its jaws are fully closed, and a student then makes many observations of a rod's diameter that scatter about a central value.
TARGET Identify the kind of error in each part and state how to reduce it.
STRATEGY Classify a consistent non-zero closed observation as systematic and irregular scatter of repeated observations as random, then apply the matching remedy to each.
EXECUTE The $0.02\ \mathrm{cm}$ closed observation is a systematic zero error, corrected by subtracting it from every observation; the scatter about the average is random error, reduced by taking a large number of observations.
REFLECT Correcting the zero error improves the accuracy while averaging improves the precision, showing that the two kinds of error demand two different remedies applied together.
SITUATION Two masses are measured on the same balance, each accurate to $\pm 0.01\ \mathrm{g}$: one reads $1.02\ \mathrm{g}$ and the other reads $9.89\ \mathrm{g}$.
TARGET Compare the two measurements by their percentage error.
STRATEGY Note the identical absolute error, then divide it by each measured value and express as a percentage to obtain the relative and percentage errors.
EXECUTE For $1.02\ \mathrm{g}$: $(\pm 0.01/1.02)\times100\% \approx \pm 1\%$; for $9.89\ \mathrm{g}$: $(\pm 0.01/9.89)\times100\% \approx \pm 0.1\%$.
REFLECT Despite an identical absolute error, the larger quantity is measured to ten times smaller percentage error, confirming that relative error, not absolute error, gauges the quality of a measurement.
Source: NCERT-derived
WE10 · Problem 1 · JEE Main 🔉⇢
SITUATION A metal cube has each side measured with a metre scale as $7.203$ m.
TARGET Report the total surface area and the volume of the cube to the appropriate number of significant figures.
STRATEGY The measured length has 4 significant figures, so any product derived from it must also be rounded to 4 s.f. Compute the raw arithmetic value first, then round at the end.
EXECUTE Surface area $=6a^2=6(7.203)^2=6\times 51.883209=311.299254$ m$^2$. Volume $=a^3=(7.203)^3=373.714754$ m$^3$. The input has 4 s.f., so round: area $=311.3$ m$^2$, volume $=373.7$ m$^3$.
REFLECT Both results carry exactly 4 s.f., matching the precision of the single measured length. The factor 6 is exact and does not limit significant figures.
Source: NCERT Example 1.1
WE11 · Problem 2 · JEE Main 🔉⇢
SITUATION A substance of mass $5.74$ g occupies a volume of $1.2$ cm$^3$.
TARGET Express the density keeping significant figures in view.
STRATEGY In division the result keeps as many s.f. as the input with the fewest s.f. Mass has 3 s.f.; volume has 2 s.f., so the answer keeps 2 s.f.
EXECUTE Density $=\dfrac{5.74\ \text{g}}{1.2\ \text{cm}^3}=4.7833\ldots$ g cm$^{-3}$. Rounding to 2 s.f. gives $\rho=4.8$ g cm$^{-3}$.
REFLECT The least-precise factor (volume, 2 s.f.) controls the precision; quoting more digits would falsely suggest better precision than the measurement allows.
Source: NCERT Example 1.2
WE12 · Problem 3 · JEE Main 🔉⇢
SITUATION An equation is written as $\tfrac12 m v^2 = m g h$, where $m$ is mass, $v$ velocity, $g$ acceleration due to gravity and $h$ height.
TARGET Check whether the equation is dimensionally correct.
STRATEGY Reduce both sides to base dimensions [M], [L], [T] and compare. Pure numbers such as $\tfrac12$ are dimensionless and ignored.
EXECUTE LHS: $[M][L T^{-1}]^2=[M\,L^2\,T^{-2}]$. RHS: $[M][L T^{-2}][L]=[M\,L^2\,T^{-2}]$. The two sides have identical dimensions.
REFLECT The equation passes the homogeneity test. Passing only proves dimensional consistency, not physical correctness, but failing would have proved it wrong.
REFLECT Dimensions eliminate (a), (c), (e) but cannot choose between (b) and (d); the pure number $\tfrac12$ vs $\tfrac{3}{16}$ needs the physical definition of $K$.
Source: NCERT Example 1.4
WE14 · Problem 5 · JEE Advanced 🔉⇢
SITUATION The period $T$ of a simple pendulum is assumed to depend on its length $l$, the bob mass $m$ and the acceleration due to gravity $g$.
TARGET Derive the form of $T$ using the method of dimensions.
STRATEGY Assume a product $T=k\,l^{x}g^{y}m^{z}$ with dimensionless $k$, write the dimensional equation, and match exponents of M, L, T.
REFLECT The mass drops out, matching experiment; dimensional analysis fixes the powers but not $k$ (actually $2\pi$). Independence of $m$ is a genuine physical prediction.
Source: NCERT Example 1.5
WE15 · Problem 6 · JEE Main 🔉⇢
SITUATION The diameter of the Earth is about $1.28\times10^7$ m and that of a hydrogen atom about $1.06\times10^{-10}$ m.
TARGET State the order of magnitude of each and how many orders of magnitude separate them.
STRATEGY Write $a\times10^b$ with $1\le a\lt10$; round $a$ to 1 if $a\le5$, else to 10, and read off the power of ten.
EXECUTE Earth: $1.28\times10^7\to 10^7$, order $7$. Atom: $1.06\times10^{-10}\to 10^{-10}$, order $-10$. Separation $=7-(-10)=17$ orders of magnitude.
REFLECT Order-of-magnitude reasoning compresses a factor of $\sim10^{17}$ into a single integer difference, ideal for quick comparison.
Source: NCERT-derived
WE16 · Problem 7 · JEE Main 🔉⇢
SITUATION Several reported measurements: (a) $0.007$ m$^2$, (b) $2.64\times10^{24}$ kg, (c) $0.2370$ g cm$^{-3}$, (d) $6.320$ J, (e) $6.032$ N m$^{-2}$, (f) $0.0006032$ m$^2$.
TARGET State the number of significant figures in each.
STRATEGY Apply the s.f. rules: leading zeros never count; captive zeros count; trailing zeros after a decimal count; powers of ten do not affect the count.
REFLECT Only the digits $6,0,3,2$ matter in (f); the four leading zeros merely locate the decimal point and are not significant.
Source: NCERT Exercise
WE17 · Problem 8 · JEE Main 🔉⇢
SITUATION A vehicle moves at $18$ km h$^{-1}$.
TARGET Find the distance covered in $1$ s, i.e. convert the speed to m s$^{-1}$.
STRATEGY Treat units as algebraic symbols: multiply by exact conversion factors $1$ km $=10^3$ m and $1$ h $=3600$ s.
EXECUTE $18\ \dfrac{\text{km}}{\text{h}}=18\times\dfrac{10^3\ \text{m}}{3600\ \text{s}}=18\times\dfrac{1}{3.6}\ \text{m s}^{-1}=5\ \text{m s}^{-1}$. In $1$ s it covers $5$ m.
REFLECT The handy shortcut $\text{km h}^{-1}\to\text{m s}^{-1}$ is division by $3.6$; the conversion factors are exact so precision is unaffected.
Source: NCERT Exercise
WE18 · Problem 9 · JEE Main 🔉⇢
SITUATION A cube has side $1$ cm.
TARGET Express its volume in m$^3$.
STRATEGY Convert length to metres first ($1$ cm $=10^{-2}$ m), then cube, so the factor $10^{-2}$ is raised to the third power.
REFLECT Cubing the length also cubes the unit-conversion factor; forgetting this is the classic error that gives $10^{-2}$ instead of $10^{-6}$.
Source: NCERT Exercise
WE19 · Problem 10 · JEE Main 🔉⇢
SITUATION A rectangular metal sheet has length $4.234$ m, breadth $1.005$ m and thickness $2.01$ cm.
TARGET Give the area of a face and the volume to correct significant figures.
STRATEGY Convert all lengths to a common unit, multiply, then round the product to the fewest s.f. among the factors (here 3 s.f. from $2.01$).
EXECUTE Thickness $=2.01$ cm $=0.0201$ m. Area $=4.234\times1.005=4.2552$ m$^2\to 4.255$ m$^2$ (4 s.f., since both factors have $\ge4$). Volume $=4.234\times1.005\times0.0201=0.08553\ldots$ m$^3\to 0.0855$ m$^3$ (3 s.f.).
REFLECT The thinnest dimension (3 s.f.) caps the volume's precision; the area, using only the two 4-figure sides, retains 4 s.f.
Source: NCERT Exercise
WE20 · Problem 11 · JEE Main 🔉⇢
SITUATION A grocer's balance reads the mass of a box as $2.30$ kg. Two gold pieces of $20.15$ g and $20.17$ g are added.
TARGET Find (a) the total mass to correct s.f. and (b) the difference of the two pieces.
STRATEGY For addition/subtraction, keep as many decimal places as the term with the fewest decimal places (in kg the box is known only to $0.01$ kg).
EXECUTE (a) Total $=2.30\ \text{kg}+0.02015\ \text{kg}+0.02017\ \text{kg}=2.34032\ \text{kg}$. The box limits to 2 decimals in kg, so total $=2.34$ kg. (b) $20.17-20.15=0.02$ g (both to 2 decimals).
REFLECT The coarse box measurement dominates the sum: adding milligram-precise gold cannot improve the kilogram-scale uncertainty of the box.
Source: NCERT Exercise
WE21 · Problem 12 · JEE Main 🔉⇢
SITUATION A student recalls the relativistic mass relation but forgets the constant $c$: $m=\dfrac{m_0}{(1-v^2)^{1/2}}$.
TARGET Use dimensions to decide where $c$ must be inserted.
STRATEGY The argument of the square root must be dimensionless, so $v^2$ must be divided by a squared speed to cancel its dimensions.
EXECUTE $[v^2]=[L^2 T^{-2}]$ is not dimensionless, so $1-v^2$ is inhomogeneous. Dividing by $c^2$ ($[L^2T^{-2}]$) fixes it: $m=\dfrac{m_0}{(1-v^2/c^2)^{1/2}}$, whose denominator is now dimensionless.
REFLECT Dimensional homogeneity of the subtraction $1-v^2/c^2$ pinpoints the missing $c^2$ without any relativity theory.
Source: NCERT Exercise
WE22 · Problem 13 · JEE Advanced 🔉⇢
SITUATION A thin rectangular sheet is measured with a metre scale as length $l=16.2$ cm and breadth $b=10.1$ cm, each to the least count $0.1$ cm.
TARGET Find the area with its uncertainty (absolute and percentage).
STRATEGY For a product $A=lb$, the fractional errors add: $\Delta A/A=\Delta l/l+\Delta b/b$. Compute each percentage, add, then convert back to absolute error.
REFLECT Percentage errors, not absolute errors, add for products; the final area is quoted with an uncertainty consistent with 3 s.f.
Source: NCERT-derived
WE23 · Problem 14 · JEE Main 🔉⇢
SITUATION A vernier calliper has a main-scale smallest division of $1$ mm and $20$ divisions on the vernier that coincide with $19$ main-scale divisions.
TARGET Find the least count.
STRATEGY Least count $=$ (value of 1 main-scale division) $-$ (value of 1 vernier division) $=$ 1 MSD $/N$ when $N$ vernier divisions span $N-1$ MSD.
EXECUTE $N=20$, 1 MSD $=1$ mm. L.C. $=\dfrac{1\ \text{MSD}}{N}=\dfrac{1\ \text{mm}}{20}=0.05\ \text{mm}=0.005$ cm.
REFLECT Twenty divisions give a $0.05$ mm resolution; increasing $N$ shrinks the least count and improves reading precision.
Source: NCERT-derived
WE24 · Problem 15 · JEE Main 🔉⇢
SITUATION A screw gauge has a pitch of $1$ mm and $100$ divisions on its circular (head) scale.
TARGET Find the least count.
STRATEGY Least count $=$ pitch $/$ (number of circular-scale divisions); pitch is the linear advance per full rotation.
EXECUTE L.C. $=\dfrac{\text{pitch}}{N}=\dfrac{1\ \text{mm}}{100}=0.01\ \text{mm}=10\ \mu\text{m}=0.001$ cm.
REFLECT The screw gauge resolves $10\,\mu$m, an order of magnitude finer than a typical vernier — suited to wire and sheet thicknesses.
Source: NCERT-derived
WE25 · Problem 16 · JEE Advanced 🔉⇢
SITUATION A screw gauge (pitch $1$ mm, 100 divisions) shows a positive zero error: with the jaws closed, the 4th circular division is above the reference line. Measuring a wire, the main scale reads $2$ mm and the 43rd circular division coincides.
TARGET Find the corrected diameter of the wire.
STRATEGY Compute least count, form the observed reading (main $+$ circular$\times$L.C.), then subtract the positive zero error (zero-error $=$ error division $\times$ L.C.).
EXECUTE L.C. $=0.01$ mm. Zero error $=+4\times0.01=+0.04$ mm. Observed $=2+43\times0.01=2.43$ mm. Corrected $=2.43-0.04=2.39$ mm.
REFLECT A positive zero error is subtracted; ignoring it would overstate the diameter by $0.04$ mm — comparable to the least count itself.
Source: NCERT-derived
WE26 · Problem 17 · JEE Advanced 🔉⇢
SITUATION A physical quantity is $Z=\dfrac{A^2 B^{3}}{\sqrt{C}\,D}$, with measured percentage errors $1\%,2\%,3\%,4\%$ in $A,B,C,D$.
TARGET Find the maximum percentage error in $Z$.
STRATEGY For products/quotients/powers, the maximum fractional error is the sum of the magnitudes of each fractional error weighted by the exponent (powers add as $|n|\times$).
REFLECT Higher powers amplify error: the $B^3$ term alone contributes $6\%$. This is the worst-case (maximum) error; errors are taken with absolute value and summed.
Source: JEE Advanced (pattern)
WE27 · Problem 18 · JEE Advanced 🔉⇢
SITUATION In a simple-pendulum experiment, $g=\dfrac{4\pi^2 l}{T^2}$. Length $l=(100.0\pm0.1)$ cm and time period $T=(2.00\pm0.01)$ s.
TARGET Find the percentage error in the measured value of $g$.
STRATEGY Since $g\propto l\,T^{-2}$, the fractional error is $\Delta l/l+2\,\Delta T/T$; $\pi$ is exact.
REFLECT The time measurement dominates because $T$ enters squared; halving $\Delta T$ would nearly halve the total error, so timing many oscillations is worthwhile.
Source: JEE Advanced (pattern)
WE28 · Problem 19 · JEE Advanced 🔉⇢
SITUATION The gravitational constant is $G=6.67\times10^{-11}$ N m$^2$ kg$^{-2}$.
TARGET Convert $G$ into CGS units, i.e. cm$^3$ s$^{-2}$ g$^{-1}$.
STRATEGY Express N m$^2$ kg$^{-2}$ in base SI, then substitute $1$ kg $=10^3$ g and $1$ m $=10^2$ cm, tracking each power.
REFLECT The numerical value changes with the unit system while the physics is unchanged; the exponent bookkeeping ($+6-3=+3$) is the whole game.
Source: NCERT Exercise
WE29 · Problem 20 · JEE Advanced 🔉⇢
SITUATION A calorie equals $4.2$ J with $1$ J $=1$ kg m$^2$ s$^{-2}$. A new system uses mass unit $\alpha$ kg, length unit $\beta$ m, time unit $\gamma$ s.
TARGET Show the magnitude of a calorie in the new units.
STRATEGY Energy has dimensions $[M L^2 T^{-2}]$. A magnitude scales by $(1/\alpha)(1/\beta)^2(1/\gamma)^{-2}$ when units are enlarged by $\alpha,\beta,\gamma$.
EXECUTE New magnitude $=4.2\times\alpha^{-1}\beta^{-2}\gamma^{2}$, because a larger mass unit ($\alpha$) divides the number, $\beta^2$ from $L^2$, and $\gamma^{2}$ multiplies from $T^{-2}$. Hence $1$ cal $=4.2\,\alpha^{-1}\beta^{-2}\gamma^{2}$ new units.
REFLECT The rule $n_2=n_1[M_1/M_2]^a[L_1/L_2]^b[T_1/T_2]^c$ underlies all unit conversion; here it produces the $\alpha^{-1}\beta^{-2}\gamma^{2}$ pattern directly.
Source: NCERT Exercise
WE30 · Problem 21 · JEE Main 🔉⇢
SITUATION Round the following to 3 significant figures: $2.745$, $2.735$, $2.746$, $1.743$.
TARGET Apply the rounding conventions, including the even-odd (banker's) rule when the dropped digit is exactly 5.
STRATEGY If the digit dropped is $\gt5$ round up; $\lt5$ round down; exactly $5$ (with nothing after) round to make the preceding digit even.
EXECUTE $2.746\to 2.75$ (drop $6\gt5$). $1.743\to 1.74$ (drop $3\lt5$). $2.745\to 2.74$ (drop exactly 5, preceding 4 is even, stays). $2.735\to 2.74$ (drop 5, preceding 3 is odd, raised to 4).
REFLECT The even-rule prevents a systematic upward bias when many 5's are rounded; both $2.745$ and $2.735$ land on the even $2.74$.
Source: NCERT-derived
WE31 · Problem 22 · JEE Main 🔉⇢
SITUATION Add the masses $436.32$ g, $227.2$ g and $0.301$ g.
TARGET Report the sum to the correct number of significant figures.
STRATEGY For addition the result keeps as many decimal places as the term with the fewest decimals; $227.2$ has only one decimal place.
EXECUTE Arithmetic sum $=436.32+227.2+0.301=663.821$ g. Least decimals $=1$ (from $227.2$), so round to $663.8$ g.
REFLECT It would be wrong to apply the multiplication rule and write $664$ g; addition/subtraction is governed by decimal places, not total s.f.
Source: NCERT-derived
WE32 · Problem 23 · JEE Main 🔉⇢
SITUATION Subtract $12.9$ g $-$ $7.06$ g, each given to 3 significant figures.
TARGET Give the difference to correct significant figures.
STRATEGY Subtraction follows the decimal-place rule; $12.9$ has one decimal place, which limits the result.
EXECUTE $12.9-7.06=5.84$ g arithmetically, but $12.9$ has one decimal place, so the answer is $5.8$ g.
REFLECT Subtraction of nearly-equal numbers loses significant figures ('loss of significance'): the 3-figure inputs yield only a 2-figure result.
Source: NCERT-derived
WE33 · Problem 24 · JEE Advanced 🔉⇢
SITUATION A new length unit is chosen so that the speed of light in vacuum is unity. Light takes $8$ min $20$ s to travel from the Sun to the Earth.
TARGET Express the Sun-Earth distance in this new unit.
STRATEGY If $c=1$ in the new units, distance (in new length units) equals the light-travel time in seconds, since distance $=c\times$time $=1\times t$.
EXECUTE $t=8\text{ min }20\text{ s}=8\times60+20=500$ s. With $c=1$, distance $=c\,t=1\times500=500$ new units.
REFLECT Choosing $c=1$ merges length and time units (as in relativity's 'light-seconds'); the AU becomes simply 500 of the new units.
Source: NCERT Exercise
WE34 · Problem 25 · JEE Advanced 🔉⇢
SITUATION A viscous drag force on a small sphere is believed to depend on the coefficient of viscosity $\eta$ ($[M L^{-1}T^{-1}]$), the radius $r$ and the velocity $v$.
TARGET Use dimensional analysis to find the form of Stokes' law $F=k\,\eta^{a}r^{b}v^{c}$.
STRATEGY Write the dimensional equation for force and match exponents of M, L, T to solve for $a,b,c$.
EXECUTE $[MLT^{-2}]=[ML^{-1}T^{-1}]^a[L]^b[LT^{-1}]^c$. M: $a=1$. T: $-a-c=-2\Rightarrow c=1$. L: $-a+b+c=1\Rightarrow -1+b+1=1\Rightarrow b=1$. So $F=k\,\eta r v$ (Stokes: $k=6\pi$).
REFLECT Dimensional analysis reproduces the celebrated $F=6\pi\eta rv$ up to the constant $6\pi$, which must come from a full hydrodynamic calculation.
Source: JEE Advanced (pattern)
WE35 · Problem 26 · JEE Advanced 🔉⇢
SITUATION For fluid flow the Reynolds number is defined as $Re=\dfrac{\rho v D}{\eta}$, with density $\rho$, speed $v$, pipe diameter $D$ and viscosity $\eta$.
TARGET Show that $Re$ is dimensionless.
STRATEGY Substitute base dimensions for each factor and confirm all powers of M, L, T cancel.
EXECUTE $[\rho v D/\eta]=\dfrac{[ML^{-3}][LT^{-1}][L]}{[ML^{-1}T^{-1}]}=\dfrac{[ML^{-1}T^{-1}]}{[ML^{-1}T^{-1}]}=[M^0L^0T^0]$.
REFLECT Being a pure number, $Re$ compares inertial to viscous forces and sets the same laminar/turbulent transition regardless of the unit system.
Source: JEE Advanced (pattern)
WE36 · Problem 27 · JEE Advanced 🔉⇢
SITUATION A distant star subtends a parallax angle of $1$ arcsecond ($\theta$) when viewed across a baseline equal to the Earth-Sun distance (1 AU).
TARGET Express the star's distance using the parallax formula $d=b/\theta$ (with $\theta$ in radians).
STRATEGY Convert the angle to radians, then divide the baseline by the angle; this is the small-angle parallax method for stellar distances.
EXECUTE $1'' =\dfrac{1}{3600}\times\dfrac{\pi}{180}=4.85\times10^{-6}$ rad. With $b=1$ AU $=1.496\times10^{11}$ m, $d=\dfrac{b}{\theta}=\dfrac{1.496\times10^{11}}{4.85\times10^{-6}}=3.09\times10^{16}$ m.
REFLECT This distance is 1 parsec by definition; the parallax method turns a tiny measured angle into astronomical distances via $d=b/\theta$.
Source: NCERT-derived
WE37 · Problem 28 · JEE Main 🔉⇢
SITUATION A vernier calliper (least count $0.01$ cm) measures a rod: main-scale reading $3.2$ cm and the 4th vernier division coincides with a main-scale mark. The instrument has no zero error.
TARGET Find the length of the rod.
STRATEGY Total reading $=$ main-scale reading $+$ (coinciding vernier division) $\times$ least count.
EXECUTE Length $=3.2\ \text{cm}+4\times0.01\ \text{cm}=3.2+0.04=3.24$ cm.
REFLECT The vernier fraction adds two-decimal precision to the millimetre main scale; with no zero error no correction is needed.
Source: NCERT-derived
WE38 · Problem 29 · JEE Advanced 🔉⇢
SITUATION Estimate the number of air molecules in an ordinary classroom of volume about $100$ m$^3$ at room temperature and pressure.
TARGET Produce an order-of-magnitude (Fermi) estimate.
STRATEGY Use that one mole ($6\times10^{23}$ molecules) occupies about $22.4$ L $=22.4\times10^{-3}$ m$^3$ at STP; scale by the room volume.
EXECUTE Moles $=\dfrac{100}{22.4\times10^{-3}}\approx4.5\times10^{3}$ mol. Molecules $\approx4.5\times10^{3}\times6\times10^{23}\approx2.7\times10^{27}$, i.e. order of magnitude $10^{27}$.
REFLECT Fermi estimates aim only for the right power of ten; the exact molar volume varies a little with temperature, which does not change the $10^{27}$ order.
Source: NCERT-derived
WE39 · Problem 30 · JEE Main 🔉⇢
SITUATION A student measures the thickness of a hair through a microscope of magnification $100$; the average apparent width in the field of view is $3.5$ mm.
EXECUTE Thickness $=\dfrac{3.5\ \text{mm}}{100}=0.035\ \text{mm}=3.5\times10^{-5}$ m $=35\ \mu\text{m}$.
REFLECT Magnification lets a coarse instrument resolve micrometre scales; the result, $\sim35\,\mu$m, is the right order for human hair.
Source: NCERT Exercise
WE40 · Problem 31 · JEE Main 🔉⇢
SITUATION The relative density (specific gravity) of lead is $11.3$.
TARGET Express its density in g cm$^{-3}$ and in kg m$^{-3}$.
STRATEGY Relative density is the ratio to water's density; multiply by water's density ($1$ g cm$^{-3}=10^3$ kg m$^{-3}$).
EXECUTE $\rho=11.3\times1\ \text{g cm}^{-3}=11.3$ g cm$^{-3}$. In SI: $11.3\ \text{g cm}^{-3}=11.3\times10^{3}$ kg m$^{-3}=1.13\times10^{4}$ kg m$^{-3}$.
