Young's modulus Y = (F·L)/(A·ΔL). Use Searle's apparatus: two parallel wires (test + reference) of equal length, incrementally load the test wire, measure ΔL with a micrometer screw on a spirit level. Plot load vs extension — slope gives Y when multiplied by L/A. Account for instrument zero error.
-- NCERT Physics Lab Manual Class 11, Part 4, p. 78Experiment 05 Young's Modulus
8 MCQs1 revision card9-step worked example
Source: NCERT Experimental SkillsOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Experiment 05 Young's Modulus, explained for NEET
The marks in this experiment are usually lost after the physics is finished. A calculator returns Y = 4.4924 × 10¹¹ N m⁻², the student writes all five digits, and the examiner marks it wrong — because the extension was read on a micrometer whose least count is 1.0 × 10⁻² mm and the diameter to two figures. No combination of measurements that coarse can support five digits. That over-reporting is a high-frequency trap here.
The experiment itself: a wire of length L and diameter D hangs from a rigid support, carries a load M, and stretches by l. Young's modulus is
Y = (Mg L) / (A l), with A = πD²/4, so Y = 4MgL / (πD² l).
Searle's apparatus makes the tiny l readable: two identical wires hang side by side from the same support, one carrying a constant dead load and one the changing load, with a spirit level and micrometer between them. The difference reading cancels temperature drift and any yielding of the support, so the micrometer records only the extension produced by the added load. The procedure and the observation table are in the NCERT Physics Lab Manual Class 11, Part 4, page 78.
The NEET link is almost always the error layer rather than the setup. D appears squared, so a 1% error in diameter puts 2% into Y — twice the weight of an equal error in L, M or l. Combining the fractional errors:
ΔY/Y = ΔM/M + ΔL/L + 2(ΔD/D) + Δl/l.
Watch-out: finish with a precision decision, not just an arithmetic one. Round Y to the number of significant figures the least precise measurement allows — usually the diameter or the extension — and round the raw extension to the micrometer's least count before it ever enters the formula.
Can you answer these Experiment 05 Young's Modulus MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In Searle's apparatus for Young's modulus, the second wire, which hangs from the same support and carries a constant dead load, is there in order to:
Show answer and why every option is right or wrong
Answer: C. C is correct: the two wires share the same support and the same surroundings, so a difference reading removes drift common to both and leaves only the load-produced extension, as described in the NCERT Physics Lab Manual Class 11, Part 4, page 78.
Why A is wrong: A is incorrect: the dead load hangs on the reference wire, not the experimental one, and contributes nothing to the stretching force whose extension is measured.
Why B is wrong: B is incorrect: the screw gauge used for the diameter is a separate instrument and is checked independently of the apparatus.
Why D is wrong: D is incorrect: verticality is secured by the clamping and the hanger itself; a second wire is not needed for it.
MCQ 2Easy RecallPractice
The SI unit of Young's modulus is:
Show answer and why every option is right or wrong
Answer: A. A is correct: Young's modulus is stress divided by strain, and since strain is a pure number the unit is that of stress, N m⁻² (pascal) — see the NCERT Physics Lab Manual Class 11, Part 4, page 78.
Why B is wrong: B is incorrect: N m is the unit of torque or work, not of a stress.
Why C is wrong: C is incorrect: N m⁻¹ is a force per unit length, the unit of a spring constant or of surface tension.
Why D is wrong: D is incorrect: only the strain is dimensionless; dividing stress by it leaves the unit of stress intact.
MCQ 3Easy RecallPractice
In this experiment the diameter of the metallic wire is measured with:
Show answer and why every option is right or wrong
Answer: D. D is correct: the wire's diameter is a fraction of a millimetre, so the screw gauge is the prescribed instrument in the NCERT Physics Lab Manual Class 11, Part 4, page 78.
Why A is wrong: A is incorrect: a metre scale resolves about 1 mm, coarser than the whole diameter being measured.
Why B is wrong: B is incorrect: a vernier's resolution of about 0.1 mm would leave the diameter known to roughly one figure, and that uncertainty then enters Y doubled.
Why C is wrong: C is incorrect: the travelling microscope belongs to the optical experiments of this unit, not to the diameter reading here.
