JEE Mains Physics · Units and Measurements

Errors in Laboratory Experiments

In a laboratory experiment each reading's error is the least count of the instrument that took it, a time is read over many oscillations to shrink its relative error, and the errors then combine by the power rule.

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Why this matters

Sixteen PYQs, eleven of them multiple choice, and four from 2026. Five time a pendulum or a spring, four of them over many oscillations; eleven take each instrument's least count as the error of its reading, in Young's modulus, Ohm's law, a lens or mirror on an optical bench, a potentiometer or a travelling microscope. The physics is in deciding what the error of each reading is; the arithmetic after that is the power rule.

Concept 1 of 2: Timing many oscillations: pendulum and spring

A stopwatch's resolution is the same whether it times one swing or fifty. Timing fifty swings and dividing by fifty spreads that error over all of them, so the relative error in the period equals the relative error in the TOTAL time. Then g or k follows from the power rule, with T squared.

Definition

Period from n oscillations: $T = t/n$ , and $\dfrac{\Delta T}{T} = \dfrac{\Delta t}{t}$ , with $\Delta t$ the watch's resolution and t the total time.
Pendulum, $g = \dfrac{4\pi^{2}l}{T^{2}}$ : $\dfrac{\Delta g}{g} = \dfrac{\Delta l}{l} + 2\dfrac{\Delta T}{T}$ .
Spring, $T = 2\pi\sqrt{m/k}$ , so $k = \dfrac{4\pi^{2}m}{T^{2}}$ : $\dfrac{\Delta k}{k} = \dfrac{\Delta m}{m} + 2\dfrac{\Delta T}{T}$ .
Length from a metre scale: $\Delta l$ = its least count, usually 1 mm.

Pendulum error

\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta t}{t} \qquad (t = \text{total time of } n \text{ oscillations})

Worked example

A pendulum's length is 50.0 cm, measured to 1 mm. Twenty-five oscillations take 50 s on a watch of resolution 0.5 s. Find the percentage error in g.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 31 January 2024 · Q93Moderate

Example 1 · Units and Measurements · Errors in Laboratory Experiments

The measured value of the length of a simple pendulum is

20\text{ }cm

with

2\text{ }mm

accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is

N\%

. The value of

N

is:

Divide the resolution by the total time, not by the period

If 40 oscillations take 80 s on a 1 s watch, ΔT/T is 1/80, not 1/2. The watch was read once over the whole run, so its error is shared by every oscillation.

The period enters g squared

In g = 4π²l/T² the period has power 2, so its relative error counts twice. Forgetting the 2 gives the 'length error + period error' distractor.

Drill 4 more on timing many oscillations: pendulum and spring

Concept 2 of 2: Least count as the error of each reading

Every reading is uncertain by the smallest division its instrument can show. So in an experiment the error of a diameter is the screw gauge's least count, the error of a length is the scale's least count, and so on. When a distance is found as the DIFFERENCE of two positions on a scale, each position carries the least count, so the distance carries twice it.

Definition

Error of one reading = least count of its instrument.
A distance found from two scale positions (an optical bench, a travelling microscope): error = 2 × least count.
Young's modulus $Y = \dfrac{4FL}{\pi d^{2}\,\Delta L}$ : $\dfrac{\Delta Y}{Y} = \dfrac{\Delta L}{L} + 2\dfrac{\Delta d}{d} + \dfrac{\Delta(\Delta L)}{\Delta L}$ ; a known load carries no error.
Ohm's law $R = V/I$ : $\dfrac{\Delta R}{R} = \dfrac{\Delta V}{V} + \dfrac{\Delta I}{I}$ , each Δ a meter's least count.
Lens or mirror, $\dfrac{1}{f} = \dfrac{1}{v} \pm \dfrac{1}{u}$ : $\dfrac{\Delta f}{f^{2}} = \dfrac{\Delta u}{u^{2}} + \dfrac{\Delta v}{v^{2}}$ .
Refractive index by travelling microscope $\mu = \dfrac{\text{real depth}}{\text{apparent depth}}$ ; potentiometer $\dfrac{E_1}{E_2} = \dfrac{l_1}{l_2}$ : relative errors add.

Errors from least counts

\Delta x = \text{LC} \qquad \Delta(x_2 - x_1) = 2\,\text{LC} \qquad \frac{\Delta f}{f^{2}} = \frac{\Delta u}{u^{2}} + \frac{\Delta v}{v^{2}}

Worked example

On an optical bench of least count 0.1 cm, the object pin is at the 10.0 cm mark, a convex lens at 40.0 cm and the sharp image at 100.0 cm. Find the focal length and its error.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q2Moderate

Example 2 · Units and Measurements · Errors in Laboratory Experiments

The diameter of a wire measured by a screw gauge of least count 0.001 cm is 0.08 cm . The length measured by a scale of least count 0.1 cm is 150 cm . When a weight of 100 N is applied to the wire, the extension in length is 0.5 cm , measured by a micrometer of least count 0.001 cm . The error in the measured Young's modulus is

\alpha \times10^{9}\text{ }N/m^{2}

. The value of

\alpha

\_\_\_\_

.(Ignore the contribution of the load to Young's modulus error calculation)

A distance read from two marks carries twice the least count

On an optical bench the object distance is (lens mark − object mark). Each mark is uncertain by one least count, so the distance is uncertain by two. Using one least count halves the error.

The diameter's error counts twice in Young's modulus

Y = 4FL/(πd²ΔL) has d squared, so 2Δd/d enters. A fine screw gauge can still give the largest term because the diameter itself is so small.

Drill 10 more on least count as the error of each reading

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Timing many oscillations: pendulum and spring
Pendulum error
$\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta t}{t} \qquad (t = \text{total time of } n \text{ oscillations})$
Least count as the error of each reading
Errors from least counts
$\Delta x = \text{LC} \qquad \Delta(x_2 - x_1) = 2\,\text{LC} \qquad \frac{\Delta f}{f^{2}} = \frac{\Delta u}{u^{2}} + \frac{\Delta v}{v^{2}}$

Watch out for (4)

Divide the resolution by the total time, not by the period
→ Timing many oscillations: pendulum and spring
The period enters g squared
→ Timing many oscillations: pendulum and spring
A distance read from two marks carries twice the least count
→ Least count as the error of each reading
The diameter's error counts twice in Young's modulus
→ Least count as the error of each reading

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