REFLECT Relative density is a pure (dimensionless) number; attaching a reference density converts it into an absolute density in either unit system.
Source: NCERT Exercise
WE41 · Problem 32 · JEE Advanced 🔉⇢
SITUATION The size of a hydrogen atom is about $0.5$ Angstrom ($1$ Å $=10^{-10}$ m).
TARGET Estimate the total atomic volume (in m$^3$) occupied by one mole of hydrogen atoms.
STRATEGY Model each atom as a sphere of radius $0.5$ Å, compute its volume, multiply by Avogadro's number $N_A=6.0\times10^{23}$.
REFLECT The atomic volume of a mole ($\sim3\times10^{-7}$ m$^3$) is far smaller than the molar gas volume ($22.4\times10^{-3}$ m$^3$), because gases are mostly empty space.
Source: NCERT Exercise
On the concept tabs
These worked examples are taught in full alongside their interactive scene:
Screw gauge reading: understand what each symbol means and when this applies — see the concept tab for the derivation.
JEE Physics — Units and Measurement
📜 Previous-Year Questions Authentic NTA · 62 questions
Every PYQ traces to an official NTA/JAB paper (year + shift + paper). Answer keys match the official key. OCR-extracted questions are flagged for SME verification.
IIT-JEE 2008 Paper 1 Q24Answer: $E_\mathrm{I}$ is minimum
Students I, II and III perform an experiment for measuring the acceleration due to gravity ($g$) using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table.
Least count for length = 0.1 cm
Least count for time = 0.1 s
| Student | Length of the pendulum (cm) | Number of oscillations ($n$) | Total time for ($n$) oscillations (s) | Time period (s) |
|---|---|---|---|---|
| I | 64.0 | 8 | 128.0 | 16.0 |
| II | 64.0 | 4 | 64.0 | 16.0 |
| III | 20.0 | 4 | 36.0 | 9.0 |
If $E_\mathrm{I}$, $E_\mathrm{II}$ and $E_\mathrm{III}$ are the percentage errors in $g$, i.e., $\left(\dfrac{\Delta g}{g}\times 100\right)$ for students I, II and III, respectively,
$E_\mathrm{I} = 0$
$E_\mathrm{I}$ is minimum
$E_\mathrm{I} = E_\mathrm{II}$
$E_\mathrm{II}$ is maximum
Solution + reasoning
Official IIT-JEE question (IIT-JEE 2008 Paper 1 Q24, source page 8). Answer per official NTA/JAB key: $E_\mathrm{I}$ is minimum.
IIT-JEE 2010 Paper 2 Q39Answer: 0.2 mm
A Vernier calipers has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale which match with 16 main scale divisions. For this Vernier calipers, the least count is
0.02 mm
0.05 mm
0.1 mm
0.2 mm
Solution + reasoning
Official IIT-JEE question (IIT-JEE 2010 Paper 2 Q39, source page 15). Answer per official NTA/JAB key: 0.2 mm.
IIT-JEE 2010 Paper 1 Q66Answer: Error $\Delta T$ in measuring $T$, the time period, is 0.05 seconds; Percentage error in the determination of $g$ is 5%
A student uses a simple pendulum of exactly 1 m length to determine $g$, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true?
Error $\Delta T$ in measuring $T$, the time period, is 0.05 seconds
Error $\Delta T$ in measuring $T$, the time period, is 1 second
Percentage error in the determination of $g$ is 5%
Percentage error in the determination of $g$ is 2.5%
Solution + reasoning
Official IIT-JEE question (IIT-JEE 2010 Paper 1 Q66, source page 19). Answer per official NTA/JAB key: Error $\Delta T$ in measuring $T$, the time period, is 0.05 seconds; Percentage error in the determination of $g$ is 5%.
IIT-JEE 2011 Paper 2 Q27Answer: 3.1%
The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is
0.9%
2.4%
3.1%
4.2%
Solution + reasoning
Official IIT-JEE question (IIT-JEE 2011 Paper 2 Q27, source page 11). Answer per official NTA/JAB key: 3.1%.
IIT-JEE 2011 Paper 1 Q38Answer: $\sqrt{\dfrac{Ne^2}{m\varepsilon_0}}$
Paragraph: A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let $N$ be the number density of free electrons, each of mass $m$. When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions. If the electric field becomes zero, the electrons begin to oscillate about the positive ions with a natural angular frequency $\omega_p$, which is called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency $\omega$, where a part of the energy is absorbed and a part of it is reflected. As $\omega$ approaches $\omega_p$, all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals.
Taking the electronic charge as $e$ and the permittivity as $\varepsilon_0$, use dimensional analysis to determine the correct expression for $\omega_p$.
$\sqrt{\dfrac{Ne}{m\varepsilon_0}}$
$\sqrt{\dfrac{m\varepsilon_0}{Ne}}$
$\sqrt{\dfrac{Ne^2}{m\varepsilon_0}}$
$\sqrt{\dfrac{m\varepsilon_0}{Ne^2}}$
Solution + reasoning
Official IIT-JEE question (IIT-JEE 2011 Paper 1 Q38, source page 18). Answer per official NTA/JAB key: $\sqrt{\dfrac{Ne^2}{m\varepsilon_0}}$.
IIT-JEE 2012 Paper 1 Q4Answer: due to the errors in the measurements of $d$ and $l$ are the same.
In the determination of Young's modulus $\left(Y=\frac{4MLg}{\pi l d^{2}}\right)$ by using Searle's method, a wire of length $L=2$ m and diameter $d=0.5$ mm is used. For a load $M=2.5$ kg, an extension $l=0.25$ mm in the length of the wire is observed. Quantities $d$ and $l$ are measured using a screw gauge and a micrometer, respectively. They have the same pitch of $0.5$ mm. The number of divisions on their circular scale is $100$. The contributions to the maximum probable error of the $Y$ measurement
due to the errors in the measurements of $d$ and $l$ are the same.
due to the error in the measurement of $d$ is twice that due to the error in the measurement of $l$.
due to the error in the measurement of $l$ is twice that due to the error in the measurement of $d$.
due to the error in the measurement of $d$ is four times that due to the error in the measurement of $l$.
Solution + reasoning
Official IIT-JEE question (IIT-JEE 2012 Paper 1 Q4, source page 3). Answer per official NTA/JAB key: due to the errors in the measurements of $d$ and $l$ are the same..
JEE Advanced 2012 Paper 1 Q1Answer: due to the errors in the measurements of d and l are the same.⚑ verify
In the determination of Young’s modulus 2 4MLg Y d = π l by using Searle’s method, a wire of length L = 2m and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension l = 0.25 mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement
due to the errors in the measurements of d and l are the same.
due to the error in the measurement of d is twice that due to the error in the measurement of l.
due to the error in the measurement of l is twice that due to the error in the measurement of d.
due to the error in the measurement of d is four times that due to the error in the measurement of l.
Solution + reasoning
JEE Advanced 2012 Paper 1 Q1 (source page 2). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Advanced 2013 Paper 1 Q1Answer: $5.124$ cm
The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between $5.10$ cm and $5.15$ cm of the main scale. The Vernier scale has $50$ divisions equivalent to $2.45$ cm. The $24^{\text{th}}$ division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is
$5.112$ cm
$5.124$ cm
$5.136$ cm
$5.148$ cm
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2013 Paper 1 Q1, source page 1). Answer per official NTA/JAB key: $5.124$ cm.
JEE Advanced 2013 Paper 2 Q1Answer: the fractional error in $d$ decreases.
Using the expression $2d\sin\theta=\lambda$, one calculates the values of $d$ by measuring the corresponding angles $\theta$ in the range $0$ to $90^\circ$. The wavelength $\lambda$ is exactly known and the error in $\theta$ is constant for all values of $\theta$. As $\theta$ increases from $0^\circ$,
the absolute error in $d$ remains constant.
the absolute error in $d$ increases.
the fractional error in $d$ remains constant.
the fractional error in $d$ decreases.
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2013 Paper 2 Q1, source page 1). Answer per official NTA/JAB key: the fractional error in $d$ decreases..
JEE Advanced 2013 Paper 2 Q17Answer: P-4, Q-2, R-1, S-3
Match List I with List II and select the correct answer using the codes given below the lists: List I -- P. Boltzmann constant; Q. Coefficient of viscosity; R. Planck constant; S. Thermal conductivity. List II -- 1. $[ML^{2}T^{-1}]$; 2. $[ML^{-1}T^{-1}]$; 3. $[MLT^{-3}K^{-1}]$; 4. $[ML^{2}T^{-2}K^{-1}]$.
P-3, Q-1, R-2, S-4
P-3, Q-2, R-1, S-4
P-4, Q-2, R-1, S-3
P-4, Q-1, R-2, S-3
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2013 Paper 2 Q17, source page 12). Answer per official NTA/JAB key: P-4, Q-2, R-1, S-3.
JEE Advanced 2014 Paper 1 Q12Answer: 4
During Searle's experiment, zero of the Vernier scale lies between $3.20 \times 10^{-2}\ \text{m}$ and $3.25 \times 10^{-2}\ \text{m}$ of the main scale. The $20^{\text{th}}$ division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of $2\ \text{kg}$ is applied to the wire, the zero of the Vernier scale still lies between $3.20 \times 10^{-2}\ \text{m}$ and $3.25 \times 10^{-2}\ \text{m}$ of the main scale but now the $45^{\text{th}}$ division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is $2\ \text{m}$ and its cross-sectional area is $8 \times 10^{-7}\ \text{m}^2$. The least count of the Vernier scale is $1.0 \times 10^{-5}\ \text{m}$. The maximum percentage error in the Young's modulus of the wire is
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2014 Paper 1 Q12, source page 7). Answer per official NTA/JAB key: 4.
JEE Advanced 2014 Paper 1 Q15Answer: 3
To find the distance $d$ over which a signal can be seen clearly in foggy conditions, a railways engineer uses dimensional analysis and assumes that the distance depends on the mass density $\rho$ of the fog, intensity (power/area) $S$ of the light from the signal and its frequency $f$. The engineer finds that $d$ is proportional to $S^{1/n}$. The value of $n$ is
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2014 Paper 1 Q15, source page 8). Answer per official NTA/JAB key: 3.
JEE Advanced 2015 Paper 2 Q2Answer: 4
The energy of a system as a function of time $t$ is given as $E(t) = A^2\exp(-\alpha t)$, where $\alpha = 0.2$ s$^{-1}$. The measurement of $A$ has an error of $1.25\%$. If the error in the measurement of time is $1.50\%$, the percentage error in the value of $E(t)$ at $t = 5$ s is
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2015 Paper 2 Q2, source page 2). Answer per official NTA/JAB key: 4.
JEE Advanced 2015 Paper 1 Q9Answer: If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.005$ mm.; If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.01$ mm.
Consider a Vernier callipers in which each $1$ cm on the main scale is divided into $8$ equal divisions and a screw gauge with $100$ divisions on its circular scale. In the Vernier callipers, $5$ divisions of the Vernier scale coincide with $4$ divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:
If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.01$ mm.
If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.005$ mm.
If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.01$ mm.
If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.005$ mm.
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2015 Paper 1 Q9, source page 5). Answer per official NTA/JAB key: If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.005$ mm.; If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.01$ mm..
Planck's constant $h$, speed of light $c$ and gravitational constant $G$ are used to form a unit of length $L$ and a unit of mass $M$. Then the correct option(s) is(are)
$M \propto \sqrt{c}$
$M \propto \sqrt{G}$
$L \propto \sqrt{h}$
$L \propto \sqrt{G}$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2015 Paper 1 Q10, source page 5). Answer per official NTA/JAB key: $M \propto \sqrt{c}$; $L \propto \sqrt{h}$; $L \propto \sqrt{G}$.
JEE Advanced 2015 Paper 2 Q10Answer: $\mu_0 I^2 = \varepsilon_0 V^2$; $I = \varepsilon_0 c V$
In terms of potential difference $V$, electric current $I$, permittivity $\varepsilon_0$, permeability $\mu_0$ and speed of light $c$, the dimensionally correct equation(s) is(are)
$\mu_0 I^2 = \varepsilon_0 V^2$
$\varepsilon_0 I = \mu_0 V$
$I = \varepsilon_0 c V$
$\mu_0 c I = \varepsilon_0 V$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2015 Paper 2 Q10, source page 5). Answer per official NTA/JAB key: $\mu_0 I^2 = \varepsilon_0 V^2$; $I = \varepsilon_0 c V$.
JEE Advanced 2016 Paper 2 Q8Answer: The error in the measurement of r is 10%⚑ verify
In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is 7 R r T 2 5g ( ) . The values of R and r are measured to be (60 1) mm and (10 1) mm, respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is(are) true?
The error in the measurement of r is 10%
The error in the measurement of T is 3.57%
The error in the measurement of T is 2%
The error in the determined value of g is 11%
Solution + reasoning
JEE Advanced 2016 Paper 2 Q8 (source page 6). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Advanced 2016 Paper 2 Q9Answer: The error in the measurement of $r$ is $10\%$; The error in the measurement of $T$ is $3.57\%$; The error in the determined value of $g$ is $11\%$
In an experiment to determine the acceleration due to gravity $g$, the formula used for the time period of a periodic motion is $T = 2\pi\sqrt{\dfrac{7(R-r)}{5g}}$. The values of $R$ and $r$ are measured to be $(60 \pm 1)$ mm and $(10 \pm 1)$ mm, respectively. In five successive measurements, the time period is found to be $0.52$ s, $0.56$ s, $0.57$ s, $0.54$ s and $0.59$ s. The least count of the watch used for the measurement of time period is $0.01$ s. Which of the following statement(s) is(are) true?
The error in the measurement of $r$ is $10\%$
The error in the measurement of $T$ is $3.57\%$
The error in the measurement of $T$ is $2\%$
The error in the determined value of $g$ is $11\%$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2016 Paper 2 Q9, source page 7). Answer per official NTA/JAB key: The error in the measurement of $r$ is $10\%$; The error in the measurement of $T$ is $3.57\%$; The error in the determined value of $g$ is $11\%$.
A length-scale ($l$) depends on the permittivity ($\varepsilon$) of a dielectric material, Boltzmann constant ($k_B$), the absolute temperature ($T$), the number per unit volume ($n$) of certain charged particles, and the charge ($q$) carried by each of the particles. Which of the following expression(s) for $l$ is(are) dimensionally correct?
Official JEE Advanced question (JEE Advanced 2016 Paper 1 Q10, source page 8). Answer per official NTA/JAB key: $l = \sqrt{\left(\dfrac{\varepsilon k_B T}{nq^2}\right)}$; $l = \sqrt{\left(\dfrac{q^2}{\varepsilon n^{1/3} k_B T}\right)}$.
JEE Advanced 2017 Paper 2 Q7Answer: $1\%$
A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is $\delta T = 0.01$ seconds and he measures the depth of the well to be $L = 20$ meters. Take the acceleration due to gravity $g = 10\ \mathrm{m\,s^{-2}}$ and the velocity of sound is $300\ \mathrm{m\,s^{-1}}$. Then the fractional error in the measurement, $\delta L/L$, is closest to
$0.2\%$
$1\%$
$3\%$
$5\%$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2017 Paper 2 Q7, source page 5). Answer per official NTA/JAB key: $1\%$.
JEE Advanced 2018 Paper 1 Q15Answer: $[E] = [B][L][T]^{-1}$
PARAGRAPH "X": In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, $[E]$ and $[B]$ stand for dimensions of electric and magnetic fields respectively, while $[\epsilon_0]$ and $[\mu_0]$ stand for dimensions of the permittivity and permeability of free space respectively. $[L]$ and $[T]$ are dimensions of length and time respectively. All the quantities are given in SI units.
The relation between $[E]$ and $[B]$ is
$[E] = [B][L][T]$
$[E] = [B][L]^{-1}[T]$
$[E] = [B][L][T]^{-1}$
$[E] = [B][L]^{-1}[T]^{-1}$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2018 Paper 1 Q15, source page 9). Answer per official NTA/JAB key: $[E] = [B][L][T]^{-1}$.
JEE Advanced 2018 Paper 1 Q16Answer: $[\mu_0] = [\epsilon_0]^{-1}[L]^{-2}[T]^2$
PARAGRAPH "X": In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, $[E]$ and $[B]$ stand for dimensions of electric and magnetic fields respectively, while $[\epsilon_0]$ and $[\mu_0]$ stand for dimensions of the permittivity and permeability of free space respectively. $[L]$ and $[T]$ are dimensions of length and time respectively. All the quantities are given in SI units.
The relation between $[\epsilon_0]$ and $[\mu_0]$ is
$[\mu_0] = [\epsilon_0][L]^2[T]^{-2}$
$[\mu_0] = [\epsilon_0][L]^{-2}[T]^2$
$[\mu_0] = [\epsilon_0]^{-1}[L]^2[T]^{-2}$
$[\mu_0] = [\epsilon_0]^{-1}[L]^{-2}[T]^2$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2018 Paper 1 Q16, source page 10). Answer per official NTA/JAB key: $[\mu_0] = [\epsilon_0]^{-1}[L]^{-2}[T]^2$.
JEE Advanced 2018 Paper 1 Q17Answer: $\frac{2\Delta a}{(1+a)^2}$
PARAGRAPH "A": If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation $z = x/y$. If the errors in $x$, $y$ and $z$ are $\Delta x$, $\Delta y$ and $\Delta z$, respectively, then $z \pm \Delta z = \frac{x \pm \Delta x}{y \pm \Delta y} = \frac{x}{y}\left(1 \pm \frac{\Delta x}{x}\right)\left(1 \pm \frac{\Delta y}{y}\right)^{-1}$. The series expansion for $\left(1 \pm \frac{\Delta y}{y}\right)^{-1}$, to first power in $\Delta y/y$, is $1 \mp (\Delta y/y)$. The relative errors in independent variables are always added. So the error in $z$ will be $\Delta z = z\left(\frac{\Delta x}{x} + \frac{\Delta y}{y}\right)$. The above derivation makes the assumption that $\Delta x/x \ll 1$, $\Delta y/y \ll 1$. Therefore, the higher powers of these quantities are neglected.
Consider the ratio $r = \frac{(1-a)}{(1+a)}$ to be determined by measuring a dimensionless quantity $a$. If the error in the measurement of $a$ is $\Delta a$ ($\Delta a/a \ll 1$), then what is the error $\Delta r$ in determining $r$?
$\frac{\Delta a}{(1+a)^2}$
$\frac{2\Delta a}{(1+a)^2}$
$\frac{2\Delta a}{(1-a^2)}$
$\frac{2a\Delta a}{(1-a^2)}$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2018 Paper 1 Q17, source page 11). Answer per official NTA/JAB key: $\frac{2\Delta a}{(1+a)^2}$.
JEE Advanced 2018 Paper 1 Q18Answer: 0.02
PARAGRAPH "A": If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation $z = x/y$. If the errors in $x$, $y$ and $z$ are $\Delta x$, $\Delta y$ and $\Delta z$, respectively, then $z \pm \Delta z = \frac{x \pm \Delta x}{y \pm \Delta y} = \frac{x}{y}\left(1 \pm \frac{\Delta x}{x}\right)\left(1 \pm \frac{\Delta y}{y}\right)^{-1}$. The series expansion for $\left(1 \pm \frac{\Delta y}{y}\right)^{-1}$, to first power in $\Delta y/y$, is $1 \mp (\Delta y/y)$. The relative errors in independent variables are always added. So the error in $z$ will be $\Delta z = z\left(\frac{\Delta x}{x} + \frac{\Delta y}{y}\right)$. The above derivation makes the assumption that $\Delta x/x \ll 1$, $\Delta y/y \ll 1$. Therefore, the higher powers of these quantities are neglected.
In an experiment the initial number of radioactive nuclei is 3000. It is found that $1000 \pm 40$ nuclei decayed in the first $1.0\ \mathrm{s}$. For $|x| \ll 1$, $\ln(1+x) = x$ up to first power in $x$. The error $\Delta \lambda$, in the determination of the decay constant $\lambda$, in $\mathrm{s^{-1}}$, is
0.04
0.03
0.02
0.01
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2018 Paper 1 Q18, source page 12). Answer per official NTA/JAB key: 0.02.
JEE Advanced 2019 Paper 1 Q8Answer: The dimension of force is L–3⚑ verify
Let us consider a system of units in which mass and angular momentum are dimensionless. If length has dimension of L, which of the following statement(s) is/are correct?
The dimension of force is L–3
The dimension of power is L–5
The dimension of linear momentum is L–1
The dimension of energy is L–2 Answer (A, C, D)
Solution + reasoning
JEE Advanced 2019 Paper 1 Q8 (source page 11). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Advanced 2019 Paper 1 Q11Answer: The dimension of linear momentum is $L^{-1}$; The dimension of energy is $L^{-2}$; The dimension of force is $L^{-3}$
Let us consider a system of units in which mass and angular momentum are dimensionless. If length has dimension of $L$, which of the following statement(s) is/are correct?
The dimension of linear momentum is $L^{-1}$
The dimension of energy is $L^{-2}$
The dimension of force is $L^{-3}$
The dimension of power is $L^{-5}$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2019 Paper 1 Q11, source page 8). Answer per official NTA/JAB key: The dimension of linear momentum is $L^{-1}$; The dimension of energy is $L^{-2}$; The dimension of force is $L^{-3}$.
JEE Advanced 2019 Paper 2 Q14Answer: 1.39
An optical bench has 1.5 m long scale having four equal divisions in each cm. While measuring the focal length of a convex lens, the lens is kept at 75 cm mark of the scale and the object pin is kept at 45 cm mark. The image of the object pin on the other side of the lens overlaps with image pin that is kept at 135 cm mark. In this experiment, the percentage error in the measurement of the focal length of the lens is ____.
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2019 Paper 2 Q14, source page 9). Answer per official NTA/JAB key: 1.39.
JEE Advanced 2020 Paper 1 Q9Answer: $\alpha + p = 2\beta$; $p + q - r = \beta$
Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are given in terms of a quantity $X$ as follows: [position] $= [X^{\alpha}]$; [speed] $= [X^{\beta}]$; [acceleration] $= [X^{p}]$; [linear momentum] $= [X^{q}]$; [force] $= [X^{r}]$. Then
$\alpha + p = 2\beta$
$p + q - r = \beta$
$p - q + r = \alpha$
$p + q + r = \beta$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2020 Paper 1 Q9, source page 6). Answer per official NTA/JAB key: $\alpha + p = 2\beta$; $p + q - r = \beta$.
JEE Advanced 2020 Paper 2 Q15Answer: 1.30
Two capacitors with capacitance values $C_1 = 2000 \pm 10\ \mathrm{pF}$ and $C_2 = 3000 \pm 15\ \mathrm{pF}$ are connected in series. The voltage applied across this combination is $V = 5.00 \pm 0.02\ \mathrm{V}$. The percentage error in the calculation of the energy stored in this combination of capacitors is ______.
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2020 Paper 2 Q15, source page 8). Answer per official NTA/JAB key: 1.30.
JEE Advanced 2021 Paper 2 Q4Answer: $\dfrac{\text{Force}}{\text{Length} \times \text{Time}}$; $\dfrac{\text{Power}}{\text{Area}}$
A physical quantity $\vec{S}$ is defined as $\vec{S} = (\vec{E} \times \vec{B})/\mu_0$, where $\vec{E}$ is electric field, $\vec{B}$ is magnetic field and $\mu_0$ is the permeability of free space. The dimensions of $\vec{S}$ are the same as the dimensions of which of the following quantity(ies)?
Official JEE Advanced question (JEE Advanced 2021 Paper 2 Q4, source page 3). Answer per official NTA/JAB key: $\dfrac{\text{Force}}{\text{Length} \times \text{Time}}$; $\dfrac{\text{Power}}{\text{Area}}$.
JEE Main 2021 (March 16 Shift 1) Paper 1 Q1Answer: 10a/n mm⚑ verify
One main scale division of a vernier callipers is 'a' cm and $n^{th}$ division of the vernier scale coincide with (n $- 1)^{th}$ division of the main scale. The least count of the callipers in mm is :
Solution + reasoning
JEE Main 2021 (March 16 Shift 1) Paper 1 Q1 (source page 1). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2021 (August 27 Shift 1) Paper 1 Q4Answer: ohm⚑ verify
If E and H represents the intensity of electric field and magnetising field respectively, then the unit of E/H will be :
Solution + reasoning
JEE Main 2021 (August 27 Shift 1) Paper 1 Q4 (source page 2). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2021 (August 26 Shift 1) Paper 1 Q15Answer: 5.15 mm⚑ verify
In a Screw Gauge, fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are 50 divisions on the circular scale, and the main scale moves by 0.5 mm on a complete rotation. For a particular observation the reading on the main scale is 5 mm and the $20^{th}$ division of the circular scale coincides with reference line. Calculate the true reading.