MCQ 4Direct ApplicationPractice
Young's modulus is obtained from Y = 4MgL/(πD²l). In one determination the percentage errors are 1.0% in the load M, 0.5% in the length L, 1.0% in the diameter D and 2.0% in the extension l. The maximum percentage error in Y is:
Show answer and why every option is right or wrong
Answer: B. B is correct: fractional errors add weighted by the magnitude of each exponent, so 1.0 + 0.5 + 2(1.0) + 2.0 = 5.5%, the combination rule set out in the NCERT Physics Lab Manual Class 11, Part 4, page 78.
Why A is wrong: A is incorrect: 4.5% comes from adding the diameter error once instead of twice, ignoring that D is squared.
Why C is wrong: C is incorrect: 3.5% drops the extension error, keeping only 1.0 + 0.5 + 2(1.0); l is a measured quantity and its 2.0% must be included.
Why D is wrong: D is incorrect: 6.5% doubles the load's error as well, 2(1.0) + 0.5 + 2(1.0) + 2.0; M appears to the first power, so only D's error is doubled.
MCQ 5Direct ApplicationPractice
The micrometer of a Searle's apparatus has a least count of 1.0 × 10⁻² mm. Averaging the loading and unloading readings gives an extension of 0.4783 mm. The extension should be recorded as:
Show answer and why every option is right or wrong
Answer: D. D is correct: the instrument resolves only to 1.0 × 10⁻² mm, so the reading is rounded to that place, giving 0.48 mm (NCERT Physics Lab Manual Class 11, Part 4, page 78).
Why A is wrong: A is incorrect: four decimal places claim a resolution a hundred times finer than the micrometer has — digits produced by averaging are not measured digits.
Why B is wrong: B is incorrect: 0.478 mm still carries one digit beyond the least count.
Why C is wrong: C is incorrect: rounding to 0.5 mm throws away a digit the micrometer genuinely resolves, losing real precision.
MCQ 6Direct ApplicationPractice
In Y = 4MgL/(πD²l), a measurement error of 1% in which single quantity produces a 2% error in Y?
Show answer and why every option is right or wrong
Answer: A. A is correct: D carries the exponent 2, so its fractional error is weighted by the factor 2 when the errors are combined, as set out in the NCERT Physics Lab Manual Class 11, Part 4, page 78.
Why B is wrong: B is incorrect: L appears to the first power, so a 1% error in it contributes 1% to Y.
Why C is wrong: C is incorrect: M appears to the first power and contributes its own 1% unchanged.
Why D is wrong: D is incorrect: l sits in the denominator but to the first power, so its 1% also enters as 1%, not 2%.
MCQ 7CalculationPractice
A wire of original length 2.0 m and diameter 4.0 × 10⁻¹ mm stretches by 5.0 × 10⁻¹ mm when a 2.0 kg load is hung from it. Taking g = 10 m s⁻² (exact), Young's modulus of the material is closest to:
Show answer and why every option is right or wrong
Answer: C. C is correct: Y = 4MgL/(πD²l) = (4 × 2.0 × 10 × 2.0) / (π × (4.0 × 10⁻⁴)² × 5.0 × 10⁻⁴) ≈ 6.4 × 10¹¹ N m⁻², following the working in the NCERT Physics Lab Manual Class 11, Part 4, page 78.
Why A is wrong: A is incorrect: 1.6 × 10¹¹ N m⁻² results from using D in place of the radius, that is, dropping the factor 4 that converts πD²/4 into the area.
Why B is wrong: B is incorrect: 2.5 × 10¹² N m⁻² is four times the correct value. It comes from putting the radius (2.0 × 10⁻⁴ m) into Y = 4MgL/(πD²l), which needs the diameter; halving D quarters D² and quadruples Y.
Why D is wrong: D is incorrect: 3.2 × 10¹¹ N m⁻² is half the correct value, the result of taking the cross-section as πD²/2.