Solution + reasoning
JEE Main 2021 (August 26 Shift 1) Paper 1 Q15 (source page 5). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2021 (August 26 Shift 2) Paper 1 Q19Answer: 3⚑ verify
If the length of the pendulum in pendulum clock increases by 0.1%, then the error in time per day is :
Solution + reasoning
JEE Main 2021 (August 26 Shift 2) Paper 1 Q19 (source page 7). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2021 (July 27 Shift 1) Paper 1 Q19Answer: Assertion A is false, Reason R is true⚑ verify
Assertion A : If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is 5 mm and there are 50 total divisions on circular scale, then least count is 0.001 cm. Reason R : Least Count = ${{Pitch} \over {Total\,divisions\,on\,circular\,scale}}$ In the light of the above statements, choose the most appropriate answer from the options given below :
Solution + reasoning
JEE Main 2021 (July 27 Shift 1) Paper 1 Q19 (source page 4). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Advanced 2022 Paper 2 Q4Answer: 4
In a particular system of units, a physical quantity can be expressed in terms of the electric charge $e$, electron mass $m_e$, Planck's constant $h$, and Coulomb's constant $k = \dfrac{1}{4\pi\epsilon_0}$, where $\epsilon_0$ is the permittivity of vacuum. In terms of these physical constants, the dimension of the magnetic field is $[B] = [e]^{\alpha}[m_e]^{\beta}[h]^{\gamma}[k]^{\delta}$. The value of $\alpha + \beta + \gamma + \delta$ is _____.
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2022 Paper 2 Q4, source page 10). Answer per official NTA/JAB key: 4.
Young's modulus of elasticity $Y$ is expressed in terms of three derived quantities, namely, the gravitational constant $G$, Planck's constant $h$ and the speed of light $c$, as $Y = c^{\alpha} h^{\beta} G^{\gamma}$. Which of the following is the correct option?
$\alpha = 7,\ \beta = -1,\ \gamma = -2$
$\alpha = -7,\ \beta = -1,\ \gamma = -2$
$\alpha = 7,\ \beta = -1,\ \gamma = 2$
$\alpha = -7,\ \beta = 1,\ \gamma = -2$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2023 Paper 2 Q2, source page 10). Answer per official NTA/JAB key: $\alpha = 7,\ \beta = -1,\ \gamma = -2$.
JEE Main 2023 (January 30 Shift 1) Paper 1 Q28Answer: 220⚑ verify
In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by $0.5 \mathrm{~mm}$ on a complete rotation of the circular scale. The zero of circular scale lies 6 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 4 linear scale divisions are clearly visible while $46^{\text {th }}$ division the circular scale coincide with the reference line. The diameter of the wire is ______________ $\times 10^{-2} \mathrm{~mm}$.
Solution + reasoning
JEE Main 2023 (January 30 Shift 1) Paper 1 Q28 (source page 8). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 8 Shift 1) Paper 1 Q31Answer: 4%⚑ verify
A cylindrical wire of mass $(0.4 \pm 0.01) \mathrm{g}$ has length $(8 \pm 0.04) \mathrm{cm}$ and radius $(6 \pm 0.03) \mathrm{mm}$. The maximum error in its density will be:
1%
5%
4%
3.5%
Solution + reasoning
JEE Main 2023 (April 8 Shift 1) Paper 1 Q31 (source page 1). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 10 Shift 2) Paper 1 Q36Answer: $3 .18 \mathrm{~cm}$⚑ verify
In an experiment with vernier callipers of least count $0.1 \mathrm{~mm}$, when two jaws are joined together the zero of vernier scale lies right to the zero of the main scale and 6th division of vernier scale coincides with the main scale division. While measuring the diameter of a spherical bob, the zero of vernier scale lies in between $3.2 \mathrm{~cm}$ and $3.3 \mathrm{~cm}$ marks, and 4th division of vernier scale coincides with the main scale division. The diameter of bob is measured as
$3 .18 \mathrm{~cm}$
$3.22 \mathrm{~cm}$
$3.26 \mathrm{~cm}$
$3.25 \mathrm{~cm}$
Solution + reasoning
JEE Main 2023 (April 10 Shift 2) Paper 1 Q36 (source page 3). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 6 Shift 1) Paper 1 Q37Answer: 4.33⚑ verify
Two resistances are given as $\mathrm{R}_{1}=(10 \pm 0.5) \Omega$ and $\mathrm{R}_{2}=(15 \pm 0.5) \Omega$. The percentage error in the measurement of equivalent resistance when they are connected in parallel is -
2.33
5.33
4.33
6.33
Solution + reasoning
JEE Main 2023 (April 6 Shift 1) Paper 1 Q37 (source page 3). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 11 Shift 2) Paper 1 Q39Answer: $\mathrm{FV}^{-4} \mathrm{~T}^{-2}$⚑ verify
If force (F), velocity (V) and time (T) are considered as fundamental physical quantity, then dimensional formula of density will be :
$\mathrm{FV}^{-2} \mathrm{~T}^{2}$
$\mathrm{FV}^{4} \mathrm{~T}^{-6}$
$\mathrm{F}^{2} \mathrm{~V}^{-2} \mathrm{~T}^{6}$
$\mathrm{FV}^{-4} \mathrm{~T}^{-2}$
Solution + reasoning
JEE Main 2023 (April 11 Shift 2) Paper 1 Q39 (source page 3). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 15 Shift 1) Paper 1 Q45Answer: $\frac{1}{2}, \frac{1}{2}, 0$⚑ verify
The speed of a wave produced in water is given by $v=\lambda^{a} g^{b} \rho^{c}$. Where $\lambda, g$ and $\rho$ are wavelength of wave, acceleration due to gravity and density of water respectively. The values of $a, b$ and $c$ respectively, are :
$\frac{1}{2}, 0, \frac{1}{2}$
$1,1,0$
$1,-1,0$
$\frac{1}{2}, \frac{1}{2}, 0$
Solution + reasoning
JEE Main 2023 (April 15 Shift 1) Paper 1 Q45 (source page 5). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 11 Shift 1) Paper 1 Q47Answer: Statements I is correct but Statements II is incorrect.⚑ verify
Given below are two statements : Statements I : Astronomical unit (Au), Parsec (Pc) and Light year (ly) are units for measuring astronomical distances. Statements II : $\mathrm{Au} < \mathrm{Parsec} (\mathrm{Pc}) < \mathrm{ly}$ In the light of the above statements, choose the most appropriate answer from the options given below:
Both Statements I and Statements II are incorrect.
Both Statements I and Statements II are correct,
Statements I is incorrect but Statements II is correct.
Statements I is correct but Statements II is incorrect.
Solution + reasoning
JEE Main 2023 (April 11 Shift 1) Paper 1 Q47 (source page 6). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2023 (April 13 Shift 1) Paper 1 Q47Answer: $(1000 \pm 140) ~\mathrm{J}$⚑ verify
A body of mass $(5 \pm 0.5) ~\mathrm{kg}$ is moving with a velocity of $(20 \pm 0.4) ~\mathrm{m} / \mathrm{s}$. Its kinetic energy will be
$(1000 \pm 140) ~\mathrm{J}$
$(500 \pm 0.14) ~\mathrm{J}$
$(1000 \pm 0.14) ~\mathrm{J}$
$(500 \pm 140) ~\mathrm{J}$
Solution + reasoning
JEE Main 2023 (April 13 Shift 1) Paper 1 Q47 (source page 7). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Advanced 2024 Paper 1 Q1Answer: $(2n,\,-n,\,-n,\,-n)$
A dimensionless quantity is constructed in terms of electronic charge $e$, permittivity of free space $\varepsilon_0$, Planck's constant $h$, and speed of light $c$. If the dimensionless quantity is written as $e^{\alpha}\,\varepsilon_0^{\beta}\,h^{\gamma}\,c^{\delta}$ and $n$ is a non-zero integer, then $(\alpha,\beta,\gamma,\delta)$ is given by
$(2n,\,-n,\,-n,\,-n)$
$(n,\,-n,\,-2n,\,-n)$
$(n,\,-n,\,-n,\,-2n)$
$(2n,\,-n,\,-2n,\,-2n)$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2024 Paper 1 Q1, source page 11). Answer per official NTA/JAB key: $(2n,\,-n,\,-n,\,-n)$.
JEE Advanced 2024 Paper 2 Q8Answer: 3
The dimensions of a cone are measured using a scale with a least count of $2$ mm. The diameter of the base and the height are both measured to be $20.0$ cm. The maximum percentage error in the determination of the volume is ______.
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2024 Paper 2 Q8, source page 14). Answer per official NTA/JAB key: 3.
JEE Main 2024 (April 6 Shift 1) Paper 1 Q21Answer: x = 71⚑ verify
While measuring diameter of wire using screw gauge the following readings were noted. Main scale reading is $1 \mathrm{~mm}$ and circular scale reading is equal to 42 divisions. Pitch of screw gauge is $1 \mathrm{~mm}$ and it has 100 divisions on circular scale. The diameter of the wire is $\frac{x}{50} \mathrm{~mm}$. The value of $x$ is :
42
71
21
142
Solution + reasoning
JEE Main 2024 (April 6 Shift 1) Paper 1 Q21 (source page 17). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
A temperature difference can generate e.m.f. in some materials. Let $S$ be the e.m.f. produced per unit temperature difference between the ends of a wire, $\sigma$ the electrical conductivity and $\kappa$ the thermal conductivity of the material of the wire. Taking $M, L, T, I$ and $K$ as dimensions of mass, length, time, current and temperature, respectively, the dimensional formula of the quantity $Z = \dfrac{S^2\sigma}{\kappa}$ is:
$[M^0 L^0 T^0 I^0 K^0]$
$[M^0 L^0 T^0 I^0 K^{-1}]$
$[M^1 L^2 T^{-2} I^{-1} K^{-1}]$
$[M^1 L^2 T^{-4} I^{-1} K^{-1}]$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2025 Paper 2 Q1, source page 8). Answer per official NTA/JAB key: $[M^0 L^0 T^0 I^0 K^{-1}]$.
JEE Advanced 2025 Paper 1 Q6Answer: $3 \times 10^{-5}$
Length, breadth and thickness of a strip having a uniform cross section are measured to be $10.5\ \mathrm{cm}$, $0.05\ \mathrm{mm}$, and $6.0\ \mu\mathrm{m}$, respectively. Which of the following option(s) give(s) the volume of the strip in $\mathrm{cm^3}$ with correct significant figures:
$3.2 \times 10^{-5}$
$32.0 \times 10^{-6}$
$3.0 \times 10^{-5}$
$3 \times 10^{-5}$
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2025 Paper 1 Q6, source page 14). Answer per official NTA/JAB key: $3 \times 10^{-5}$.
JEE Advanced 2025 Paper 1 Q12Answer: 75.6
A single slit diffraction experiment is performed to determine the slit width using the equation, $\dfrac{bd}{D} = m\lambda$, where $b$ is the slit width, $D$ the shortest distance between the slit and the screen, $d$ the distance between the $m$th diffraction maximum and the central maximum, and $\lambda$ is the wavelength. $D$ and $d$ are measured with scales of least count of $1$ cm and $1$ mm, respectively. The values of $\lambda$ and $m$ are known precisely to be $600$ nm and $3$, respectively. The absolute error (in $\mu$m) in the value of $b$ estimated using the diffraction maximum that occurs for $m = 3$ with $d = 5$ mm and $D = 1$ m is ___
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2025 Paper 1 Q12, source page 18). Answer per official NTA/JAB key: 75.6.
JEE Advanced 2026 Paper 2 Q14Answer: 25
In a new system of units, the units of mass, length, time and current are $5\ \mathrm{kg}$, $5\ \mathrm{m}$, $5\ \mathrm{s}$ and $5\ \mathrm{A}$, respectively. If $\mu_0$ and $\epsilon_0$ are the permeability and permittivity of free space, respectively, then in this new system of units, the magnitude of one SI unit of $\sqrt{\mu_0/\epsilon_0}$, is:
Solution + reasoning
Official JEE Advanced question (JEE Advanced 2026 Paper 2 Q14, source page 16). Answer per official NTA/JAB key: 25.
JEE Main 2026 (April 2 Shift 1) Paper 1 Q26Answer: 2a − b + c = 1⚑ verify
The dimensional formula of $\frac{1}{2} \epsilon_0 E^2$ ($\epsilon_0$ = permittivity of vacuum and $E$ = electric field) is $M^a L^b T^c$. The value of $2a - b + c =$ ________.
0
1
-1
2
Solution + reasoning
JEE Main 2026 (April 2 Shift 1) Paper 1 Q26 (source page 10). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 2 Shift 2) Paper 1 Q26Answer: $[h T^{-1} L M^{-2}]$⚑ verify
Dimensions of universal gravitational constant ($G$) in terms of Planck's constant ($h$), distance ($L$), mass ($M$) and time ($T$) are _______.
$[h T L M^{-2}]$
$[h T^{-1} L M^{-2}]$
$[h T L^2 M^{-2}]$
$[h^{-1} T^{-1} L M^{-2}]$
Solution + reasoning
JEE Main 2026 (April 2 Shift 2) Paper 1 Q26 (source page 10). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 4 Shift 1) Paper 1 Q26Answer: $ 5 \times 10^{-3}$⚑ verify
In a screw gauge when the circular scale is given five complete rotations it moves linearly by 2.5 mm . If the circular scale has 100 divisions, the least count of screw gauge is $\_\_\_\_$ mm.
$1 \times 10^{-2}$
$ 1 \times 10^{-3}$
$5 \times 10^{-2}$
$ 5 \times 10^{-3}$
Solution + reasoning
JEE Main 2026 (April 4 Shift 1) Paper 1 Q26 (source page 9). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 5 Shift 1) Paper 1 Q26Answer: 0.07 cm positive zero error⚑ verify
In a Vernier calipers, when both jaws touch each other, zero of the Vernier scale is shifted to the right of zero of the main scale and $7^{\text {th }}$ Vernier division coincides with a main scale reading. If the value of 1 main scale division is 1 mm and there are 10 Vernier scale divisions, then the Vernier caliper has
0.07 cm negative zero error
0.7 cm negative zero error
0.07 cm positive zero error
0.7 cm positive zero error
Solution + reasoning
JEE Main 2026 (April 5 Shift 1) Paper 1 Q26 (source page 9). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 6 Shift 1) Paper 1 Q26Answer: 0.82%⚑ verify
The density $\rho$ of a uniform cylinder is determined by measuring its mass $m$, length $l$ and diameter $d$. The measured values of $m, l$ and $d$ are $97.42 \pm 0.02 \mathrm{~g}$, $8.35 \pm 0.05 \mathrm{~mm}$ and $20.20 \pm 0.02 \mathrm{~mm}$, respectively. Calculated percentage fractional error in $\rho$ is $\_\_\_\_$ .
0.63%
0.82%
0.72%
0.25%
Solution + reasoning
JEE Main 2026 (April 6 Shift 1) Paper 1 Q26 (source page 10). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 8 Shift 2) Paper 1 Q26Answer: $400 \alpha$⚑ verify
A new unit ( $\alpha$ ) of length is chosen such that it is equal to the speed of light in vacuum. What is the distance between Venus and Earth in terms of $\alpha$ units if light takes 6 min. 40 s to cover this distance?
$200 \alpha$
$400 \alpha$
$300 \alpha$
$500 \alpha$
Solution + reasoning
JEE Main 2026 (April 8 Shift 2) Paper 1 Q26 (source page 11). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 5 Shift 1) Paper 1 Q27Answer: $\frac{R}{\sqrt{L C}}$⚑ verify
$L, C$ and $R$ represents physical quantities inductance, capacitance and resistance respectively. The dimensional formula $\mathrm{ML}^2 \mathrm{~T}^{-4} \mathrm{~A}^{-2}$ corresponds to $\_\_\_\_$ .
$\frac{R}{\sqrt{L C}}$
$\frac{R}{L C}$
$\frac{C}{\sqrt{L R}}$
$\frac{1}{R} \sqrt{\frac{L}{C}}$
Solution + reasoning
JEE Main 2026 (April 5 Shift 1) Paper 1 Q27 (source page 9). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 6 Shift 1) Paper 1 Q27Answer: $\left[\mathrm{M}^1 \mathrm{~L}^{7 / 2} \mathrm{~T}^{-2}\right]$⚑ verify
The potential energy of a particle changes with distance $x$ from a fixed origin as $V=\frac{A \sqrt{x}}{x+B}$, where $A$ and $B$ are constant with appropriate dimensions. The dimensions of $A B$ are $\_\_\_\_$
JEE Main 2026 (April 6 Shift 1) Paper 1 Q27 (source page 10). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 8 Shift 2) Paper 1 Q27Answer: $1, 1, 1, 1$⚑ verify
Consider the equation $H=\frac{x^p \epsilon^q E^r}{t^s}$ Where $H=$ magnetic field; $E=$ electric field, $\epsilon=$ permittivity, $x=$ distance, $t=$ time The values of $p, q, r$ and $s$ respectively are :
$1, 1, 1, 1$
$-1,1,2,1$
$1, -1, -2, 1$
$-1,-2,-2,1$
Solution + reasoning
JEE Main 2026 (April 8 Shift 2) Paper 1 Q27 (source page 11). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
JEE Main 2026 (April 2 Shift 2) Paper 1 Q32Answer: 5.045⚑ verify
In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are 100 divisions in circular scale and pitch of screw gauge is 0.1 mm. When diameter of a sphere is measured, the reading of main scale is 5 mm and $50^{th}$ division of circular scale coincides with the reference line of main scale. The diameter of sphere is ______ mm.
5.045
5.055
5.450
5.550
Solution + reasoning
JEE Main 2026 (April 2 Shift 2) Paper 1 Q32 (source page 12). Answer per published-compilation key — flagged for SME verification (⚑). A compilation key is not authoritative; verify against the physics before relying on it.
🎯 Question Bank 117 MCQs · graded
Distribution — advanced: 22 · easy: 31 · hard: 25 · medium: 39. Every question carries a source trace; each ends in an SME-verify solution.
Q1 How many base (fundamental) quantities are there in the SI system? easy
Step solution + source
The SI recognises seven base quantities: length, mass, time, electric current, thermodynamic temperature, amount of substance and luminous intensity. Every other physical quantity is derived from these seven, so their units combine to express all derived units. This is why only a limited number of base units suffices. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q2 What is the SI base unit for the amount of substance? easy
Step solution + source
The mole (symbol mol) is the SI base unit of amount of substance; one mole contains exactly the Avogadro number of elementary entities. Candela measures luminous intensity, kelvin measures thermodynamic temperature and ampere measures electric current, so those three are the wrong base units here. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q3 Which of the following is NOT a base quantity in the SI system? easy
Step solution + source
Force is a derived quantity because it equals mass times acceleration, giving the derived unit newton. Length, electric current and luminous intensity are three of the seven base quantities. A derived quantity is always expressible as a combination of the seven base units, unlike a genuine base quantity. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q4 The SI unit of luminous intensity is the: medium
Step solution + source
Luminous intensity has the SI base unit candela (symbol cd). The lumen is a derived unit of luminous flux, the mole measures amount of substance and the kelvin measures thermodynamic temperature. Only the candela is one of the seven base units, defined via the luminous efficacy of a specified monochromatic radiation. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q5 The kelvin is the SI base unit associated with which physical quantity? medium
Step solution + source
The kelvin measures thermodynamic temperature and is now fixed through the Boltzmann constant. Heat energy is measured in joules and is a derived quantity, amount of substance uses the mole, and pressure is derived with the unit pascal. Hence only thermodynamic temperature is the correct base quantity for the kelvin. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q6 The radian and the steradian, used for plane angle and solid angle, are best described as: medium
Step solution + source
A plane angle is a ratio of arc length to radius and a solid angle is a ratio of area to the square of radius, so both are ratios of like quantities. Being ratios of similar magnitudes, radian and steradian are dimensionless, which is why they were not counted among the seven base units. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q7 In the revised SI, the kilogram is defined by fixing the numerical value of which constant? hard
Step solution + source
The modern kilogram is fixed by assigning the Planck constant $h$ the exact value $6.62607015\times10^{-34}$ J s, where J s equals kg m$^2$ s$^{-1}$. The Avogadro constant now fixes the mole, the speed of light fixes the metre, and the caesium hyperfine frequency fixes the second, so those do not define mass. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q8 The SI second is defined by fixing the numerical value of the caesium-133 hyperfine transition frequency at: hard
Step solution + source
The second is defined by taking the unperturbed ground-state hyperfine transition frequency of the caesium-133 atom, $\Delta\nu_{cs}$, to be exactly $9192631770$ Hz, where Hz equals s$^{-1}$. The value $299792458$ belongs to the speed of light, while the charge and Planck-constant figures define the ampere and the kilogram respectively. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q9 The candela is defined using the luminous efficacy $K_{cd}=683$ lm W$^{-1}$ of monochromatic radiation. That radiation has a frequency of: advanced
Step solution + source
The candela fixes the luminous efficacy of monochromatic radiation of frequency $540\times10^{12}$ Hz to be exactly $683$ lm W$^{-1}$, which equals cd sr W$^{-1}$. The other figures are the caesium hyperfine frequency defining the second and the fixed speed of light defining the metre, so neither of them is the reference frequency for the candela. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q10 The SI prefix that denotes a factor of $10^{-9}$ is: easy
Step solution + source
The prefix nano stands for $10^{-9}$. In contrast micro is $10^{-6}$, pico is $10^{-12}$ and milli is $10^{-3}$. Prefixes let us write very small or very large measured magnitudes compactly in the decimal SI scheme, keeping the numerical part convenient while preserving the significant figures of the measurement. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q11 The prefix 'mega' represents a multiplying factor of: easy
Step solution + source
Mega denotes $10^{6}$, as in megawatt or megahertz. The factor $10^{3}$ is kilo, $10^{9}$ is giga and $10^{-6}$ is micro. Because SI uses a decimal system, these prefixes make conversions within the system simple and convenient, which is one of the practical advantages of SI over older unit systems. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q12 One kilometre expressed in metres equals: easy
Step solution + source
The prefix kilo means $10^{3}$, so one kilometre is $10^{3}$ metres, i.e. one thousand metres. The choice $10^{2}$ would be a hectometre and $10^{6}$ would be a megametre. Such decimal relationships between multiples and sub-multiples make unit conversion straightforward within the SI framework. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q13 Written in proper scientific notation $a\times10^{b}$ (with $1\le a\lt 10$), the length $0.00045$ m is: medium
Step solution + source
Scientific notation requires the coefficient $a$ to satisfy $1\le a\lt 10$. Shifting the decimal point four places to the right gives $4.5$, so $0.00045=4.5\times10^{-4}$ m. The forms $45\times10^{-5}$ and $0.45\times10^{-3}$ have coefficients outside the allowed range, and $4.5\times10^{-3}$ is off by a power of ten. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q14 The diameter of the earth is about $1.28\times10^{7}$ m. Its order of magnitude is: medium
Step solution + source
To find the order of magnitude, the coefficient $a=1.28$ is rounded to $1$ because $a\le 5$, leaving $10^{7}$. Hence the order of magnitude is the exponent $7$. This estimation lets us compare scales quickly; the earth being of order $10^{7}$ m is a standard NCERT illustration of order-of-magnitude reasoning. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q15 Which pairing of SI prefix and factor is correct? medium
Step solution + source
Giga correctly denotes $10^{9}$. The other statements are wrong: micro is $10^{-6}$ not $10^{-3}$, kilo is $10^{3}$ not $10^{6}$, and nano is $10^{-9}$ not $10^{-6}$. Knowing the exact powers of ten attached to each prefix is essential for correct scientific-notation conversions and order-of-magnitude estimates. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q16 Using the rounding rule (round $a$ to $1$ for $a\le 5$ and to $10$ for $5\lt a\le 10$), the order of magnitude of $6.4\times10^{6}$ is: hard
Step solution + source
Here $a=6.4$ lies in the range $5\lt a\le 10$, so it is rounded up to $10$, giving $10\times10^{6}=10^{7}$. Therefore the order of magnitude is $7$, not $6$. This subtlety, that coefficients above five round up a power of ten, is exactly what the NCERT order-of-magnitude convention is designed to capture. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q17 A speed of $3.0$ m s$^{-1}$ converted to kilometres per hour equals: hard
Step solution + source
Multiply by the conversion chain: $3.0\ \text{m s}^{-1}\times\dfrac{1\ \text{km}}{1000\ \text{m}}\times\dfrac{3600\ \text{s}}{1\ \text{h}}=3.0\times3.6=10.8$ km h$^{-1}$. The factor $3.6$ converts m s$^{-1}$ to km h$^{-1}$. The distractor $3.6$ forgets to multiply by the given magnitude, and the others invert or misplace the conversion factor. 🔉⇢
Source: JEE Main (pattern)
Q18 The diameter of the earth is of order $10^{7}$ m and that of a hydrogen atom of order $10^{-10}$ m. By how many orders of magnitude is the earth larger? advanced
Step solution + source
The number of orders of magnitude between two scales is the difference of their exponents: $7-(-10)=17$. Thus the earth's diameter is seventeen orders of magnitude greater than a hydrogen atom's. Subtracting a negative exponent adds, which is why the answer is far larger than either individual order of magnitude taken alone. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q19 The dimensional formula of force is: easy
Step solution + source
Force equals mass times acceleration, and acceleration has dimensions of length per time squared, namely $[L\,T^{-2}]$. Multiplying by mass $[M]$ therefore gives $[M\,L\,T^{-2}]$ for force. The option $[M\,L^{2}\,T^{-2}]$ is actually energy, $[M\,L^{-1}\,T^{-2}]$ is pressure and $[M\,L\,T^{-1}]$ is momentum, so only the first choice correctly represents force. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q20 The dimensions of velocity are: easy
Step solution + source
Velocity is displacement divided by time, so its dimension is length over time, $[L\,T^{-1}]$. The choice $[L\,T^{-2}]$ is acceleration, while the remaining options do not correspond to any standard kinematic quantity. Since magnitudes are ignored in dimensional work, all velocities and speeds share this same dimensional formula. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q21 The dimensional formula of area is: easy
Step solution + source
Area is the product of two lengths, so its dimension is $[L]\times[L]=[L^{2}]$, independent of mass and time. Volume, being three lengths, is $[L^{3}]$, a single length is $[L]$, and $[M\,L^{2}]$ mixes in mass. Area therefore has zero dimension in both mass and time and two dimensions in length. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q22 The dimensional formula of mass density is: medium
Step solution + source
Density is mass per unit volume, so its dimension is $[M]/[L^{3}]=[M\,L^{-3}\,T^{0}]$. It has one dimension in mass, negative three in length and zero in time. The option $[M\,L^{3}]$ inverts the length power, and adding a time power as in $[M\,L^{-3}\,T^{-1}]$ is dimensionally inconsistent with a static density. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q23 The dimensional formula of pressure is: medium
Step solution + source
Pressure is defined as force per unit area, so its dimension is $[M\,L\,T^{-2}]/[L^{2}]=[M\,L^{-1}\,T^{-2}]$, with a negative single power of length. The bare force dimension $[M\,L\,T^{-2}]$ forgets to divide by area, $[M\,L^{2}\,T^{-2}]$ is energy, and $[M\,L^{-1}\,T^{-1}]$ is the coefficient of viscosity, so only the first choice correctly represents pressure. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q24 The dimensional formula of energy (work) is: medium
Step solution + source
Work equals force times displacement, so its dimension is $[M\,L\,T^{-2}]\times[L]=[M\,L^{2}\,T^{-2}]$. Kinetic energy $\tfrac12 mv^{2}$ yields exactly the same result, confirming dimensional consistency between different forms of energy. The option $[M\,L\,T^{-2}]$ is force alone, and $[M\,L^{2}\,T^{-1}]$ is angular momentum, so neither of those distractors correctly represents energy or work. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q25 Using $F=\dfrac{G m_1 m_2}{r^{2}}$, the dimensional formula of the gravitational constant $G$ is: hard
Step solution + source
Rearranging, $G=\dfrac{F r^{2}}{m_1 m_2}$, so its dimension is $\dfrac{[M\,L\,T^{-2}][L^{2}]}{[M^{2}]}=[M^{-1}\,L^{3}\,T^{-2}]$. The negative mass power comes from dividing force by a product of two masses. Errors in the distractors arise from mishandling either the mass power or the time power during this algebra. 🔉⇢
Source: JEE Main (pattern)
Q26 The coefficient of viscosity $\eta$, defined through $F=\eta A\dfrac{dv}{dx}$, has the dimensional formula: hard
Step solution + source
From $\eta=\dfrac{F}{A(dv/dx)}$, the velocity gradient $dv/dx$ has dimension $[T^{-1}]$, so $\eta=\dfrac{[M\,L\,T^{-2}]}{[L^{2}][T^{-1}]}=[M\,L^{-1}\,T^{-1}]$. Note that this coincides with the dimension of pressure multiplied by time, a useful cross-check. The distractor $[M\,L^{-1}\,T^{-2}]$ is simply pressure itself, obtained by wrongly forgetting to include the velocity-gradient factor in the denominator. 🔉⇢
Source: JEE Main (pattern)
Q27 From $E=h\nu$, where $\nu$ is frequency, the dimensional formula of Planck's constant $h$ is: advanced
Step solution + source
Since $h=E/\nu$ and frequency has dimension $[T^{-1}]$, we get $h=\dfrac{[M\,L^{2}\,T^{-2}]}{[T^{-1}]}=[M\,L^{2}\,T^{-1}]$. This is identical to the dimension of angular momentum, which is why $h$ is often called the quantum of action. The energy dimension $[M\,L^{2}\,T^{-2}]$ is the distractor obtained by ignoring the frequency division. 🔉⇢
Source: JEE Advanced (pattern)
Q28 According to the principle of homogeneity of dimensions, two physical quantities can be added only if they: easy
Step solution + source
The principle of homogeneity states that only quantities with identical dimensions may be added or subtracted; for example velocity cannot be added to force. Magnitude, scalar nature and choice of unit system are irrelevant to this rule. This principle is the basis for checking the dimensional consistency of any physical equation. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q29 A dimensionally correct equation is: easy
Step solution + source
Dimensional consistency is a necessary but not sufficient condition for correctness. An equation that passes the dimensional test may still be wrong by a dimensionless factor or an added dimensionless term, whereas a dimensionally inconsistent equation is certainly wrong. Thus passing the test does not prove an equation, but failing it disproves the equation. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q30 For the kinematic relation $x=x_0+v_0 t+\tfrac12 a t^{2}$, each term on the right has the dimension: medium
Step solution + source
Checking each term individually: $[x_0]=[L]$, $[v_0 t]=[L\,T^{-1}][T]=[L]$, and $[\tfrac12 a t^{2}]=[L\,T^{-2}][T^{2}]=[L]$. Every term reduces to length, matching the left-hand side $[x]=[L]$, so the equation is dimensionally consistent and passes the homogeneity test. The numerical factor $\tfrac12$ is dimensionless and therefore does not affect this dimensional analysis at all. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q31 In any physical equation, the argument of a trigonometric, logarithmic or exponential function must be: medium
Step solution + source
Special functions such as $\sin$, $\log$ and $\exp$ can only accept pure numbers, because their series expansions add together powers of the argument, which requires the argument to be dimensionless. For instance the phase $\omega t$ in $\sin(\omega t)$ is dimensionless since $[\omega]=[T^{-1}]$ cancels the time dimension. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q32 On dimensional grounds, which proposed formula for kinetic energy $K$ can be ruled out immediately? medium
Step solution + source
Kinetic energy has dimension $[M\,L^{2}\,T^{-2}]$. The expression $m^{2}v^{3}$ has dimension $[M^{2}][L\,T^{-1}]^{3}=[M^{2}\,L^{3}\,T^{-3}]$, which does not match energy at all, so it is ruled out immediately. Each $mv^{2}$ form gives the correct $[M\,L^{2}\,T^{-2}]$; dimensional analysis cannot distinguish among those forms differing only by a dimensionless numerical factor such as one half. 🔉⇢
Source: NCERT Example 1.4
Q33 The period $T$ of a simple pendulum is assumed to depend on length $l$, mass $m$ and $g$. Dimensional analysis gives $T=k\,l^{x}g^{y}m^{z}$ with: hard
Step solution + source
Equating dimensions $[T]=[L]^{x}[L\,T^{-2}]^{y}[M]^{z}$ leads to the equations $x+y=0$ for length, $-2y=1$ for time and $z=0$ for mass. Solving these gives $y=-\tfrac12$, $x=\tfrac12$ and $z=0$, so $T=k\sqrt{l/g}$, completely independent of the bob mass. The dimensionless constant $k$ turns out to be $2\pi$, a value that dimensional analysis alone can never supply. 🔉⇢
Source: NCERT Example 1.5
Q34 A body falls from rest through height $h$; its speed $v$ at the bottom is assumed to depend on $g$ and $h$ as $v=k\,g^{a}h^{b}$. Dimensional analysis gives: hard
Step solution + source
Writing $[L\,T^{-1}]=[L\,T^{-2}]^{a}[L]^{b}$ gives $a+b=1$ for the length powers and $-2a=-1$ for the time powers, so $a=\tfrac12$ and $b=\tfrac12$. Hence $v\propto\sqrt{gh}$, which matches the exact kinematic result $v=\sqrt{2gh}$ up to the dimensionless numerical factor $\sqrt2$ that dimensional reasoning alone can never determine. 🔉⇢
Source: JEE Main (pattern)
Q35 The time period $\tau$ of oscillation of a liquid drop depends on its density $\rho$, radius $r$ and surface tension $S$ (dimension $[M\,T^{-2}]$). Dimensional analysis gives $\tau\propto$: advanced
Step solution + source
Assume $\tau=k\,\rho^{a}r^{b}S^{c}$. With $[\rho]=[M\,L^{-3}]$, $[r]=[L]$ and $[S]=[M\,T^{-2}]$, matching the dimension $[T]$ gives mass $a+c=0$, length $-3a+b=0$ and time $-2c=1$. Solving these simultaneously yields $c=-\tfrac12$, $a=\tfrac12$ and $b=\tfrac32$, so $\tau\propto\sqrt{\rho r^{3}/S}$. The remaining options either invert this ratio or assign the wrong powers to the variables. 🔉⇢
Source: JEE Advanced (pattern)
Q36 In the van der Waals equation $\left(P+\dfrac{a}{V^{2}}\right)(V-b)=RT$, the dimensional formula of the constant $a$ is: advanced
Step solution + source
By the principle of homogeneity, the term $a/V^{2}$ must have the dimension of pressure $[M\,L^{-1}\,T^{-2}]$ so that it can be added to $P$. Therefore $a=[\text{pressure}]\times[V^{2}]=[M\,L^{-1}\,T^{-2}][L^{6}]=[M\,L^{5}\,T^{-2}]$, since the volume-squared term supplies six extra powers of length. Option $[M\,L^{-1}\,T^{-2}]$ is merely pressure, obtained by wrongly forgetting to multiply by $V^{2}$. 🔉⇢
Source: JEE Advanced (pattern)
Q37 How many significant figures are there in the measurement $2.308$ cm? easy
Step solution + source
All four digits $2,3,0,8$ are significant: non-zero digits always count and the zero lies between two non-zero digits, so it counts too. Expressing the same value as $0.02308$ m or $23.08$ mm does not change the count, confirming that a change of units never alters the number of significant figures. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q38 The number of significant figures in $0.007$ m$^{2}$ is: easy
Step solution + source
For a number less than one, the zeros to the right of the decimal point but to the left of the first non-zero digit are not significant; they merely fix the decimal place. Hence in $0.007$ only the digit $7$ is significant, giving exactly one significant figure. The leading zeros carry no information about precision. 🔉⇢
Source: NCERT Exercise
Q39 How many significant figures does the reported value $6.320$ J contain? easy
Step solution + source
Trailing zeros in a number that carries a decimal point are significant, because they convey the precision of the measurement. Thus $6.320$ has four significant figures: $6,3,2$ and the trailing $0$. Had the value been written $6.32$, it would carry only three, so the explicit final zero is meaningful. 🔉⇢
Source: NCERT Exercise
Q40 The number of significant figures in $0.2370$ g cm$^{-3}$ is: medium
Step solution + source
The leading zero before the decimal is never significant, and the zero to the right of the decimal before the $2$ would not count, but here it is a trailing zero after non-zero digits with a decimal present. The significant digits are $2,3,7,0$, giving four, since a trailing zero after the decimal indicates measured precision. 🔉⇢
Source: NCERT Exercise
Q41 How many significant figures are there in $2.64\times10^{24}$ kg? medium
Step solution + source
In scientific notation, only the digits of the coefficient $a$ count as significant; the power of ten merely fixes the magnitude. The coefficient $2.64$ has three significant figures, so the whole value has three. The exponent $24$ plays no part in the significant-figure count, a key advantage of scientific notation. 🔉⇢
Source: NCERT Exercise
Q42 The number of significant figures in $0.0006032$ m$^{2}$ is: medium
Step solution + source
The four leading zeros only locate the decimal point and are not significant. The significant digits are $6,0,3,2$, where the middle zero counts because it lies between non-zero digits. Hence the value has four significant figures. Rewriting it as $6.032\times10^{-4}$ makes this immediately clear from the coefficient. 🔉⇢
Source: NCERT Exercise
Q43 A length is reported as $4.700$ m. Expressed as $470.0$ cm, $4700$ mm and $0.004700$ km, the number of significant figures is: hard
Step solution + source
A mere change of units cannot change the number of significant figures. The value $4.700$ m has four significant figures, and each equivalent form must also have four. Naively applying the trailing-zero rule to $4700$ mm would wrongly suggest two, which is exactly the ambiguity that scientific notation, $4.700\times10^{3}$ mm, removes. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q44 In the relation $s=2\pi r$ or the factor $2$ in $r=d/2$, the exact number $2$ has: hard
Step solution + source
Multiplying or dividing factors that are not measured values, such as the $2$ in $r=d/2$ or the $n$ in a mean, are exact numbers with an infinite number of significant figures. They can be written as $2.0$ or $2.0000$ as needed and never limit the significant figures of a calculated result, unlike measured quantities. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q45 A measurement of $1.02$ g and another of $9.89$ g are each accurate to $\pm0.01$ g. Regarding relative uncertainty and significant figures, which statement is correct? advanced
Step solution + source
Both readings display three significant figures. However relative error depends on the number itself: $0.01/1.02\approx1\%$ while $0.01/9.89\approx0.1\%$. So the same absolute uncertainty produces a ten-fold larger relative error for the smaller quantity, showing that significant-figure count alone does not fix relative precision. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q46 The number $2.746$ rounded off to three significant figures is: easy
Step solution + source
The digit to be dropped is $6$, which is greater than $5$, so the preceding digit is raised by one: $4$ becomes $5$, giving $2.75$. Rounding preserves the precision consistent with three significant figures. Keeping the discarded $6$ would overstate the precision of the reported value. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q47 The number $1.743$ rounded off to three significant figures is: easy
Step solution + source
The digit being dropped is $3$, which is less than $5$, so by the rounding convention the preceding digit is left unchanged. Hence $1.743$ becomes $1.74$ to three significant figures. Raising the last retained digit would be incorrect because the dropped digit does not reach the halfway mark of five. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q48 Using the even-odd convention for a trailing $5$, the number $2.745$ rounded to three significant figures is: medium
Step solution + source
When the digit to be dropped is exactly $5$, the convention keeps the preceding digit unchanged if it is even and raises it if it is odd. Here the preceding digit $4$ is even, so it stays, giving $2.74$. This even-rounding rule avoids a systematic upward bias when many such values are rounded. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q49 Using the same convention, the number $2.735$ rounded to three significant figures becomes: medium
Step solution + source
The digit to be dropped is $5$ and the preceding digit is $3$, which is odd. The convention then raises the odd preceding digit by one, so $2.735$ becomes $2.74$. This mirrors the previous case: even preceding digits stay, odd ones increase, so that rounding of exact halves is unbiased on average. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q50 If mass $=4.237$ g and volume $=2.51$ cm$^{3}$, the density reported to the correct number of significant figures is: medium
Step solution + source
In division the result keeps as many significant figures as the input with the fewest. Mass has four and volume has three, so the answer must have three. The raw quotient $1.68804...$ rounds to $1.69$ g cm$^{-3}$. Reporting more digits would falsely imply a precision the $2.51$ cm$^{3}$ measurement does not support. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q51 The sum of $436.32$ g, $227.2$ g and $0.301$ g, reported to the correct number of decimal places, is: hard
Step solution + source
For addition the result retains as many decimal places as the least precise term. The value $227.2$ g has only one decimal place, so although the arithmetic sum is $663.821$ g, it must be rounded to one decimal, giving $663.8$ g. The addition-subtraction rule is stated in decimal places, not significant figures. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q52 The subtraction $12.9$ g $-\,7.06$ g, reported to the correct precision, gives: hard
Step solution + source
In subtraction the result keeps as many decimal places as the term with the fewest. Here $12.9$ g has one decimal place, so even though $12.9-7.06=5.84$, the answer is rounded to one decimal place, giving $5.8$ g. Subtraction can thus reduce the number of significant figures relative to the inputs. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q53 Each side of a cube is measured as $7.203$ m. Its total surface area, to the correct number of significant figures, is: advanced
Step solution + source
The surface area is $6(7.203)^{2}=311.299254$ m$^{2}$. Since the measured side has four significant figures, the area must be rounded to four, giving $311.3$ m$^{2}$. Keeping extra digits would overstate precision, while rounding to only three would discard information genuinely present in the four-figure measurement. 🔉⇢
Source: NCERT Example 1.1
Q54 A substance of mass $5.74$ g occupies $1.2$ cm$^{3}$. Its density to the correct number of significant figures is: advanced
Step solution + source
The mass has three significant figures but the volume has only two, so the quotient must be reported to two significant figures. The raw division $5.74/1.2=4.783...$ therefore rounds to $4.8$ g cm$^{-3}$. The least precise factor, the volume, controls the final precision of the multiplication-or-division result. 🔉⇢
Source: NCERT Example 1.2
Q55 The parallax method is most suitable for measuring: easy
Step solution + source
Parallax uses the apparent shift of a distant object viewed from two separated points to infer its distance, and it excels for astronomical distances like those to planets and nearby stars. It is unsuited to microscopic thicknesses or atomic diameters, which need indirect optical methods, and it measures distance, not mass. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q56 One angstrom, a convenient unit of length on the atomic scale, equals: easy
Step solution + source
The angstrom, symbol Å, is defined as $1\ \text{Å}=10^{-10}$ m and is handy for expressing atomic sizes; a hydrogen atom is about $0.5$ Å across. The value $10^{-9}$ m is a nanometre and $10^{-15}$ m is a femtometre, so those are the wrong magnitudes for the angstrom. 🔉⇢
Source: NCERT Exercise
Q57 In the parallax method, if $b$ is the baseline and $\theta$ the small parallax angle (in radians), the distance $D$ of the object is: medium
Step solution + source
For a small angle subtended by the baseline $b$ at the far object, $\theta\approx b/D$ with $\theta$ in radians, so $D=b/\theta$. A smaller parallax angle therefore means a larger distance. The product $b\theta$ and the inverse $\theta/b$ have the wrong dimensions and behaviour to represent a distance. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q58 A microscope of magnification $100$ shows the average width of a hair as $3.5$ mm. The actual thickness of the hair is about: medium
Step solution + source
Magnification multiplies the apparent size, so the true size is the observed size divided by the magnification: $3.5\ \text{mm}/100=0.035$ mm. This indirect method lets a metre-scale-limited eye estimate microscopic thicknesses. Multiplying rather than dividing would give an absurdly large hair, so the observed width must be scaled down. 🔉⇢
Source: NCERT Exercise
Q59 To estimate the diameter of a thin thread with only a metre scale, the best practical method is to: medium
Step solution + source
A single thread diameter is far below the least count of a metre scale, so it cannot be read directly. Winding $n$ tight, non-overlapping turns spreads the diameter across a measurable width $w$; then the diameter is $w/n$. Averaging over many turns effectively multiplies the small quantity to a scale the instrument can resolve. 🔉⇢
Source: NCERT Exercise
Q60 The distance of a planet is found by parallax with baseline $b=3.0\times10^{11}$ m and parallax angle $\theta=2.0\times10^{-5}$ rad. The distance is: hard
Step solution + source
Using the parallax relation $D=b/\theta=\dfrac{3.0\times10^{11}}{2.0\times10^{-5}}=1.5\times10^{16}$ m. Dividing the baseline by a very small angle produces an enormously large distance, exactly as expected for astronomical scales. The distractor $6.0\times10^{6}$ comes from multiplying instead of dividing, which wrongly reverses the physical dependence of distance on the parallax angle. 🔉⇢
Source: JEE Main (pattern)
Q61 In the oleic-acid film method, a drop of volume $V$ spreads into a nearly monomolecular circular film of area $A$. The molecular size (film thickness) is estimated as: hard
Step solution + source