MCQ 8CalculationPractice
The same wire material, the same load and the same micrometer are used, but the experimental wire is replaced by one of twice the original length and the same diameter. Compared with the first trial, the extension and the percentage error in the measured extension respectively:
Show answer and why every option is right or wrong
Answer: B. B is correct: l is proportional to L for a fixed load and cross-section, so the extension doubles, while the absolute uncertainty is still one least count, making the fractional error half as large — the reason a long wire is specified in the NCERT Physics Lab Manual Class 11, Part 4, page 78.
Why A is wrong: A is incorrect: Y is the material constant that stays fixed; the extension itself scales with the original length.
Why C is wrong: C is incorrect: the absolute uncertainty is set by the micrometer's least count and does not grow with the reading, so a larger extension carries a smaller percentage error.
Why D is wrong: D is incorrect: doubling the length increases the extension for the same stress; it does not halve it.
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Experiment 05 Young's Modulus: quick recall before you leave
How do you solve a Experiment 05 Young's Modulus question? A worked example
- 1
Given.
Original length L = 1.50 m; diameter D = 5.0 × 10⁻¹ mm = 5.0 × 10⁻⁴ m (screw gauge, two significant figures); extension l = 2.5 × 10⁻¹ mm = 2.5 × 10⁻⁴ m (Searle's micrometer, least count 1.0 × 10⁻² mm); load M = 1.5 kg; g = 9.8 m s⁻².
- 2
Required.
Young's modulus Y of the wire material, reported to a defensible precision.
- 3
Concept.
The stress on the wire is the load's weight spread over the circular cross-section; the strain is the extension divided by the original length. Within the elastic limit their ratio is Y.
- 4
Formula.
Y = (Mg L)/(A l) with A = πD²/4, hence Y = 4MgL/(πD² l).
- 5
Substitution.
Y = (4 × 1.5 × 9.8 × 1.50) / (π × (5.0 × 10⁻⁴)² × 2.5 × 10⁻⁴).
- 6
Calculation.
Numerator = 88.2 N m. Denominator = π × 2.5 × 10⁻⁷ × 2.5 × 10⁻⁴ = 1.96 × 10⁻¹⁰ m³. Y = 88.2 ÷ (1.96 × 10⁻¹⁰) = 4.492 × 10¹¹ N m⁻². The 4 arising from the area conversion and the constant π are exact and contribute nothing to the significant-figure count; only the measured M, L, D and l do.
- 7
Final answer.
The diameter and the extension are each known to two significant figures, so Y = 4.5 × 10¹¹ N m⁻².
- 8
Common trap.
Copying 4.492 × 10¹¹ — or the calculator's full 4.49237 × 10¹¹ — straight off the screen. Four or six digits assert a precision the screw gauge and micrometer never delivered. Round at the end to the figures of the least precise measurement, and round the raw extension to the micrometer's least count before it enters the formula.
- 9
Similar NEET-style question.
The same wire is loaded so that the extension becomes 5.0 × 10⁻¹ mm while every other quantity is unchanged. State the new value of Y and the new maximum percentage error in Y, given errors of 1.0% in M, 0.5% in L, 2.0% in D and 2.0% in l.
What to remember before solving Experiment 05 Young's Modulus questions
Which Experiment 05 Young's Modulus formulas do you need for NEET?
1 formula — click to collapse
Percentage error in derived quantity
Combines fractional errors of independent measurements. Adds in absolute value (worst-case).
| Symbol | Quantity | SI Unit |
|---|---|---|
| Z | derived quantity | - |
| x_i | measured | - |
| n_i | exponent | - |
Valid when
- Independent measurements
- Maximum-error analysis
Where do students lose marks on Experiment 05 Young's Modulus?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
Category: Overthinking
Reporting a measured length to more decimal places than the least count permits, e.g. writing 2.345 cm when LC = 0.01 cm.
When it triggers
Calculator output gives 4-5 decimals while the instrument's least count limits resolution to 2-3.
How to avoid
Round the final answer to the precision of the least-precise input or the limiting least count, whichever is coarser.
Root cause: concept gap
Correction
Final reading precision = least count of the limiting instrument; do NOT carry extra digits from a calculator.
More in Experimental Skills: 4 exam traps and mistakes · 2 formulas · 1 question pattern from its other lessons.
Experiment 05 Young's Modulus questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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