If the film is one molecule thick, its volume equals area times thickness, $V=A t$, so $t=V/A$. This indirect method estimates a molecular diameter far below any scale's least count by exploiting a measurable macroscopic volume and area. The ratio $A/V$ and the product $VA$ have the wrong dimensions for a length. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q62 A house occupies $1.75$ cm$^{2}$ on a slide and $1.55$ m$^{2}$ on the projected screen. The linear magnification of the projector is approximately (note $1.55$ m$^{2}=1.55\times10^{4}$ cm$^{2}$): advanced
Step solution + source
Areal magnification is the ratio of areas: $\dfrac{1.55\times10^{4}}{1.75}\approx8857$. Linear magnification is the square root of the areal magnification because area scales as the square of length: $\sqrt{8857}\approx94$. Reporting $8857$ confuses areal with linear magnification, the classic error this problem is designed to expose. 🔉⇢
Source: NCERT Exercise
Q63 The size of a hydrogen atom is about $0.5$ Å. The total atomic volume of one mole of hydrogen atoms (take $N_A\approx6\times10^{23}$, atom volume $\tfrac{4}{3}\pi r^{3}$) is of order: advanced
Step solution + source
With $r=0.5\ \text{Å}=0.5\times10^{-10}$ m, the volume of one atom is $\tfrac{4}{3}\pi r^{3}\approx5\times10^{-31}$ m$^{3}$. Multiplying by the Avogadro number $N_A\approx6\times10^{23}$ gives roughly $3\times10^{-7}$ m$^{3}$ for a whole mole of atoms. This is vastly smaller than the $22.4\times10^{-3}$ m$^{3}$ molar volume of a gas, which vividly illustrates how mostly empty a gas really is. 🔉⇢
Source: NCERT Exercise
Q64 The SI base unit used to measure time intervals is the: easy
Step solution + source
Time is a base quantity whose SI unit is the second, defined through the caesium hyperfine transition frequency. The hertz is a derived unit of frequency equal to inverse seconds, while the minute and the day are non-SI units retained for convenience. Only the second is the SI base unit of time. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q65 The masses of planets and stars are most appropriately determined using: easy
Step solution + source
Enormous masses like those of planets cannot be placed on any balance, so they are inferred from gravitational effects, for example from the orbital motion of satellites or moons via Newton's law of gravitation. Balances and callipers are limited to laboratory-scale masses and lengths and are wholly unsuited to astronomical bodies. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q66 On the atomic scale, mass is conveniently expressed in the: easy
Step solution + source
The unified atomic mass unit, defined as one twelfth of the mass of a carbon-12 atom, is convenient for atomic and molecular masses because the kilogram is far too large a unit at that scale. The newton is a unit of force and the candela is luminous intensity, so neither measures mass at all. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q67 A laboratory balance that compares an unknown mass with standard masses measures: medium
Step solution + source
A common beam balance compares the gravitational pull on the unknown with that on standard masses; when they balance, the gravitational masses are equal. This differs from measuring inertial mass through acceleration under a known force. The balance reads mass, not force in newtons, and it tells nothing directly about the object's volume. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q68 Extremely accurate modern time standards are maintained using: medium
Step solution + source
The second is realised by caesium atomic clocks, which count the fixed hyperfine transition frequency of caesium-133. Pendulum and quartz devices, though useful, drift far more and cannot match this precision, and a sundial is only a coarse indicator. Atomic clocks provide the internationally accepted, highly reproducible time standard. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q69 To measure the period of a simple pendulum accurately, one should time: medium
Step solution + source
Timing one oscillation gives a large relative error because human reaction time is comparable to the period. Timing, say, twenty oscillations and dividing by twenty spreads that fixed reaction-time error over many periods, reducing the random error in the mean period. Averaging over many events is a standard way to improve time measurements. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q70 A box of mass $2.30$ kg has two gold pieces of masses $20.15$ g and $20.17$ g added. The total mass to correct significant figures is: hard
Step solution + source
Convert the pieces: $20.15\ \text{g}+20.17\ \text{g}=40.32\ \text{g}=0.04032$ kg. Adding to $2.30$ kg gives $2.34032$ kg. In addition the result keeps the least number of decimal places, and $2.30$ kg has two decimals, so the total is rounded to $2.34$ kg, matching the precision of the least precise term. 🔉⇢
Source: NCERT Exercise
Q71 For the same two gold pieces of masses $20.15$ g and $20.17$ g, the difference in their masses to correct significant figures is: advanced
Step solution + source
Both masses have two decimal places, so $20.17-20.15=0.02$ g, keeping two decimal places. Although each original mass has four significant figures, the subtraction leaves a result with only one significant figure, showing how subtraction of nearly equal quantities can drastically reduce the significant figures and hence the relative precision. 🔉⇢
Source: NCERT Exercise
Q72 A time interval is measured five times as $2.42$, $2.44$, $2.41$, $2.45$, $2.43$ s. The mean period reported with its mean absolute error is: advanced
Step solution + source
The mean is $(2.42+2.44+2.41+2.45+2.43)/5=2.43$ s. The individual deviations are $0.01,0.01,0.02,0.02$ and $0.00$ s; their mean is $0.06/5=0.012$ s, which rounds to $0.01$ s to remain consistent with two decimal places. Hence the period is reported as $2.43\pm0.01$ s, that is the mean value together with its mean absolute error. 🔉⇢
Source: JEE Main (pattern)
Q73 Systematic errors are best described as errors that: easy
Step solution + source
Systematic errors have a definite sign and cause: instrument miscalibration, imperfect technique or personal bias push readings consistently one way. Because their direction is known, they can often be identified and corrected. This contrasts with random errors, which scatter unpredictably about the true value in both directions. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q74 A zero error in a screw gauge or vernier calliper is an example of a: easy
Step solution + source
A zero error is a fixed instrumental offset present before any measurement, so it biases every reading by the same amount in the same direction, which is the hallmark of a systematic error. Once identified it is simply subtracted from each reading. It is not random, since it does not fluctuate between measurements. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q75 Random errors in a set of repeated measurements can be reduced by: medium
Step solution + source
Random errors scatter symmetrically about the true value, so averaging many independent readings lets positive and negative deviations cancel, and the mean approaches the true value. Using a single reading retains the full random error, aggressive rounding discards information, and changing units does nothing to the underlying scatter of the data. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q76 An error arising from a wrongly calibrated instrument scale is classified as: medium
Step solution + source
Faulty calibration shifts every reading by a predictable amount, so it is a systematic instrumental error that can, in principle, be corrected by recalibration or by applying a known correction. Random errors, by contrast, come from unpredictable fluctuations, and personal errors arise from the observer's habits rather than the instrument itself. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q77 The least-count error of an instrument is: medium
Step solution + source
The least count is the smallest value an instrument can resolve, and the uncertainty of the last recorded digit is the least-count error. It is tied to the instrument's precision and cannot be removed merely by taking more readings, unlike purely random error. Using a finer instrument with a smaller least count reduces it. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q78 Which of the following errors is most effectively reduced by repeating the measurement many times? hard
Step solution + source
Only random error, which fluctuates in sign from trial to trial, diminishes on averaging because opposite deviations cancel. A constant zero error, a calibration offset and a one-directional parallax bias are all systematic: they shift every reading the same way, so repeating and averaging leaves them completely unchanged. Systematic errors need correction, not repetition. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q79 An observer who consistently reads a scale from an angle, introducing a one-sided parallax bias, is committing a: hard
Step solution + source
A habitual viewing angle produces a parallax shift that is always in the same direction, so it is a personal error, a subclass of systematic error caused by the observer's technique. It can be removed by viewing the scale perpendicularly. Because it is one-directional and reproducible, averaging repeated readings will not eliminate it. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q80 Five readings of a length are $5.10, 5.12, 5.11, 5.13, 5.14$ cm, while the true value is $5.00$ cm. This pattern most strongly indicates: advanced
Step solution + source
All five readings lie well above the true value $5.00$ cm, with a mean of about $5.12$ cm, so they are biased consistently upward by roughly $0.12$ cm. That one-directional offset is a systematic error, while the small $\pm0.02$ cm spread about the mean reflects random scatter. Pure random error would straddle the true value. 🔉⇢
Source: JEE Main (pattern)
Q81 Regarding accuracy and precision, a set of measurements that is precise but not accurate is characterised by readings that are: advanced
Step solution + source
Precision refers to reproducibility, the closeness of repeated readings to one another, whereas accuracy refers to closeness to the true value. Readings tightly clustered but offset from the true value are precise yet inaccurate, a signature of a systematic error. High accuracy with low precision would instead give scattered readings averaging near the truth. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q82 The absolute error in a single measurement is defined as: easy
Step solution + source
Absolute error $\Delta a$ is the magnitude of the difference between an individual measured value and the true or best (mean) value: $\Delta a=|a_{\text{measured}}-a_{\text{mean}}|$. It carries the same units as the quantity. Dividing by the true value gives relative error, not absolute error, and the least count is a separate instrumental notion. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q83 The relative error of a measurement is given by: easy
Step solution + source
Relative error is the ratio of the mean absolute error to the mean value, $\Delta\bar a/\bar a$, and being a ratio of like quantities it is dimensionless. It expresses the uncertainty as a fraction of the quantity itself. The product or the inverted ratio would not represent a fractional uncertainty and are dimensionally meaningless here. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q84 The percentage error is obtained from the relative error by: medium
Step solution + source
Percentage error is simply the relative error expressed as a percentage: $\left(\dfrac{\Delta\bar a}{\bar a}\right)\times100\%$. Multiplying the dimensionless fraction by one hundred rescales it to a per-hundred basis. Dividing by one hundred or subtracting would give a physically meaningless quantity, and a square root has no place in this simple definition. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q85 A mass is measured as $1.02$ g with an absolute error of $\pm0.01$ g. Its relative (percentage) error is about: medium
Step solution + source
The percentage error is $\dfrac{0.01}{1.02}\times100\%\approx0.98\%$, which rounds to about $\pm1\%$. This shows how a small absolute error becomes a modest fractional error for a small quantity. Contrast this with the same $\pm0.01$ g on a $9.89$ g mass, where the percentage error is only about one tenth as large. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q86 For repeated readings, the mean absolute error $\Delta\bar a$ is calculated as: medium
Step solution + source
The mean absolute error is the average of the absolute deviations $|a_i-\bar a|$ over all readings, $\Delta\bar a=\tfrac1n\sum|a_i-\bar a|$. Taking only the largest deviation overstates the typical error, the plain mean of readings gives the best value not the error, and the range (max minus min) is a cruder spread measure. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q87 Four readings of a length are $2.63, 2.65, 2.64, 2.62$ cm. The mean value and mean absolute error are: hard
Step solution + source
The mean value is $(2.63+2.65+2.64+2.62)/4=2.635$ cm. The individual deviations from this mean are $0.005,0.015,0.005$ and $0.015$ cm; their average is $0.04/4=0.01$ cm. Hence the result is reported as $2.635\pm0.01$ cm. The mean absolute error is the average of the individual absolute deviations, and it is not the same as the full range of the data. 🔉⇢
Source: JEE Main (pattern)
Q88 A mass of $9.89$ g has an absolute error of $\pm0.01$ g. Its percentage error is about: hard
Step solution + source
The percentage error is $\dfrac{0.01}{9.89}\times100\%\approx0.10\%$. Because the same absolute error is now shared over a much larger quantity, the fractional uncertainty is about ten times smaller than for the $1.02$ g case. This demonstrates that relative error depends on the size of the measured quantity, not just the absolute error. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q89 Two lengths are measured as $L_1=25.0\pm0.1$ cm and $L_2=5.0\pm0.1$ cm. Which has the smaller percentage error, and roughly how small? advanced
Step solution + source
For $L_1$, the percentage error is $\dfrac{0.1}{25.0}\times100\%=0.4\%$; for $L_2$ it is $\dfrac{0.1}{5.0}\times100\%=2\%$. The same absolute uncertainty is a smaller fraction of the larger length, so $L_1$ is measured with the smaller relative error. This is why measuring larger quantities, where possible, improves fractional precision. 🔉⇢
Source: JEE Main (pattern)
Q90 A quantity is recorded as $g=9.80\pm0.05$ m s$^{-2}$. The best statement of its percentage uncertainty is: advanced
Step solution + source
The percentage error is $\dfrac{0.05}{9.80}\times100\%\approx0.51\%$, so the result may be written as $g=9.80$ m s$^{-2}\pm0.5\%$. Expressing an uncertainty as a percentage is convenient when the quantity later enters a product or quotient, where percentage errors combine additively rather than the absolute errors. 🔉⇢
Source: JEE Main (pattern)
Q91 When two quantities are added or subtracted, the error in the result is obtained by: easy
Step solution + source
For a sum or difference $Z=A\pm B$, the maximum absolute error is the sum of the individual absolute errors, $\Delta Z=\Delta A+\Delta B$. Relative or percentage errors combine additively only for products and quotients, not for addition or subtraction. This is why subtracting nearly equal quantities can badly inflate the relative error of the result. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q92 When two quantities are multiplied or divided, the maximum relative error in the result equals: easy
Step solution + source
For $Z=AB$ or $Z=A/B$, the maximum relative error is $\dfrac{\Delta Z}{Z}=\dfrac{\Delta A}{A}+\dfrac{\Delta B}{B}$, so fractional errors add. Absolute errors add only for sums and differences. Multiplying the relative errors would give a negligible second-order term, which is why the correct rule uses their sum instead. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q93 If a quantity $A$ is raised to the power $n$, the relative error in $A^{n}$ is: medium
Step solution + source
For $Z=A^{n}$, the fractional error scales with the power: $\dfrac{\Delta Z}{Z}=|n|\dfrac{\Delta A}{A}$. A higher power magnifies the relative error proportionally, so quantities entering with large exponents dominate the total uncertainty. This is why, for example, a radius measured to one percent gives a volume uncertain by about three percent. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q94 For $Z=A\times B$, if $A$ has $2\%$ error and $B$ has $3\%$ error, the maximum percentage error in $Z$ is: medium
Step solution + source
For a product the percentage errors add: $\dfrac{\Delta Z}{Z}\times100\%=2\%+3\%=5\%$. Multiplying the percentages would wrongly give $6\%$, and taking a difference or an average has no physical justification. The additive rule gives the maximum possible error, assuming the individual errors could reinforce each other in the worst case. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q95 A rectangular sheet has length $16.2\pm0.1$ cm and breadth $10.1\pm0.1$ cm. The percentage error in its area is about: medium
Step solution + source
The percentage error in length is $\dfrac{0.1}{16.2}\times100\%\approx0.6\%$ and in breadth $\dfrac{0.1}{10.1}\times100\%\approx1.0\%$. For the area, a product, these add to about $1.6\%$. Thus the area $163.62$ cm$^{2}$ carries an uncertainty of roughly $\pm2.6$ cm$^{2}$, and is reported consistent with this precision. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q96 For $Z=\dfrac{A^{2}B^{3}}{C}$, the maximum relative error $\dfrac{\Delta Z}{Z}$ equals: hard
Step solution + source
Each factor contributes its relative error multiplied by the magnitude of its exponent, and all contributions add regardless of whether the power is positive or negative. Hence $\dfrac{\Delta Z}{Z}=2\dfrac{\Delta A}{A}+3\dfrac{\Delta B}{B}+\dfrac{\Delta C}{C}$. Using a minus sign for $C$ is wrong because we seek the maximum error, where deviations add rather than cancel. 🔉⇢
Source: JEE Main (pattern)
Q97 The density of a sphere is found from mass ($2\%$ error) and radius ($1\%$ error) via $\rho=\dfrac{m}{\frac43\pi r^{3}}$. The maximum percentage error in $\rho$ is: hard
Step solution + source
Since $\rho\propto m\,r^{-3}$, the fractional errors combine as $\dfrac{\Delta\rho}{\rho}=\dfrac{\Delta m}{m}+3\dfrac{\Delta r}{r}=2\%+3\times1\%=5\%$. The radius enters with power three, so its one-percent error is tripled. The constant $\tfrac43\pi$ is exact and contributes no error, which is why only the mass and radius uncertainties matter. 🔉⇢
Source: JEE Main (pattern)
Q98 The value of $g$ from a pendulum is $g=\dfrac{4\pi^{2}l}{T^{2}}$. If $l$ is measured to $0.5\%$ and $T$ to $1\%$, the maximum percentage error in $g$ is: advanced
Step solution + source
Since $g\propto l\,T^{-2}$, the fractional error is $\dfrac{\Delta g}{g}=\dfrac{\Delta l}{l}+2\dfrac{\Delta T}{T}=0.5\%+2\times1\%=2.5\%$. The period enters squared, so its error is doubled and dominates. This is why, in a pendulum experiment, timing many oscillations to reduce the error in $T$ is more important than extra care with $l$. 🔉⇢
Source: JEE Advanced (pattern)
Q99 A resistance is found from $R=\dfrac{V}{I}$ with $V=100\pm5$ V and $I=10\pm0.2$ A. The percentage error in $R$ is: advanced
Step solution + source
The percentage error in voltage is $\dfrac{5}{100}\times100\%=5\%$ and in current $\dfrac{0.2}{10}\times100\%=2\%$. For the quotient $R=V/I$ these add: $\dfrac{\Delta R}{R}\times100\%=5\%+2\%=7\%$. Errors in a division combine by addition of relative errors, exactly as for a product, giving the maximum uncertainty of seven percent. 🔉⇢
Source: JEE Advanced (pattern)
Q100 The least count of a vernier calliper is defined as: easy
Step solution + source
The vernier least count equals one main-scale division (MSD) minus one vernier-scale division (VSD): $\text{L.C.}=1\ \text{MSD}-1\ \text{VSD}$. This tiny difference is the smallest length the instrument can resolve. Adding the divisions or taking the main-scale length has no meaning as a resolution, and the mere count of divisions is not itself a length. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q101 A common vernier calliper has a least count of: easy
Step solution + source
A standard vernier with a $1$ mm main-scale division and $10$ vernier divisions covering $9$ mm has least count $1\ \text{mm}-0.9\ \text{mm}=0.1$ mm, i.e. $0.01$ cm. This makes it about ten times finer than a plain millimetre scale. The value $0.01$ mm belongs to a typical screw gauge, not a vernier calliper. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q102 If $10$ vernier divisions coincide with $9$ main-scale divisions and one main-scale division is $1$ mm, the least count is: medium
Step solution + source
One vernier division is $\dfrac{9\ \text{mm}}{10}=0.9$ mm, so the least count is $1\ \text{MSD}-1\ \text{VSD}=1.0-0.9=0.1$ mm. Equivalently, L.C. equals one main-scale division divided by the number of vernier divisions, $1\ \text{mm}/10=0.1$ mm. This is the smallest length the calliper can read reliably. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q103 The total reading of a vernier calliper is computed as: medium
Step solution + source
The final reading is the main-scale reading just before the vernier zero, plus the vernier-scale contribution, which is the number of the vernier division that best coincides with a main-scale line multiplied by the least count. Multiplying the main-scale reading by the least count, or subtracting, does not correspond to how the two scales combine. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q104 A vernier scale has $20$ divisions coinciding with $19$ main-scale divisions, each main division being $1$ mm. The least count is: medium
Step solution + source
The least count is one main-scale division divided by the number of vernier divisions: $\dfrac{1\ \text{mm}}{20}=0.05$ mm. Equivalently one vernier division is $\dfrac{19}{20}=0.95$ mm, and $1.00-0.95=0.05$ mm. Increasing the number of vernier divisions to twenty makes this calliper twice as fine as the usual ten-division type. 🔉⇢
Source: JEE Main (pattern)
Q105 A vernier calliper reads a positive zero error when the jaws are closed and the vernier zero lies to the right of the main-scale zero. This zero error should be: hard
Step solution + source
A positive zero error means the instrument reads too high by a fixed amount even with the jaws shut, so it is a systematic offset that must be subtracted from each observed reading to obtain the true value. Adding it would double the error. Because it is constant and one-directional, it cannot be reduced by averaging repeated readings. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q106 A vernier calliper (L.C. $=0.01$ cm) shows a main-scale reading of $1.2$ cm and its $4$th vernier division coincides. With no zero error, the measured length is: hard
Step solution + source
The reading is main-scale reading plus vernier contribution: $1.2\ \text{cm}+(4\times0.01\ \text{cm})=1.2+0.04=1.24$ cm. The coinciding division number times the least count gives the fractional part beyond the main scale. Writing $1.204$ misplaces the vernier contribution by a decimal place, and $4.8$ cm wrongly multiplies unrelated quantities. 🔉⇢
Source: JEE Main (pattern)
Q107 A vernier has a least count of $0.01$ cm and a positive zero error of $0.03$ cm. If an observed reading is $3.56$ cm, the corrected length is: advanced
Step solution + source
A positive zero error is subtracted from the observed reading: $3.56\ \text{cm}-0.03\ \text{cm}=3.53$ cm. This removes the fixed systematic offset that the instrument adds even when the jaws are closed. Adding the zero error would give $3.59$ cm, which double-counts the bias, while ignoring it leaves the reading systematically too large. 🔉⇢
Source: JEE Advanced (pattern)
Q108 In a vernier, $50$ vernier divisions coincide with $49$ main-scale divisions, and one main-scale division equals $0.5$ mm. The least count is: advanced
Step solution + source
The least count equals one main-scale division divided by the number of vernier divisions: $\dfrac{0.5\ \text{mm}}{50}=0.01$ mm. Equivalently, one vernier division is $\dfrac{49\times0.5}{50}=0.49$ mm, and $0.50-0.49=0.01$ mm. Such a fine vernier rivals a screw gauge in resolution, reading to one hundredth of a millimetre. 🔉⇢
Source: JEE Advanced (pattern)
Q109 The least count of a screw gauge is given by: easy
Step solution + source
The screw-gauge least count is $\text{L.C.}=\dfrac{\text{pitch}}{\text{number of circular-scale divisions}}$. The pitch is the linear advance per full rotation, and dividing it among the circular divisions gives the smallest measurable length. Multiplying, or dividing the divisions by the pitch, gives the wrong dimensions, and the pitch alone is far coarser than the true least count. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q110 The pitch of a screw gauge is defined as: easy
Step solution + source
The pitch is the axial (linear) distance moved by the spindle for one full turn of the thimble; for a typical screw gauge this is $1.0$ mm or $0.5$ mm. It is a property of the screw thread. The number of circular divisions is a separate quantity, and dividing the pitch by that number yields the least count. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q111 A screw gauge has a pitch of $1$ mm and $100$ divisions on its circular scale. Its least count is: medium
Step solution + source
The least count is $\dfrac{\text{pitch}}{\text{circular divisions}}=\dfrac{1\ \text{mm}}{100}=0.01$ mm. Thus each circular-scale division corresponds to a spindle advance of one hundredth of a millimetre, giving the screw gauge its high precision. Dividing by ten or by a thousand would misplace the decimal point of this standard result. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q112 The total reading of a screw gauge is: medium
Step solution + source
The reading is the linear (pitch) scale value plus the circular-scale contribution, which is the coinciding circular division multiplied by the least count. This combines the coarse axial advance with the fine rotational fraction. Multiplying the circular reading directly by the pitch, or subtracting the least count, does not reflect how the two scales are read together. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q113 A screw gauge has a pitch of $0.5$ mm and $50$ divisions on the circular scale. Its least count is: medium
Step solution + source
The least count is $\dfrac{0.5\ \text{mm}}{50}=0.01$ mm. Even though the pitch is half a millimetre, distributing it over fifty circular divisions still yields a resolution of one hundredth of a millimetre. This shows that both a smaller pitch and more circular divisions increase precision, and here they combine to the common $0.01$ mm least count. 🔉⇢
Source: JEE Main (pattern)
Q114 A screw gauge shows a positive zero error, its circular-scale zero lying below the reference line when fully closed. To correct a reading, this zero error must be: hard
Step solution + source
A positive zero error means the gauge reads too high by a fixed amount when the studs just touch, so it is a systematic offset that must be subtracted from every observed reading. Adding it would compound the bias. Since it is constant and one-directional, repeating and averaging readings cannot remove it, unlike random error. 🔉⇢
Source: NCERT Class XI Physics — Units and Measurement
Q115 A screw gauge (pitch $1$ mm, $100$ circular divisions) shows a main-scale reading of $2$ mm and a circular reading of $45$, with no zero error. The measured diameter is: hard
Step solution + source
The least count is $1/100=0.01$ mm. The reading is $2\ \text{mm}+(45\times0.01\ \text{mm})=2+0.45=2.45$ mm. The circular reading times the least count supplies the fractional part beyond the millimetre scale. The value $2.045$ mm misplaces a decimal, and treating millimetres as centimetres changes the magnitude entirely. 🔉⇢
Source: JEE Main (pattern)
Q116 Regarding increasing the accuracy of a screw gauge by adding more divisions on the circular scale, the correct statement is: advanced
Step solution + source
Although more circular divisions lower the least count and improve nominal resolution, accuracy cannot be improved without bound: backlash, imperfect screw threads, temperature effects and the observer's ability to read closely spaced marks all set practical limits. Beyond a point, extra divisions add apparent precision without real gains in the trustworthiness of the measurement. 🔉⇢
Source: NCERT Exercise
Q117 A screw gauge (pitch $1$ mm, $100$ divisions) has a negative zero error of $3$ divisions. If the observed reading is $4.20$ mm, the corrected diameter is: advanced
Step solution + source
A negative zero error of $3$ divisions equals $3\times0.01=0.03$ mm, and a negative zero error is added to the observed reading: $4.20\ \text{mm}+0.03\ \text{mm}=4.23$ mm. The gauge was reading too low, so the correction restores the true value. Subtracting instead would wrongly worsen the underestimate to $4.17$ mm. 🔉⇢
Source: JEE Advanced (pattern)
⏱️ Mock Test 30 Q · 60 min · +4 correct, -1 incorrect (JEE Main pattern)
Rules: ['Attempt all questions within the 60-minute limit; there is no sectional time restriction.', 'Negative marking applies, so avoid random guessing when uncertain about a question.', 'Report every numerical answer with the correct number of significant figures.']
--:--
🎬 Video Lectures clip-indexed · NPTEL / IIT / MIT
MIT OCW / Walter Lewin, 3Blue1Brown, Khan, Veritasium. Each video is broken into clips — click a clip to play just that segment.
मात्रक और मापन (भाग 1)🔉⇢
NCERT - PM eVidya Class 11
👁 Observe: PM eVidya का NCERT-आधारित पाठ: मात्रक, मूल एवं व्युत्पन्न मात्रक और SI पद्धति का परिचय।
📚 Teaches: SI पद्धति, मूल और व्युत्पन्न मात्रक — हिंदी में NCERT स्तर पर।
📑 Clips (3)
0:00–12:30अध्याय परिचय एवं मापन की भूमिकामात्रक एवं मापन अध्याय का परिचय, मापन की भूमिका तथा मापन में सदैव विद्यमान अनिश्चितता की चर्चा।general
12:30–27:58यथार्थता एवं परिशुद्धतायथार्थता (accuracy) और परिशुद्धता (precision) का अंतर घड़ी तथा द्रव्यमान के उदाहरणों से समझाया गया है।errors-in-measurement
27:58–45:32त्रुटियों के प्रकारक्रमबद्ध (systematic) एवं यादृच्छिक (random) त्रुटियों के प्रकार तथा उन्हें न्यूनतम करने के उपाय।errors-types
मात्रक और मापन (भाग 2)🔉⇢
NCERT - PM eVidya Class 11
👁 Observe: भौतिक राशियों की विमाएँ और विमीय सूत्र, तथा विमीय समांगता का सिद्धांत समझाया गया है।
📚 Teaches: विमाएँ, विमीय सूत्र और विमीय विश्लेषण — हिंदी माध्यम।
📑 Clips (3)
0:00–12:50निरपेक्ष, आपेक्षिक एवं प्रतिशत त्रुटिनिरपेक्ष, आपेक्षिक एवं प्रतिशत त्रुटि की परिभाषा तथा अमीटर पाठ्यांकों के उदाहरण से गणना।absolute-relative-percentage-error
12:50–27:58त्रुटियों का संयोजनयोग, गुणनफल एवं घातों में त्रुटियों के संयोजन के नियम तथा दोलक से g निकालने का उदाहरण।combination-of-errors
27:58–47:13सार्थक अंक एवं पूर्णांकनसार्थक अंकों के नियम, वैज्ञानिक संकेतन तथा पूर्णांकन (rounding) के नियमों की चर्चा।significant-figures
मात्रक और मापन (भाग 3)🔉⇢
NCERT - PM eVidya Class 11
👁 Observe: सार्थक अंक, मापन में त्रुटियाँ और त्रुटियों का संचरण NCERT उदाहरणों के साथ।
📚 Teaches: सार्थक अंक और मापन की त्रुटियाँ — हिंदी में।
📑 Clips (3)
0:00–10:10भौतिक राशियाँ, समतलीय एवं घन कोणभौतिक राशियों का परिचय तथा समतलीय कोण (radian) एवं घन कोण (steradian) की संकल्पना।unit-of-physical-quantities
10:10–22:58लंबाई, द्रव्यमान एवं समय के परिसरलंबन विधि (parallax) सहित लंबाई, द्रव्यमान एवं समय के परिसर तथा कोटि (order of magnitude) की चर्चा।measurement-of-length
22:58–36:21SI मूल मात्रक एवं 2019 पुनर्परिभाषासात SI मूल मात्रक तथा 2019 में नियतांकों (Planck constant आदि) पर आधारित पुनर्परिभाषा।si-base-units
Dimensional Analysis🔉⇢
MIT OpenCourseWare
👁 Observe: An MIT lecture segment reasons out a physical relation purely from the dimensions of the quantities involved.
📚 Teaches: Using dimensional homogeneity to guess the form of a physical law up to a dimensionless constant.
📑 Clips (3)
0:02–2:35Mars Curiosity landing motivates dimensional analysisUses the parachute-slowed Mars descent to introduce why dimensional reasoning matters.dimensional-analysis
2:35–7:40Writing dimensions of velocity and energy and choosing relevant variablesDenotes dimensions in square brackets and identifies the quantities a terminal-velocity model depends on.dimensions-of-physical-quantities
7:40–18:06Forming dimensionless variables to recover the terminal-velocity lawConstructs dimensionless groups and fits data to find the functional form of the descent speed.dimensional-analysis
SI Base Units and Derived Units🔉⇢
The Organic Chemistry Tutor
👁 Observe: The narrator lists the seven SI base quantities and builds familiar derived units (newton, joule, pascal) from them.
📚 Teaches: The seven SI base units and how every derived unit is a product of powers of the base units.
📑 Clips (3)
0:02–5:02The seven SI base units: metre, kilogram, second, kelvin, mole, ampere, candelaIntroduces each base quantity with its standard SI unit and clarifies that a light-year measures distance, not time.si-base-units
5:02–12:39Deriving units for velocity, acceleration, force and volume from base unitsShows how derived units like m/s^2 and the newton are built by combining base units through defining equations.unit-of-physical-quantities
12:39–22:40Derived units for energy, power and frequency (joule, watt, hertz)Links work, energy and power to their joule and watt units and defines frequency in hertz as inverse seconds.unit-of-physical-quantities
The 7 Base SI Units🔉⇢
Physics Lab
👁 Observe: A short clip naming each base quantity with its unit and symbol — metre, kilogram, second, ampere, kelvin, mole, candela.
📚 Teaches: The one-to-one correspondence between the seven base physical quantities and their SI units and symbols.
📑 Clips (1)
0:00–1:23Quick roll-call of the seven base SI units and their symbolsLists length, mass, time, temperature, amount, current and luminous intensity with their standard symbols.si-base-units
S.I. base units and derived units🔉⇢
Cowen Physics
👁 Observe: Cowen distinguishes base from derived units and shows how to express a derived unit in base units.
📚 Teaches: Base vs derived units and expressing quantities like force and energy in fundamental SI units.
📑 Clips (2)
0:00–2:33Distinguishing SI base units from derived unitsExplains which units are fundamental and which SI units to use in physics calculations.si-base-units
2:33–6:32Building derived units and unit-symbol capitalisation conventionsDerives the unit of force from F=ma and covers why unit names are lowercase while person-named symbols are capitalised.unit-of-physical-quantities
Scientific Notation and Dimensional Analysis🔉⇢
Professor Dave Explains
👁 Observe: Dave converts units by chaining conversion factors so unwanted units cancel, then writes the result in scientific notation.
📚 Teaches: Unit conversion by dimensional analysis and expressing measured values in scientific notation.
📑 Clips (2)
0:00–2:31Scientific notation for very large and very small numbersShows how powers of ten compactly express magnitudes in a base-10 counting system.units-prefixes-notation
2:31–5:13Metric prefixes and unit conversion by dimensional analysisIntroduces prefixes such as nano and converts metres to nanometres using conversion factors.dimensional-analysis
Dimensional Analysis🔉⇢
The Organic Chemistry Tutor
👁 Observe: Worked unit-conversion problems where the narrator cancels units step by step to reach the target unit.
📚 Teaches: The factor-label (unit-cancellation) method for converting between systems of units.
📑 Clips (3)
0:00–5:04Chained unit conversion: seconds in a year and mph to m/sDemonstrates cancelling units through successive conversion factors to change compound units.dimensional-analysis
5:04–11:47Converting density units from kg/m^3 to g/mLHandles cubed length units while converting the density of aluminium into g/mL.dimensional-analysis
11:47–15:45Converting squared units (area) and solving a rate problemSquares conversion factors to convert square inches to square feet and works a reading-rate problem.dimensional-analysis
Using Dimensional Analysis to Find Units of a Variable🔉⇢
The Organic Chemistry Tutor
👁 Observe: Given an equation, the narrator solves for the unknown quantity's units by balancing dimensions on both sides.
📚 Teaches: Finding the SI unit or dimensional formula of an unknown constant from an equation.
📑 Clips (3)
0:01–5:06Finding the units of the gravitational constant G from F=Gm1m2/r^2Rearranges Newton's law of gravitation to solve for the units of G.dimensional-formulae
5:06–7:40Units of E=mc^2 and the gas constant RDerives the joule for energy and the units of R from familiar equations.dimensional-formulae
7:40–12:20Units of the Coulomb constant k and rate constantsIsolates k in Coulomb's law and simplifies units for reaction-rate expressions.dimensional-formulae
Measurement and Significant Figures🔉⇢
Professor Dave Explains
👁 Observe: Dave explains why measurements carry uncertainty and states the rules for counting significant figures.
📚 Teaches: The meaning of significant figures and the rules for identifying them in a measured value.
📑 Clips (1)
0:00–3:32Why we measure and how to round to significant figuresMotivates units of measurement and states the round-up-at-five rule for reporting figures.significant-figures
Significant Figures - A Fast Review🔉⇢
The Organic Chemistry Tutor
👁 Observe: A fast run through the sig-fig counting rules with many quick examples including trailing zeros and leading zeros.
📚 Teaches: Counting significant figures, including the treatment of zeros in different positions.
📑 Clips (3)
0:01–5:03Counting significant figures: non-zero digits, captive and trailing zerosGives the rules for which digits count, including trailing zeros only when a decimal point is present.significant-figures
5:03–10:06Rounding products and quotients to the least significant-figure countShows multiplication and division results are rounded to the fewest significant figures of the inputs.rounding-and-arithmetic
10:06–15:07Rounding sums to the least precise decimal placeExplains that addition keeps the fewest decimal places, using a vertical-line trick.rounding-and-arithmetic
Unit Conversion & Significant Figures: Crash Course🔉⇢
CrashCourse
👁 Observe: Crash Course works conversions and then rounds answers to the correct number of significant figures.
📚 Teaches: Combining unit conversion with the significant-figure rules for multiplication and rounding.
📑 Clips (3)
0:01–5:03Units are arbitrary: the platinum kilogram and defining the secondExplores how the IPK defined the kilogram and how Earth's rotation defined the second.si-base-units
5:03–7:40Sanity-checking answers with dimensional reasoningApplies the 'does this make sense' test to a light-year-per-second calculation.dimensional-analysis
7:40–11:24Preserving precision: significant figures and scientific notationArgues digits beyond measured precision are meaningless and shows scientific-notation reporting.significant-figures
👁 Observe: The narrator applies the decimal-place rule for addition/subtraction and the sig-fig rule for multiplication/division.
📚 Teaches: The two distinct rounding rules: least decimal places for +/-, least significant figures for x/div.
📑 Clips (4)
0:01–7:30Counting significant figures across tricky zero casesWorks many examples of leading, captive and trailing zeros with and without decimal points.significant-figures
7:30–15:06Addition and subtraction: rounding to the least precise decimal placeAligns decimals and rounds sums and differences to the correct place value.rounding-and-arithmetic
17:38–27:45Multiplication and division: rounding to the fewest significant figuresRounds products and quotients to the smallest significant-figure count of the operands.rounding-and-arithmetic
32:18–45:23Combined operations and significant figures in scientific notationHandles mixed add/multiply problems and applies the rules to numbers in scientific notation.rounding-and-arithmetic
Random and systematic error explained🔉⇢
Fizzics Organisation
👁 Observe: The clip contrasts random scatter about a mean with a systematic shift, using a target-and-arrows analogy.
📚 Teaches: The difference between random and systematic errors and how each affects precision and accuracy.
📑 Clips (2)
0:00–2:33Random errors from human timing in a drop experimentShows how reaction time causes scatter that must be estimated and reduced.errors-types
2:33–3:30Systematic errors: forgetting the bowl mass and a non-zeroed balanceIllustrates avoidable one-directional errors such as a zero offset on a scale.errors-types
Percent Error Made Easy🔉⇢
The Organic Chemistry Tutor
👁 Observe: Worked examples computing percentage error from an experimental value and an accepted value.
📚 Teaches: Absolute, relative and percentage error and how to compute them from measurements.
📑 Clips (2)
0:01–2:35Percent error from measured versus accepted densityDefines percent error using an aluminium density measurement.absolute-relative-percentage-error
2:35–6:26Worked percent-error examples using the absolute valueApplies the formula to iron density, taking the absolute value of the deviation.absolute-relative-percentage-error
Uncertainty - Addition and Subtraction🔉⇢
The Organic Chemistry Tutor
👁 Observe: The narrator propagates uncertainty through sums and differences by adding absolute uncertainties.
📚 Teaches: Error propagation for sums and differences: absolute uncertainties add.
📑 Clips (2)
0:01–5:06Adding uncertainties: sum the absolute uncertaintiesShows that when adding measured values their absolute uncertainties add.combination-of-errors
5:06–11:24Subtracting uncertain values and the max/min range checkDemonstrates absolute uncertainties still add on subtraction, verified by extreme-value ranges.combination-of-errors
A Level Physics: All of Uncertainties🔉⇢
ZPhysics
👁 Observe: A full revision of uncertainty types and their propagation through products and powers.
📚 Teaches: Combining fractional uncertainties for products, quotients and powers of measured quantities.
📑 Clips (2)
0:00–5:01Absolute versus percentage uncertaintyDefines both forms of uncertainty and converts between them for a voltage reading.absolute-relative-percentage-error
5:01–11:10Combining uncertainties for sums, products and powersSummarises adding absolute uncertainties for sums, percentages for products, and multiplying by the power.combination-of-errors
How to Read a Metric Vernier Caliper🔉⇢
WeldNotes
👁 Observe: A close-up demonstration reading main-scale and vernier-scale divisions to take a precise measurement.
📚 Teaches: Reading a vernier caliper: main-scale reading plus the coinciding vernier division times the least count.
📑 Clips (2)
0:00–2:32Metric vernier caliper with a least count of 0.02 mmExplains how a 1/50 vernier scale gives a 0.02 mm resolution.least-count-vernier-calipers
2:32–7:02Reading a vernier caliper by finding the coinciding markLocates which vernier line best aligns with the main scale to read the measurement.vernier-calipers
Vernier Calliper the intuitive way🔉⇢
FloatHeadPhysics
👁 Observe: An intuitive derivation of why the vernier scale works, building the least-count idea from first principles.
📚 Teaches: The principle behind the vernier scale and how the least count arises from the MSD minus VSD.
📑 Clips (3)
0:00–5:00Reinventing the vernier caliper from first principlesBuilds the instrument's logic without formulas to show how it actually works.vernier-calipers
5:00–12:30Reading a caliper logically using spacing differencesComputes the fractional reading from main- and vernier-scale spacings with no formula.vernier-calipers
12:30–20:56Deriving least count and solving a JEE 10-VSD-equals-11-MSD scaleGeneralises the least-count idea to the unusual vernier scales seen in JEE problems.least-count-vernier-calipers
Screw Gauge - MeitY OLabs🔉⇢
amritacreate
👁 Observe: The Amrita/MeitY virtual-lab animation shows the pitch and circular scale of a screw gauge and takes a reading.
📚 Teaches: Least count of a screw gauge as pitch divided by number of circular-scale divisions (virtual lab).
📑 Clips (2)
0:00–2:32Screw gauge parts and the meaning of pitchIdentifies stud, thimble, ratchet and the pitch and circular scales, and how pitch is measured.screw-gauge
2:32–6:34Measuring wire, lead-shot and lamina dimensions with a screw gaugeUses the gauge to find diameters and compute the volume and area of sample objects.screw-gauge
How to Read a Metric Micrometer🔉⇢
WeldNotes
👁 Observe: A demonstration reading the sleeve (linear) and thimble (circular) scales of a metric micrometer.
📚 Teaches: Reading a micrometer screw gauge: linear-scale reading plus circular-scale reading times least count.
📑 Clips (2)
0:00–2:30Metric micrometer resolution of 0.01 mm and its partsIntroduces the micrometer and how turning the thimble opens the measuring faces.screw-gauge
2:30–6:15Reading barrel and thimble scales, including barely-showing marksCombines 0.5 mm barrel marks with 0.01 mm thimble marks and handles edge cases.least-count-screw-gauge
Micrometer Screw Gauge - Zero Error🔉⇢
myhometuition
👁 Observe: The clip shows positive and negative zero error on a screw gauge and how to correct the observed reading.
📚 Teaches: Identifying and correcting positive and negative zero error in a screw-gauge reading.
📑 Clips (2)
0:00–2:30Positive and negative zero error on a micrometer screw gaugeShows how the thimble-zero offset relative to the reference line defines zero error.screw-gauge-and-error-propagation
2:30–3:35Correcting a reading by subtracting the zero errorCombines main and thimble readings, then removes zero error to get the true value.screw-gauge-and-error-propagation
Class 11 Physics One Shot | मात्रक और मापन🔉⇢
MP Board 11th and 12th
👁 Observe: पूरे अध्याय का एक-शॉट हिंदी पुनरावलोकन — मात्रक, विमाएँ, सार्थक अंक और त्रुटियाँ।
📚 Teaches: अध्याय का सम्पूर्ण हिंदी पुनरावलोकन (one-shot).
📑 Clips (3)
0:00–30:05भौतिक राशियाँ एवं मूल-व्युत्पन्न मात्रकमूल एवं व्युत्पन्न राशियों, उनके मात्रकों तथा अनुपूरक कोणों का विस्तृत परिचय।unit-of-physical-quantities
30:05–53:31मात्रक पद्धतियाँ एवं उपसर्गFPS, CGS, MKS एवं SI मात्रक पद्धतियाँ तथा उपसर्गों (prefixes) की चर्चा।systems-of-units
53:31–70:13विमीय सूत्र एवं विमीय विश्लेषणविमीय सूत्र लिखना, विमीय विश्लेषण द्वारा सूत्र प्राप्त करना तथा इसकी सीमाएँ।dimensional-analysis
Significant figures🔉⇢
Khan Academy
👁 Observe: Sal identifies which digits in several measurements are significant and explains the reasoning for each.
📚 Teaches: Which digits count as significant and why, with Khan Academy's step-by-step reasoning.
📑 Clips (2)
0:00–2:33Why significant figures prevent over-stating precisionExplains that a computed result cannot be more precise than the measurements behind it.significant-figures
2:33–5:03Rules for leading, captive and trailing zerosClarifies that leading zeros never count and trailing zeros count only with a decimal point.significant-figures
💬 Doubt Solving Ask-any-doubt RAG + FAQ
🤖 Ask any doubt RAG · NCERT-grounded, cited
Answers are grounded in tier-1 material (NCERT / JEE PYQ) and cited — the AI will say so if a doubt isn't in the indexed syllabus.
Frequently-asked doubts
Why do we even need units? Is a number not enough?
A measurement compares a quantity against an internationally accepted reference standard, so the number alone is meaningless without saying which standard. Reporting a length as 'five' begs the question, five what: metres, centimetres or feet. The unit fixes the standard of comparison, and the number tells you how many of those standards fit. That is why every physical quantity is written as a number accompanied by a unit.
What is the difference between base units and derived units?
Base units belong to the fundamental quantities chosen as independent, such as the metre for length and the kilogram for mass. Derived units are built by combining base units according to a defining relation, like the newton being kg m s$^{-2}$. There are only seven base units in the SI, yet every other unit in physics is some product or ratio of them. The complete collection of base and derived units is called a system of units.
Why did the world switch to the SI system?
Earlier the CGS, FPS and MKS systems coexisted, so the same experiment could be reported in incompatible units and conversions were error-prone. The SI gave one internationally accepted scheme of symbols and units, and because it is decimal, conversions within it are simple. It was most recently revised in 2018 to tie base units to constants of nature. This makes measurements reproducible anywhere in the world.
Do I have to memorise the exact definitions of the metre and kilogram?
No. NCERT explicitly states that the precise numerical values in the base-unit definitions need not be remembered or asked in a test. They are given only to show the accuracy to which the standards are known. What matters is that you understand each base unit is now fixed through a constant of nature, such as the speed of light for the metre. The definitions are revised over time to allow greater precision.
What exactly are dimensions of a physical quantity?
Dimensions are the powers to which the base quantities must be raised to represent a given quantity, written in square brackets. For example, force has dimensions $[M L T^{-2}]$, meaning one power of mass, one of length and minus two of time. Dimensions describe the nature or quality of a quantity, not its magnitude. That is why velocity, average velocity and speed all share the dimension $[L T^{-1}]$.
What is the difference between a dimensional formula and a dimensional equation?
The dimensional formula is the expression showing which base quantities, and to what powers, make up a quantity, such as $[M^{0} L^{3} T^{0}]$ for volume. The dimensional equation is obtained by equating the physical quantity itself to that formula, for instance $[V] = [M^{0} L^{3} T^{0}]$. So the formula is the right-hand side and the equation is the full statement. They carry the same information in slightly different form.
How does the principle of homogeneity let me check an equation?
The principle says magnitudes can be added or subtracted only if they have the same dimensions, so every term in a valid equation must share one dimensional formula. To check an equation you reduce each term to base dimensions and confirm they all match, on both sides. If any term differs, the equation is definitely wrong. This is the fastest sanity check you can run on a formula you half-remember.
If an equation is dimensionally correct, is it definitely correct?
No, and this is the trap examiners love. Dimensional consistency is necessary but not sufficient: a dimensionally wrong equation must be wrong, but a dimensionally right one need not be exactly correct. Dimensions cannot see dimensionless factors, so they cannot distinguish $\tfrac{1}{2}mv^{2}$ from $\tfrac{3}{16}mv^{2}$. Both pass the dimensional test, yet only one is the true formula for kinetic energy.
Can dimensional analysis derive any formula I want?
Only relations of the product type, and only when the quantity depends on at most three other quantities. You assume the form $Q = k\,a^{x} b^{y} c^{z}$ and solve for the exponents by matching dimensions. The catch is that the dimensionless constant $k$ never comes out of this method. For the simple pendulum it correctly gives $T = k\sqrt{l/g}$, but the fact that $k = 2\pi$ must come from theory or experiment.
Why must the argument of $\sin$, $\log$ or $e^{x}$ be dimensionless?
These special functions are defined by power series in which their argument is added to its own higher powers, and you can only add quantities of the same dimension. If the argument had dimensions, terms like $x$ and $x^{2}$ could never be summed. So arguments of trigonometric, logarithmic and exponential functions must be pure numbers. This is a common way JEE tests whether a proposed expression is even sensible.
What are significant figures really telling me?
Significant figures are the reliably known digits plus the first uncertain digit of a measurement. Their count signals the precision of the measurement, which in turn depends on the least count of the instrument. Writing a period as $1.62$ s says the 1 and 6 are certain and the 2 is doubtful. Reporting more digits than are significant is misleading because it overstates the precision you actually achieved.
Does changing units change the number of significant figures?
No, and this is a key NCERT point. The quantity $2.308$ cm has four significant figures, and rewriting it as $0.02308$ m, $23.08$ mm or $23080$ mm keeps exactly those four. The location of the decimal point is irrelevant to the count. A mere change of units cannot change how precisely something was measured.
Are zeros significant or not? I keep getting confused.
Non-zero digits are always significant, and zeros between non-zero digits are always significant. Leading zeros to the left of the first non-zero digit in a number less than one are never significant. Trailing zeros are the tricky case: not significant in a number without a decimal point, but significant when a decimal point is present. So $0.06900$ has four significant figures while $12300$ (no decimal) has three.
Why is scientific notation recommended for reporting measurements?
Because it removes all ambiguity about trailing zeros. Once you write a value as $a \times 10^{b}$, every zero in the base number $a$ is significant, and the power of 10 plays no role in the count. So $4700$ mm, which looks like two significant figures, is unambiguously written $4.700 \times 10^{3}$ mm to show four. Scientific notation is therefore the ideal format for stating precision.
What is order of magnitude and how do I find it?
Order of magnitude is the exponent $b$ when a quantity is written as roughly $10^{b}$. You round the coefficient $a$ up to 10 if it exceeds 5 and down to 1 otherwise, then read off the power. The Earth's diameter $1.28 \times 10^{7}$ m is of order $10^{7}$, so its order of magnitude is 7. It is a fast way to compare quantities that differ enormously in scale.
In multiplication, how many significant figures should the answer keep?
The result of a multiplication or division keeps as many significant figures as the factor with the fewest. If mass 4.237 g (four figures) is divided by volume 2.51 cm cubed (three figures), the density is reported to three figures as $1.69$ g cm$^{-3}$. The extra digits your calculator shows are not meaningful. The least precise input always caps the precision of the output.
Why is the rule for addition different from the rule for multiplication?
In addition and subtraction the answer keeps as many decimal places as the term with the fewest decimal places, not significant figures. For $436.32 + 227.2 + 0.301$, the term $227.2$ has only one decimal place, so the sum $663.821$ is rounded to $663.8$ g. This is because uncertainty in a sum is governed by the coarsest decimal position present. Using the multiplication rule here would give a wrong precision.
How do I round off a number ending in exactly 5?
When the digit to be dropped is a lone 5, the convention is the even rule: raise the preceding digit by 1 if it is odd, and leave it unchanged if it is even. So $2.745$ rounds to $2.74$ because 4 is even, while $2.735$ rounds to $2.74$ because 3 is odd. This avoids a systematic bias that always-rounding-up would introduce. For any other dropped digit, you round up if it exceeds 5 and down if it is less.
Why keep an extra guard digit in the middle of a long calculation?
Because rounding at every intermediate step lets small rounding errors accumulate and distort the final answer. NCERT shows that the reciprocal of 9.58 rounded to three figures, then reciprocated again, drifts away from 9.58, but keeping one more digit recovers it. So you retain one digit beyond the significant figures through the working, and round only at the very end. This keeps the final result honest.
Do exact numbers like $2\pi$ or a counting factor have significant figures?
Exact numbers that are pure counts or defined multiplying factors have an infinite number of significant figures, so they never limit the precision of a result. The 2 in $s = 2\pi r$ can be treated as $2.0000$ as needed, and the number of oscillations $n$ in $T = t/n$ is exact. Similarly $\pi$ is known to as many figures as you like, and you use $3.14$ or $3.142$ as required. Only measured quantities constrain significant figures.
What is the difference between accuracy and precision?
Accuracy is how close a measurement is to the true value, while precision is how finely the measurement is resolved, which is set by the least count of the instrument. A clock that is fast by five minutes but reads to the second is precise but not accurate. Precision relates to significant figures and random scatter, whereas accuracy relates mainly to systematic error. JEE questions often hinge on keeping these two ideas separate.
What is the difference between systematic and random errors?
Systematic errors shift results consistently in one direction and come from causes like a faulty instrument, a zero offset or a flawed method, so they can often be identified and corrected. Random errors scatter results unpredictably from one trial to the next and are reduced by taking many readings and averaging. A systematic error affects accuracy, while a random error affects precision. Good experimenters attack each with a different strategy.
Why does taking 100 readings beat taking 5?
Random errors are as likely to push a reading high as low, so averaging many readings lets them partly cancel and drives the mean closer to the true value. A set of 100 measurements of a rod's diameter therefore yields a more reliable estimate than a set of 5, because the spread of the mean shrinks as readings increase. It does nothing, however, for a systematic error, which no amount of averaging removes. That is why both strategies matter.
What is the difference between absolute, relative and percentage error?
The absolute error $\Delta a$ is the magnitude of the difference between a reading and the true or mean value, carrying the same units as the quantity. The relative error is the ratio $\Delta a / a$, a pure number that says how large the error is compared with the quantity itself. The percentage error is simply $(\Delta a / a)\times 100\%$. Relative and percentage errors let you compare the quality of measurements of completely different quantities.
Why can the same absolute error give very different relative errors?
Because relative error depends on the size of the quantity, not just the error. A balance accurate to $\pm 0.01$ g gives a relative error of about $1\%$ on a $1.02$ g mass but only about $0.1\%$ on a $9.89$ g mass. The larger the measured value, the smaller the fraction the fixed error represents. This is why measuring larger quantities, when possible, improves relative precision.
How do errors combine when I multiply or divide quantities?
For a product or quotient, the relative errors add. If a rectangle's length has $0.6\%$ error and its breadth has $1\%$ error, the area carries about $1.6\%$ relative error. You compute the area normally, then attach the summed relative error to get the absolute uncertainty. This additive rule for relative errors is one of the most frequently tested ideas in the chapter.
What happens to the error when a quantity is raised to a power?
For a quantity raised to a power $n$, the relative error is multiplied by the magnitude of that power. So if a radius has a $2\%$ relative error, the volume $\propto r^{3}$ carries about a $6\%$ relative error. Powers therefore amplify errors, which is why a small mistake in a squared or cubed quantity matters a lot. Always check whether a formula involves high powers of the least precise quantity.
What does least count mean?
The least count is the smallest measurement an instrument can resolve, and it fixes the precision of any reading taken with it. A metre scale typically has a least count of 1 mm, a vernier callipers around 0.1 mm, and a screw gauge around 0.01 mm. The significant figures you can honestly report are limited by this least count. Choosing an instrument with a smaller least count is how you increase precision.
How does a vernier callipers achieve a fine least count?
The sliding vernier scale has divisions slightly shorter than the main-scale divisions, so its least count equals one main-scale division minus one vernier division. Equivalently it is the value of one main-scale division divided by the number of vernier divisions. You read the main scale up to the vernier zero, then find the vernier line that best coincides with a main-scale line and multiply by the least count. Adding the two gives the final reading.
What is the difference between pitch and least count of a screw gauge?
The pitch is the linear distance the spindle advances in one complete rotation of the circular scale, often 0.5 mm or 1 mm. The least count is the pitch divided by the number of divisions on the circular scale, so a pitch of 0.5 mm with 50 divisions gives a least count of 0.01 mm. Pitch is a property of the screw thread, while least count is the finest length you can read. Confusing the two is a classic exam slip.
What is zero error and how do I correct for it?
Zero error is a systematic error present when the instrument does not read zero even though the jaws or faces are fully closed. A positive zero error is subtracted from every observed reading, and a negative zero error is added, to get the true value. Both vernier callipers and screw gauges can carry it, so you must always check the zero before measuring. Ignoring zero error shifts all your readings by the same amount.
Can I make a screw gauge infinitely accurate by adding more circular divisions?
No. Increasing the number of circular divisions reduces the least count and improves the resolution, but only up to a point. Beyond a limit the divisions become too crowded to read reliably and other errors, such as those from the screw threads and from applying inconsistent pressure, dominate. So accuracy cannot be raised arbitrarily just by subdividing the scale. Real instruments are limited by mechanical quality, not only by division count.
How much weight does this chapter carry in JEE, and how should I prepare?
Units and Measurement typically contributes one to two objective questions in JEE Main, usually on dimensional analysis, significant figures and error propagation, with vernier and screw-gauge numericals appearing regularly. These are high-value marks because the material is small and the questions are formula-light but concept-heavy. Prepare by drilling dimensional-formula recall, practising error-combination sums, and solving instrument-reading problems until zero-error and least-count handling is automatic. The payoff extends to every physics practical and to sanity-checking formulas across the whole syllabus.
🚪 Dive Deeper Mystery room · 73 discoveries
Discovered 0 / 73
Until 2019 the kilogram was defined by a single platinum-iridium cylinder kept near Paris since 1889. Comparisons showed the artefact and its copies drifted by tens of micrograms over a century. The 2019 SI revision replaced it with a fixed value of the Planck constant, $h=6.62607015\times10^{-34}$ J s, realised through a Kibble (watt) balance.
bipm.org
Since 1967 the second has been defined by the hyperfine transition frequency of caesium-133, fixed at $9\,192\,631\,770$ Hz. This ties the world's timekeeping to an unchanging atomic property rather than the Earth's uneven rotation. Modern caesium fountain clocks realise it to better than one part in $10^{16}$.
nist.gov
In 1983 the metre was redefined as the distance light travels in vacuum in $1/299\,792\,458$ of a second, fixing $c$ exactly. This means the speed of light is no longer measured but defined, and length measurement reduces to a time measurement. It replaced the earlier krypton-86 wavelength standard of 1960.
bipm.org
The 2019 SI fixed the Avogadro constant at exactly $6.02214076\times10^{23}$ per mole. The mole is thereby defined as that exact number of entities, decoupling it from the kilogram and the old carbon-12 mass reference.
bipm.org
The 2019 redefinition fixed the elementary charge at $e=1.602176634\times10^{-19}$ C. The ampere is now one coulomb per second in terms of this exact charge, replacing the old force-between-two-wires definition that was hard to realise precisely.
bipm.org
Before 2019 the kelvin was defined by assigning $273.16$ K to the triple point of water. The revision instead fixed the Boltzmann constant at $k=1.380649\times10^{-23}$ J K$^{-1}$, linking temperature directly to energy and removing dependence on a specific substance's properties.
bipm.org
The candela is defined by fixing the luminous efficacy of monochromatic radiation of frequency $540\times10^{12}$ Hz at $683$ lm W$^{-1}$. That frequency (about 555 nm, yellow-green) is where the human eye is most sensitive in daylight.
bipm.org
A Kibble (watt) balance equates mechanical power to electrical power, letting mass be realised from the Planck constant via quantum electrical standards. It uses the Josephson effect and the quantum Hall effect, whose quantised values depend on $h$ and $e$.
nist.gov
A key design goal of the 2019 SI overhaul was continuity: a kilogram, metre and second still mean essentially what they meant before, to well within measurement uncertainty. Only the definitions - the way the units are realised - changed, not their everyday size.
bipm.org
Optical lattice and single-ion clocks tick at optical frequencies hundreds of THz, far above caesium's microwave 9.2 GHz. The best reach fractional uncertainties near $10^{-18}$, and future SI seconds may be redefined around them.
nist.gov
General relativity predicts that clocks deeper in a gravitational field run slower. Optical clocks are now so precise that they measure this shift over a height change of just a centimetre, effectively using time to survey elevation.
nist.gov
Satellite clocks run faster than ground clocks by about $38$ microseconds per day once special and general relativity are combined. Uncorrected, this would introduce navigation errors of about $10$ km per day, so relativistic corrections are engineered into the system.
nasa.gov
Laser interferometers compare path lengths using the wavelength of light as a ruler. Gravitational-wave detectors sense length changes of order $10^{-19}$ m across kilometre-long arms - a fraction of the width of a proton.
ligo.caltech.edu
Using only dimensional analysis and declassified time-lapse photos of a nuclear fireball, physicist G. I. Taylor estimated the blast energy from the relation $R\sim(E t^2/\rho)^{1/5}$. His estimate closely matched the still-secret yield.
en.wikipedia.org
The Buckingham $\pi$ theorem states that a physical relation among $n$ variables using $k$ independent dimensions can be rewritten with $n-k$ dimensionless groups. It is the formal backbone of dimensional analysis used across fluid mechanics and engineering.
en.wikipedia.org
The dimensionless Reynolds number compares inertial to viscous forces. Below roughly 2000 in a pipe, flow is smooth (laminar); above about 4000 it becomes turbulent - the same number predicting this regardless of the fluid or units.
en.wikipedia.org
Lord Rayleigh argued that scattered light intensity scales as the inverse fourth power of wavelength. That $1/\lambda^4$ dependence, obtainable by dimensional reasoning about scatterers much smaller than the wavelength, makes short-wavelength blue light scatter far more than red.
en.wikipedia.org
At the first nuclear test, Enrico Fermi dropped scraps of paper and watched how far the blast wave pushed them. From that crude displacement he estimated the explosion's energy to the right order of magnitude, exemplifying 'Fermi estimation'.
en.wikipedia.org
Pierre Vernier described his auxiliary sliding scale in 1631. By making its divisions slightly smaller than the main scale's, it lets the eye judge fractions of a division precisely - the principle behind today's vernier callipers.
en.wikipedia.org
A screw gauge converts one full rotation of a finely threaded screw into a small, known linear advance (the pitch). Dividing the pitch among many circular-scale divisions yields least counts of $10\,\mu$m or better.
en.wikipedia.org
Several familiar scales are order-of-magnitude scales in disguise: each unit step of the Richter magnitude, the decibel (in factors of ten of power), and the pH corresponds to a power-of-ten change in the underlying quantity. This compresses vast ranges into manageable numbers.
en.wikipedia.org
In November 2022 the General Conference on Weights and Measures added the prefixes ronna ($10^{27}$), quetta ($10^{30}$) and their small counterparts ronto ($10^{-27}$) and quecto ($10^{-30}$). They were introduced partly to express ever-growing quantities of digital data.
bipm.org
The parsec is the distance at which one astronomical unit subtends an angle of one arcsecond, about $3.09\times10^{16}$ m or $3.26$ light-years. Its name is literally 'parallax of one arcsecond'.
en.wikipedia.org
Retroreflectors left by the Apollo astronauts and Soviet landers let lunar laser ranging measure the Earth-Moon distance to a few millimetres. The data show the Moon receding at about $3.8$ cm per year.
nasa.gov
Around 240 BCE Eratosthenes compared noon shadow angles at Syene and Alexandria and, knowing the distance between them, estimated Earth's circumference. His result was within a small percentage of the modern value using only geometry.
en.wikipedia.org
In 1676 Ole Romer noticed that eclipses of Jupiter's moon Io ran early or late depending on Earth's distance from Jupiter. Interpreting the delay as light-travel time gave the first finite estimate of the speed of light.
en.wikipedia.org
In 1798 Henry Cavendish used a delicate torsion balance to measure the tiny gravitational attraction between lead spheres. The experiment effectively determined the gravitational constant $G$ and hence Earth's mass and density.
en.wikipedia.org
Robert Millikan balanced charged oil droplets against gravity in an electric field to find the elementary charge around 1909-1913. The value he measured is the same $e$ now fixed exactly in the redefined SI.
nobelprize.org
The Committee on Data for Science and Technology (CODATA) periodically publishes recommended values and uncertainties for the fundamental physical constants. These adjustments underpin the numbers used in the SI and in precision physics.
nist.gov
The Guide to the Expression of Uncertainty in Measurement (GUM) standardises how scientists combine and report uncertainties. It distinguishes Type A (statistical) from Type B (other) evaluations and defines the combined standard uncertainty.
bipm.org
Carl Friedrich Gauss developed the method of least squares and used it to predict where the newly discovered asteroid Ceres would reappear after being lost near the Sun. Least squares remains the standard way to fit data and estimate random error.
en.wikipedia.org
Averaging $N$ independent measurements reduces the random error of the mean by a factor of $\sqrt{N}$. This is why taking 100 readings of a diameter gives a more reliable estimate than taking only a few - a point NCERT stresses for vernier work.
NCERT-derived
Unlike random errors, systematic errors bias every reading in the same direction and cannot be reduced by averaging. Zero errors in a screw gauge, a miscalibrated scale, or parallax in reading a meniscus are classic examples.
NCERT-derived
Combining $G$, $\hbar$ and $c$ by dimensional analysis yields natural units such as the Planck length $\sqrt{\hbar G/c^3}\approx1.6\times10^{-35}$ m and the Planck time $\approx5.4\times10^{-44}$ s. These mark scales where known physics is expected to break down.
en.wikipedia.org
In natural units, choosing $c=\hbar=1$ collapses mass, energy, momentum and inverse length into the same dimension. This dramatically simplifies equations, and dimensional factors of $c$ and $\hbar$ are restored at the end.
en.wikipedia.org
Joseph Fourier, in his 1822 work on heat, articulated that every term in a physical equation must have the same dimensions. This principle of homogeneity is exactly what lets us check equations dimensionally today.
en.wikipedia.org
The pre-2019 kelvin used the triple point of Vienna Standard Mean Ocean Water, a defined isotopic mixture, held in sealed cells. Isotopic composition mattered enough to be specified, a subtlety the Boltzmann-constant definition sidesteps.
bipm.org
Periodic verifications showed the International Prototype Kilogram and its official copies had drifted apart by up to about 50 micrograms over a century. No one could say which had changed, a discomfort that motivated the constant-based redefinition.
bipm.org
One angstrom is $10^{-10}$ m, conveniently about the size of an atom and the spacing in crystals. Though not an SI unit, it remains widely used in atomic physics, chemistry and X-ray crystallography.
en.wikipedia.org
A light-year is how far light travels in one year in vacuum, about $9.46\times10^{15}$ m. Using $c$ as a fixed conversion, astronomers routinely trade travel time for distance across the cosmos.
en.wikipedia.org
In 2012 the International Astronomical Union redefined the astronomical unit as exactly $149\,597\,870\,700$ m. It no longer depends on measuring the Earth's orbit, making it a clean conversion constant.
en.wikipedia.org
Bouncing radar signals off Venus in the 1960s measured the Earth-Venus distance directly from the round-trip time. Combined with orbital mechanics, this sharpened the value of the astronomical unit enormously.
nasa.gov
Very Long Baseline Interferometry links radio dishes on different continents so their combined baseline can span the Earth. This gives angular resolution fine enough to help define reference frames and measure continental drift.
nasa.gov
The cosmic distance ladder chains methods - parallax, standard candles like Cepheids and supernovae - each calibrating the next to ever larger distances. Each rung carries its own uncertainty that propagates up the ladder.
nasa.gov
Reporting a result with the right significant figures communicates how precisely it was measured. Writing extra digits, NCERT warns, is not just superfluous but misleading about the instrument's least count.
NCERT-derived
Writing a measurement as $a\times10^{b}$ with $1\le a\lt10$ makes the number of significant figures unambiguous, because every digit shown in $a$ is significant. NCERT recommends it precisely to settle trailing-zero confusion.
NCERT-derived
Counting numbers and defined factors - like the 2 in $2\pi r$ or the $n$ in $T=t/n$ - are exact and never limit the significant figures of a result. Only measured quantities constrain precision.
NCERT-derived
Dimensional consistency is necessary but not sufficient: an equation can balance dimensionally yet be physically incorrect, since dimensionless factors are invisible to the test. But any dimensionally inconsistent equation is certainly wrong.
NCERT-derived
The radian (plane angle) and steradian (solid angle) are ratios of like quantities - arc to radius, area to radius squared - so they carry no dimension. This is why angles slip freely through dimensional checks.
NCERT-derived
Before SI, the CGS, FPS and MKS systems competed, differing in their base units of length and mass. NCERT notes SI won out partly because its decimal structure makes conversions within the system simple.
NCERT-derived
The Bureau International des Poids et Mesures, founded by the 1875 Metre Convention, coordinates measurement standards worldwide. The General Conference on Weights and Measures, its governing body, ratified the 2018/2019 SI revision.
bipm.org
The caesium second is tied to the tiny energy difference between two hyperfine states of the atom's ground level, produced by the electron and nuclear spins. Driving that transition with microwaves gives the fixed 9.19 GHz reference.
nist.gov
In a caesium fountain clock, laser-cooled atoms are launched upward and fall back under gravity, giving a long interrogation time. The extra time sharpens the frequency measurement to around one part in $10^{16}$.
nist.gov
The arguments of sine, log and exponential functions must be dimensionless, because their series expansions add different powers of the argument. Spotting a dimensional argument inside such a function reveals a formula error.
NCERT-derived
NCERT notes the Earth's diameter ($\sim10^{7}$ m) exceeds a hydrogen atom's ($\sim10^{-10}$ m) by 17 orders of magnitude. Order-of-magnitude thinking makes such staggering ratios comprehensible.
NCERT-derived
The smallest division an instrument can resolve - its least count - fixes the precision of every reading it gives. NCERT ties the number of significant figures directly to this instrumental least count.
NCERT-derived
An uncertainty of $\pm0.01$ g is a $1\%$ relative error on a $1.02$ g mass but only $0.1\%$ on a $9.89$ g mass. Relative (percentage) error, not absolute error, tells you how good a measurement really is.
NCERT-derived
When two close measurements are subtracted, leading digits cancel and few significant figures survive. This 'loss of significance' is why NCERT flags subtraction as reducing significant figures differently from multiplication.
NCERT-derived
While worst-case error analysis simply sums fractional errors, independent random errors more realistically combine as the square root of the sum of squares. Both views appear in JEE problems, the sum giving the maximum possible error.
en.wikipedia.org
Because visible light has wavelengths of a few hundred nanometres, optical instruments can in principle measure lengths to within a wavelength. NCERT lists such an optical instrument as more precise than either vernier or screw gauge.
NCERT-derived
Quantities like refractive index (ratio of two speeds) and strain (ratio of two lengths) are ratios of like quantities and thus dimensionless. They pass through any dimensional check untouched.
NCERT-derived
By fixing $h=6.62607015\times10^{-34}$ J s exactly, the SI ties mass to a quantum constant. Since $\text{J s}=\text{kg m}^2\text{s}^{-1}$, and the metre and second are already defined by $c$ and the caesium frequency, the kilogram follows.
bipm.org
Dimensional analysis reconstructs the powers in a relation - such as $T\propto\sqrt{l/g}$ for a pendulum - but never the dimensionless constant like $2\pi$. NCERT states this limitation plainly.
NCERT-derived
Kinetic, potential, thermal and electrical energy all reduce to $[M L^2 T^{-2}]$. This shared dimension is why they can be added in an energy balance and why the joule serves every kind of energy.
NCERT-derived
Since 1983 the metre is no longer a scratch on a platinum bar. The BIPM fixed the speed of light in vacuum at exactly $c=299\,792\,458$ m/s, so the metre is defined as the distance light travels in $1/299\,792\,458$ of a second. Length is therefore built on top of the second: once you can measure time superbly, you get length for free, and every laboratory can realise the metre without shipping a physical artefact around the world.
bipm.org (SI Brochure, 9th ed.)
The second is defined by taking the caesium-133 ground-state hyperfine transition frequency to be exactly $\Delta\nu_{Cs}=9\,192\,631\,770$ Hz. A caesium fountain clock counts these oscillations and would drift by less than a second in about 100 million years. Because so many other units (the metre, the volt, the ampere) are now tied to the second, the atomic clock quietly sits at the foundation of the entire measurement system.
bipm.org (SI Brochure, 9th ed.)
On 20 May 2019 the SI was rebuilt so that four fundamental constants have exact numerical values: the Planck constant $h=6.626\,070\,15\times10^{-34}$ J s, the elementary charge $e=1.602\,176\,634\times10^{-19}$ C, the Boltzmann constant $k=1.380\,649\times10^{-23}$ J/K, and the Avogadro constant $N_A=6.022\,140\,76\times10^{23}$ mol$^{-1}$. The kilogram, ampere, kelvin and mole are now derived from these fixed numbers rather than from any physical object.
bipm.org (26th CGPM, 2018)
The old ampere was defined by the force between two infinite parallel wires one metre apart. Since 2019 it is defined by fixing the elementary charge at $e=1.602\,176\,634\times10^{-19}$ coulomb: one ampere is the flow of $1/e$ elementary charges per second. This also fixed the vacuum permeability $\mu_0$, which used to be exactly $4\pi\times10^{-7}$ but is now a measured quantity with a tiny experimental uncertainty.
bipm.org (SI Brochure, 9th ed.)
The candela is the only base unit built around human physiology. It fixes the luminous efficacy of monochromatic 540 THz green light at exactly $K_{cd}=683$ lm/W. That frequency is where the human eye is most sensitive, so the candela translates raw radiant power in watts into perceived brightness. It reminds us that a 'base quantity' can be chosen for practical human reasons, not only for mathematical elegance.
bipm.org (SI Brochure, 9th ed.)
For centuries the astronomical unit (AU) was tied to the Earth-Sun geometry and had to be re-measured as constants improved. In 2012 the International Astronomical Union simply defined it as exactly $1\text{ AU}=149\,597\,870\,700$ m. The parsec then follows geometrically: it is the distance at which one AU subtends one arc-second, about $3.086\times10^{16}$ m. Astronomy thus adopted the same 'fix the number' philosophy the SI later used in 2019.
iau.org (IAU 2012 Resolution B2)
In 1999 NASA's Mars Climate Orbiter fired its thrusters using numbers that one team had computed in pound-force-seconds while the navigation software expected newton-seconds. The unconverted factor of about 4.45 accumulated over months, the probe dipped too deep into the Martian atmosphere and was lost. A $327$-million-dollar mission ended because two groups skipped a single unit conversion — the most expensive dimensional-analysis error ever recorded.
NASA Mars Climate Orbiter Mishap Investigation Board (1999)
Dimensional analysis is more than checking equations. The Buckingham $\pi$ theorem states that a physical relation involving $n$ variables built from $k$ independent base dimensions can be rewritten using exactly $n-k$ independent dimensionless groups. This is how engineers collapse messy experiments — pipe flow, drag, heat transfer — into a handful of universal curves in variables like the Reynolds number, saving enormous experimental effort.
NCERT Exemplar Physics XI, Ch. 2 (Units and Measurements)
When a result is a product or quotient of measured quantities, the fractional (relative) errors add, not the absolute errors. For $Z=A^p B^q C^r$ the rule is $\dfrac{\Delta Z}{Z}=|p|\dfrac{\Delta A}{A}+|q|\dfrac{\Delta B}{B}+|r|\dfrac{\Delta C}{C}$. That is why a quantity raised to a high power dominates the final uncertainty: measuring a radius to $1\%$ gives a $3\%$ error in a volume that depends on $r^3$. Precision effort should go where the exponent is largest.
The frequency $\nu$ of a vibrating string may depend on the tension $F$, the length $l$ and the mass per unit length $\mu$. Use dimensional analysis to find $\nu$ up to a constant.
Attempt, then reveal full solution
Assume $\nu=k\,F^{a}l^{b}\mu^{c}$. Dimensions: $[\nu]=[T^{-1}]$, $[F]=[MLT^{-2}]$, $[l]=[L]$, $[\mu]=[ML^{-1}]$. Then $[T^{-1}]=[MLT^{-2}]^a[L]^b[ML^{-1}]^c$. M: $a+c=0$. T: $-2a=-1\Rightarrow a=\tfrac12$, so $c=-\tfrac12$. L: $a+b-c=0\Rightarrow \tfrac12+b+\tfrac12=0\Rightarrow b=-1$. Hence $\nu=\dfrac{k}{l}\sqrt{\dfrac{F}{\mu}}$ (the true constant is $\tfrac12$ for the fundamental).
JEE Advanced (pattern)
In an experiment $Y=\dfrac{4MgL}{\pi d^2 l}$ (Young's modulus) is found from $M,g,L,d,l$. Measured fractional errors: $L\ (0.5\%),\ d\ (1\%),\ l\ (2\%)$; $M$ and $g$ are exact. Find the maximum percentage error in $Y$.
Attempt, then reveal full solution
$Y\propto L\,d^{-2}\,l^{-1}$. So $\dfrac{\Delta Y}{Y}=\dfrac{\Delta L}{L}+2\dfrac{\Delta d}{d}+\dfrac{\Delta l}{l}=0.5\%+2(1\%)+2\%=0.5+2+2=4.5\%$. The $d^{-2}$ term doubles the diameter's contribution, so measuring $d$ precisely matters most.
JEE Advanced (pattern)
The period of a satellite orbiting a planet may depend on the orbital radius $r$, the planet's mass $M$ and the gravitational constant $G$. Derive the form of the period $T$.
Attempt, then reveal full solution
Let $T=k\,r^{a}M^{b}G^{c}$. $[G]=[M^{-1}L^{3}T^{-2}]$. $[T]=[L]^a[M]^b[M^{-1}L^{3}T^{-2}]^c$. M: $b-c=0$. L: $a+3c=0$. T: $-2c=1\Rightarrow c=-\tfrac12$, so $b=-\tfrac12$, $a=\tfrac32$. Thus $T=k\,r^{3/2}M^{-1/2}G^{-1/2}=k\sqrt{\dfrac{r^3}{GM}}$, reproducing Kepler's third law up to the constant $2\pi$.
JEE Advanced (pattern)
A quantity is measured as $x=(3.50\pm0.05)$ and $y=(2.00\pm0.02)$. Find $z=x-y$ with its absolute error and the resulting percentage error.
Attempt, then reveal full solution
For subtraction, absolute errors add: $\Delta z=\Delta x+\Delta y=0.05+0.02=0.07$. $z=3.50-2.00=1.50$, so $z=1.50\pm0.07$. Percentage error $=\dfrac{0.07}{1.50}\times100=4.7\%$, far larger than the $\sim1.4\%$ and $1\%$ inputs, illustrating amplification of error under subtraction.
JEE Advanced (pattern)
Check dimensionally whether $v=\sqrt{\dfrac{2GM}{R}}$ (escape speed) is consistent, and state whether that proves it correct.
Attempt, then reveal full solution
$[GM/R]=\dfrac{[M^{-1}L^3T^{-2}][M]}{[L]}=\dfrac{[L^3T^{-2}]}{[L]}=[L^2T^{-2}]$. Its square root is $[LT^{-1}]$, the dimension of speed, so the relation is dimensionally consistent. However, consistency does not prove correctness: the factor 2 (and the absence of other dimensionless factors) cannot be verified dimensionally and must come from energy conservation.
JEE Advanced (pattern)
A screw gauge has pitch $0.5$ mm and $50$ divisions on the circular scale, with a negative zero error of $3$ divisions. A wire gives main-scale reading $1.5$ mm and circular reading $27$. Find the corrected diameter.
Attempt, then reveal full solution
L.C. $=\dfrac{0.5}{50}=0.01$ mm. Negative zero error $=-3\times0.01=-0.03$ mm. Observed $=1.5+27\times0.01=1.77$ mm. Corrected $=$ observed $-$ (zero error) $=1.77-(-0.03)=1.80$ mm. A negative zero error is added back to the reading.
JEE Advanced (pattern)
Using dimensional analysis, estimate how the speed of a surface wave (deep-water gravity wave) depends on wavelength $\lambda$ and $g$, assuming it does not depend on water density.
Attempt, then reveal full solution
Let $v=k\,\lambda^{a}g^{b}$. $[LT^{-1}]=[L]^a[LT^{-2}]^b$. L: $a+b=1$. T: $-2b=-1\Rightarrow b=\tfrac12$, so $a=\tfrac12$. Thus $v=k\sqrt{g\lambda}$: longer-wavelength ocean swells travel faster, which is why they outrun a storm and arrive at a coast first.
JEE Advanced (pattern)
The density of a cube is found from mass $m=(200\pm2)$ g and side $a=(5.00\pm0.01)$ cm. Find the density and its maximum percentage error.
Attempt, then reveal full solution
$\rho=\dfrac{m}{a^3}=\dfrac{200}{5.00^3}=\dfrac{200}{125}=1.60$ g cm$^{-3}$. $\dfrac{\Delta\rho}{\rho}=\dfrac{\Delta m}{m}+3\dfrac{\Delta a}{a}=\dfrac{2}{200}+3\times\dfrac{0.01}{5.00}=1\%+0.6\%=1.6\%$. So $\rho=1.60$ g cm$^{-3}\pm1.6\%$, i.e. $\pm0.026$ g cm$^{-3}$.
JEE Advanced (pattern)
Which is more precise for measuring length: (a) a vernier with 20 divisions on the sliding scale (main-scale 1 mm), (b) a screw gauge of pitch 1 mm and 100 divisions, (c) an optical instrument reading to within a wavelength of light ($\sim5\times10^{-7}$ m)? Justify with least counts.
Attempt, then reveal full solution
(a) L.C. $=\dfrac{1\ \text{mm}}{20}=0.05$ mm $=5\times10^{-5}$ m. (b) L.C. $=\dfrac{1\ \text{mm}}{100}=0.01$ mm $=1\times10^{-5}$ m. (c) resolution $\sim5\times10^{-7}$ m. Smallest least count wins, so the optical instrument (c) is most precise, then the screw gauge, then the vernier.
NCERT-derived
The number $N$ of oscillations counted in a stopwatch experiment is $50$ (exact) and the total time is $(100.0\pm0.1)$ s. The length is $(1.000\pm0.001)$ m. Find $g=\dfrac{4\pi^2 l}{T^2}$ and its percentage error.
Attempt, then reveal full solution
$T=\dfrac{100.0}{50}=2.000$ s; $\Delta T=\dfrac{0.1}{50}=0.002$ s. $g=\dfrac{4\pi^2(1.000)}{(2.000)^2}=\dfrac{39.478}{4.000}=9.87$ m s$^{-2}$. $\dfrac{\Delta g}{g}=\dfrac{\Delta l}{l}+2\dfrac{\Delta T}{T}=\dfrac{0.001}{1.000}+2\times\dfrac{0.002}{2.000}=0.1\%+0.2\%=0.3\%$.
JEE Advanced (pattern)
Convert a pressure of $1$ atm $=1.013\times10^{5}$ Pa (Pa $=$ kg m$^{-1}$ s$^{-2}$) into the CGS unit dyn cm$^{-2}$.
Attempt, then reveal full solution
$1$ Pa $=1$ kg m$^{-1}$ s$^{-2}$. Convert: kg $=10^3$ g, m$^{-1}=10^{-2}$ cm$^{-1}$. So $1$ Pa $=10^3\times10^{-2}=10$ g cm$^{-1}$ s$^{-2}=10$ dyn cm$^{-2}$. Hence $1$ atm $=1.013\times10^{5}\times10=1.013\times10^{6}$ dyn cm$^{-2}$.
JEE Advanced (pattern)
A physical quantity $P=\dfrac{a^3 b^2}{\sqrt{c}\,d}$. If the percentage errors in $a,b,c,d$ are $1\%,3\%,4\%,2\%$, find the percentage error in $P$ and identify the dominant contributor.
Attempt, then reveal full solution
$\dfrac{\Delta P}{P}=3(1\%)+2(3\%)+\tfrac12(4\%)+1(2\%)=3+6+2+2=13\%$. The $b^2$ term contributes $6\%$, the largest single share, because its $3\%$ error is doubled by the power 2.
JEE Advanced (pattern)
Show, using dimensions, that the combination $\sqrt{\dfrac{\hbar G}{c^3}}$ has the dimension of length (the Planck length), given $[\hbar]=[ML^2T^{-1}]$, $[G]=[M^{-1}L^3T^{-2}]$, $[c]=[LT^{-1}]$.
Attempt, then reveal full solution
$\dfrac{\hbar G}{c^3}=\dfrac{[ML^2T^{-1}][M^{-1}L^3T^{-2}]}{[LT^{-1}]^3}=\dfrac{[L^5T^{-3}]}{[L^3T^{-3}]}=[L^2]$. Its square root is $[L]$, a length. Numerically it is about $1.6\times10^{-35}$ m.
JEE Advanced (pattern)
A calorimetry result uses $\theta=(t_2-t_1)$ where $t_1=(30.5\pm0.1)^\circ$C and $t_2=(80.5\pm0.1)^\circ$C. Find $\theta$ and its percentage error, and comment.
Attempt, then reveal full solution
$\theta=80.5-30.5=50.0^\circ$C. Absolute errors add: $\Delta\theta=0.1+0.1=0.2^\circ$C. Percentage error $=\dfrac{0.2}{50.0}\times100=0.4\%$. Because the difference here is large, the subtraction does not badly inflate the relative error - unlike subtraction of nearly equal quantities.
JEE Advanced (pattern)
The refractive index of a material is $\mu=1.514\pm0.002$. Compute the percentage error, and explain why $\mu$ carries no unit.
Attempt, then reveal full solution
Percentage error $=\dfrac{0.002}{1.514}\times100\approx0.13\%$. The refractive index is the ratio of two speeds (speed of light in vacuum to that in the medium), $[LT^{-1}]/[LT^{-1}]=[M^0L^0T^0]$, hence dimensionless and unitless.
NCERT-derived
Estimate the order of magnitude of the number of seconds in a human lifetime of about 70 years (a Fermi estimate).
Attempt, then reveal full solution
Seconds per year $\approx 3.15\times10^{7}$ (since $365\times24\times3600\approx3.15\times10^{7}$). For 70 years: $70\times3.15\times10^{7}\approx2.2\times10^{9}$ s. Order of magnitude $\approx10^{9}$: a human lifetime is roughly two billion seconds.
NCERT-derived
Given the speed of light $c=3.00\times10^{8}$ m s$^{-1}$ (3 s.f.) and one year $=3.1557\times10^{7}$ s (5 s.f.), compute one light-year to the correct significant figures.
Attempt, then reveal full solution
$1$ ly $=c\times(1\text{ year})=3.00\times10^{8}\times3.1557\times10^{7}=9.4671\times10^{15}$ m. The least precise factor ($c$, 3 s.f.) limits the result: $1$ ly $=9.47\times10^{15}$ m.
NCERT-derived
The van der Waals equation is $\left(P+\dfrac{a}{V^2}\right)(V-b)=nRT$. Find the dimensions of the constants $a$ and $b$ for one mole ($n=1$).
Attempt, then reveal full solution
$b$ is subtracted from volume, so $[b]=[V]=[L^3]$. The term $a/V^2$ must have the dimension of pressure $[ML^{-1}T^{-2}]$: $[a]=[P][V^2]=[ML^{-1}T^{-2}][L^6]=[ML^{5}T^{-2}]$. Thus $[a]=[M L^{5}T^{-2}]$ and $[b]=[L^3]$.
JEE Advanced (pattern)
A stopwatch measures the time for 20 oscillations five times: $40.1, 40.3, 39.9, 40.2, 40.0$ s. Find the mean period, the mean absolute error and the percentage error in the period.
Attempt, then reveal full solution
Mean total time $=\dfrac{40.1+40.3+39.9+40.2+40.0}{5}=\dfrac{200.5}{5}=40.10$ s. Deviations: $0.00,0.20,0.20,0.10,0.10$; mean absolute error $=\dfrac{0.60}{5}=0.12$ s. Period $T=\dfrac{40.10}{20}=2.005$ s, $\Delta T=\dfrac{0.12}{20}=0.006$ s. Percentage error $=\dfrac{0.12}{40.10}\times100\approx0.3\%$ (same for total time and period).
JEE Advanced (pattern)
Deduce, by dimensions, how the terminal-drag energy dissipated per second (power) by a sphere depends on force $F$ and velocity $v$, then verify $P=Fv$.
Attempt, then reveal full solution
Let $P=k\,F^{a}v^{b}$. $[P]=[ML^2T^{-3}]$, $[F]=[MLT^{-2}]$, $[v]=[LT^{-1}]$. M: $a=1$. So $[ML^2T^{-3}]=[MLT^{-2}][LT^{-1}]^b=[ML^{1+b}T^{-2-b}]$. L: $1+b=2\Rightarrow b=1$. T: $-2-b=-3\Rightarrow b=1$ (consistent). Hence $P=k\,Fv$, and physically $k=1$, so $P=Fv$.
JEE Advanced (pattern)
The energy of a photon is $E=h\nu$. Given $[E]=[ML^2T^{-2}]$ and frequency $[\nu]=[T^{-1}]$, find the dimensions of Planck's constant $h$ and state its SI unit.
Attempt, then reveal full solution
$[h]=\dfrac{[E]}{[\nu]}=\dfrac{[ML^2T^{-2}]}{[T^{-1}]}=[ML^2T^{-1}]$. The SI unit is J s (joule-second), consistent with the value $6.62607015\times10^{-34}$ J s that now defines the kilogram.
NCERT-derived
A rectangular plate is measured with vernier callipers (L.C. $0.01$ cm): length $2.34$ cm, breadth $1.56$ cm, each with uncertainty one least count. Find the area with its percentage error.
Attempt, then reveal full solution
$A=2.34\times1.56=3.6504\approx3.65$ cm$^2$. $\dfrac{\Delta A}{A}=\dfrac{0.01}{2.34}+\dfrac{0.01}{1.56}=0.43\%+0.64\%=1.07\%\approx1.1\%$. So $A=3.65$ cm$^2\pm1.1\%$, i.e. $\Delta A\approx0.04$ cm$^2$; $A=(3.65\pm0.04)$ cm$^2$.
JEE Advanced (pattern)
Establish, by dimensional analysis, the time period of a liquid drop oscillating under surface tension $S$ ($[MT^{-2}]$), given it depends on $S$, the drop radius $r$ and density $\rho$.
Attempt, then reveal full solution
Let $T=k\,S^{a}r^{b}\rho^{c}$. $[S]=[MT^{-2}]$, $[\rho]=[ML^{-3}]$. $[T]=[MT^{-2}]^a[L]^b[ML^{-3}]^c$. M: $a+c=0$. T: $-2a=1\Rightarrow a=-\tfrac12$, so $c=\tfrac12$. L: $b-3c=0\Rightarrow b=\tfrac32$. Thus $T=k\sqrt{\dfrac{\rho r^3}{S}}$.
JEE Advanced (pattern)
A sphere of radius $r$ moving with speed $v$ through a fluid of viscosity $\eta$ feels a resistive force $F$. Assuming $F$ depends only on $\eta$, $r$ and $v$, use dimensional analysis to find how $F$ scales.
Attempt, then reveal full solution
Write $F = k\,\eta^{a} r^{b} v^{c}$. With $[\eta] = \mathrm{ML^{-1}T^{-1}}$, $[r]=\mathrm{L}$, $[v]=\mathrm{LT^{-1}}$, matching $[F]=\mathrm{MLT^{-2}}$ gives, from M: $a=1$; from T: $-a-c=-2\Rightarrow c=1$; from L: $-a+b+c=1\Rightarrow b=1$. Hence $F = k\,\eta r v$ — exactly the form of Stokes' law, with the dimensionless constant $k=6\pi$ supplied by experiment, not by dimensions.
JEE Advanced (pattern)
The period of a simple pendulum is $T=2\pi\sqrt{L/g}$, so $g=4\pi^{2}L/T^{2}$. In an experiment $L$ is measured to $0.1\%$ and $T$, timed over $100$ oscillations, to $0.05\%$. What is the percentage error in $g$?
Attempt, then reveal full solution
For a product/quotient the fractional errors add, each weighted by its power: $\dfrac{\Delta g}{g} = \dfrac{\Delta L}{L} + 2\dfrac{\Delta T}{T} = 0.1\% + 2(0.05\%) = 0.2\%$. Note that timing many oscillations is what makes $\Delta T/T$ so small — a single swing timed by hand would dominate the whole error budget.
JEE Main (pattern)
A vernier callipers has $20$ vernier divisions coinciding with $19$ main-scale divisions, and $1$ MSD $=1\,\mathrm{mm}$. It shows a positive zero error of $4$ divisions. While measuring a rod the main scale reads $3.5\,\mathrm{cm}$ and the $12^{\text{th}}$ vernier division coincides. Find the true length.
Attempt, then reveal full solution
Least count $=1\,\mathrm{MSD}-1\,\mathrm{VSD}=1-\tfrac{19}{20}=0.05\,\mathrm{mm}$. Observed reading $=35\,\mathrm{mm}+(12\times0.05)=35.60\,\mathrm{mm}$. A positive zero error is subtracted: correction $=4\times0.05=0.20\,\mathrm{mm}$. True length $=35.60-0.20=35.40\,\mathrm{mm}=3.540\,\mathrm{cm}$.
JEE Advanced (pattern)
A screw gauge has pitch $0.5\,\mathrm{mm}$ and $50$ divisions on its circular scale, with a negative zero error of $5$ divisions. Measuring a wire, the linear scale shows $4\,\mathrm{mm}$ and the $28^{\text{th}}$ circular division is on the reference line. Find the diameter.
Attempt, then reveal full solution
Least count $=\dfrac{\text{pitch}}{\text{no. of divisions}}=\dfrac{0.5}{50}=0.01\,\mathrm{mm}$. A negative zero error is added back: effective circular reading $=(28+5)=33$ divisions. Diameter $=4\,\mathrm{mm}+(33\times0.01)=4+0.33=4.33\,\mathrm{mm}$.
JEE Main (pattern)
A block has mass $m=4.237\,\mathrm{g}$ (measured to four significant figures) and volume $V=2.5\,\mathrm{cm^{3}}$ (two significant figures). Report its density to the correct number of significant figures.
Attempt, then reveal full solution
$\rho=m/V=4.237/2.5=1.6948\,\mathrm{g\,cm^{-3}}$ on the calculator. In multiplication/division the result keeps the least number of significant figures among the inputs — here two, set by the volume. So $\rho=1.7\,\mathrm{g\,cm^{-3}}$; quoting more digits would falsely claim precision the volume never had.
NCERT-derived
Check whether the equation $E=\tfrac{1}{2}mv^{2}+mgh$ is dimensionally homogeneous.
Attempt, then reveal full solution
$[\tfrac{1}{2}mv^{2}]=\mathrm{M(LT^{-1})^{2}}=\mathrm{ML^{2}T^{-2}}$ and $[mgh]=\mathrm{M(LT^{-2})(L)}=\mathrm{ML^{2}T^{-2}}$. Both additive terms carry the dimension of energy, $\mathrm{ML^{2}T^{-2}}$, so the equation is homogeneous. Homogeneity is necessary but not sufficient — it cannot catch a missing dimensionless factor such as the $\tfrac{1}{2}$.
NCERT-derived
A derived quantity is $Z=\dfrac{A^{2}B^{3}}{\sqrt{C}}$. The measured fractional errors are $\Delta A/A=1\%$, $\Delta B/B=2\%$ and $\Delta C/C=4\%$. Find the maximum percentage error in $Z$.
Attempt, then reveal full solution
For a power law the fractional errors add, each scaled by the magnitude of its exponent: $\dfrac{\Delta Z}{Z}=2\dfrac{\Delta A}{A}+3\dfrac{\Delta B}{B}+\tfrac{1}{2}\dfrac{\Delta C}{C}=2(1)+3(2)+\tfrac{1}{2}(4)=10\%$. The quantity raised to the highest power ($B$) dominates the error, which tells you where to measure most carefully.
JEE Advanced (pattern)
Using dimensional analysis, find how the time period $T$ of a satellite in a circular orbit of radius $r$ around a planet of mass $M$ depends on $r$, $M$ and the gravitational constant $G$.
Attempt, then reveal full solution
Set $T = k\,G^{a} M^{b} r^{c}$. With $[G]=\mathrm{M^{-1}L^{3}T^{-2}}$, $[M]=\mathrm{M}$, $[r]=\mathrm{L}$, and $[T]=\mathrm{T}$: from M, $-a+b=0$; from L, $3a+c=0$; from T, $-2a=1\Rightarrow a=-\tfrac12$. Then $b=-\tfrac12$ and $c=\tfrac32$. So $T = k\sqrt{r^{3}/(GM)}$ — Kepler's third law recovered, up to the factor $2\pi$ dimensions cannot supply.
JEE Advanced (pattern)
A student adds three length measurements $2.1\,\mathrm{cm}$, $0.036\,\mathrm{cm}$ and $12.13\,\mathrm{cm}$. Report the sum with the correct precision.
Attempt, then reveal full solution
Raw sum $=14.266\,\mathrm{cm}$. In addition and subtraction the result is limited by the term with the fewest decimal places — here $2.1\,\mathrm{cm}$, with one decimal place. Rounding $14.266$ to one decimal gives $14.3\,\mathrm{cm}$. Precision, not the number of significant figures, controls the outcome of a sum.
NCERT-derived
The resistance is found from $R = V/I$ with $V = (100 \pm 5)\,\mathrm{V}$ and $I = (10 \pm 0.2)\,\mathrm{A}$. Find $R$ and its percentage and absolute error.
In a screw gauge experiment the pitch is $1\,\mathrm{mm}$ and there are $100$ divisions on the circular scale. To improve resolution, a student proposes halving the pitch. By what factor does the least count change, and what is the new value?
Attempt, then reveal full solution
Least count $=\text{pitch}/(\text{circular divisions})$. Originally $1/100 = 0.01\,\mathrm{mm}$. Halving the pitch to $0.5\,\mathrm{mm}$ halves the least count to $0.5/100 = 0.005\,\mathrm{mm}$ — a factor of $2$ finer. Resolution improves because each full rotation now advances the spindle by less, so one division represents a smaller length.
JEE Advanced (pattern)
The Planck length is formed from the constants $G$, $\hbar$ and $c$. Use dimensional analysis to find the combination $\ell_P=G^{a}\hbar^{b}c^{d}$ that has the dimension of length, and state the exponents.
Attempt, then reveal full solution
Dimensions: $[G]=M^{-1}L^{3}T^{-2}$, $[\hbar]=ML^{2}T^{-1}$, $[c]=LT^{-1}$. Require $M^{0}L^{1}T^{0}$. Mass: $-a+b=0\Rightarrow a=b$. Time: $-2a-b-d=0$. Length: $3a+2b+d=1$. From $a=b$: time gives $-3a-d=0\Rightarrow d=-3a$; length gives $5a+d=1\Rightarrow 5a-3a=1\Rightarrow a=\tfrac12$. So $a=b=\tfrac12$ and $d=-\tfrac32$, giving $\ell_P=\sqrt{\dfrac{G\hbar}{c^{3}}}\approx1.6\times10^{-35}$ m. This is the scale where quantum gravity is expected to matter.
NIST / dimensional-analysis (Planck units)
In a screw-gauge experiment the pitch is $1\text{ mm}$, the circular scale has $100$ divisions, and there is a positive zero error of $4$ divisions. The main scale reads $3\text{ mm}$ and the circular scale reads $47$ divisions when measuring a wire. Find the least count and the corrected diameter.
Attempt, then reveal full solution
Least count $=\dfrac{\text{pitch}}{\text{no. of divisions}}=\dfrac{1}{100}=0.01\text{ mm}$. Observed reading $=$ main-scale $+$ circular $\times$ LC $=3+47\times0.01=3.47\text{ mm}$. A positive zero error of $4$ divisions equals $+0.04\text{ mm}$ and must be subtracted: corrected diameter $=3.47-0.04=3.43\text{ mm}$. The lesson is that the zero error is systematic — it shifts every reading by the same amount and is removed by subtraction, not by averaging.
JEE Main (pattern)
📊 Rank Predictor JoSAA/MCC-calibrated
Disclaimer: These bands are an indicative, pattern-based estimate for orientation only and are not an official cut-off or a guaranteed result.
What this does: Units and Measurement contributes roughly one to two objective questions to JEE Main every year, most often on dimensional analysis, significant figures and error propagation. Because these marks are quick to secure and the concepts underpin the entire physics paper, they are exactly the kind of reliable scoring that separates competitive ranks. Getting them wrong through careless rounding or a misread instrument, by contrast, is a needless leak in an otherwise strong attempt. How to read it: enter your score on a full chapter mock below. The tool maps it — via historical JEE marks→percentile→JoSAA closing-rank data — to the percentile and All-India-Rank band a student at that level typically lands in. It is a calibration signal for THIS chapter's mastery, not a full-exam rank.
Chapter-mock score
Percentile band
Projected AIR band
281-300
99.9+
1-100
251-280
99.5-99.9
100-2000
211-250
99.0-99.5
2000-9000
171-210
98.0-99.0
9000-25000
121-170
95.0-98.0
25000-70000
81-120
88.0-95.0
70000-170000
41-80
70.0-88.0
170000-420000
JEE Main (pattern)
🔖 Bookmarks & Notes Saved to this browser
Bookmark any question or concept card (click the ☆ that appears on hover), and jot notes below. Everything is saved locally in your browser.
Bookmarked items
No bookmarks yet.
Authoritative & comprehensive JEE Main + Advanced resource · sources traced Tier 1–3 · SME-review state (append ?review=1)