Formula
Young's modulus
Y = (longitudinal stress)/(longitudinal strain) = (F·L)/(A·ΔL). Higher Y means stiffer material. Steel: Y ≈ 2 × 10¹¹ Pa. Rubber: Y ≈ 5 × 10⁶ Pa.
-- NCERT Class 11 Physics, Ch. 8, p. 170Young's Modulus
8 MCQs4 revision cards9-step worked example
Source: NCERT Properties of Solids and LiquidsPYQ coverage: NEET 2024Official key: NTA-verifiedLast updated: 25 Sep 2026
Young's Modulus, explained for NEET
A wire is stretched by a force. NEET asks you to find the elongation, the stress, or Young's modulus itself — and a common trap is grabbing the bulk modulus formula instead.
Young's modulus (Y) measures a material's resistance to longitudinal stretching or compression. It is the ratio of longitudinal (tensile or compressive) stress to longitudinal strain, valid only within the elastic limit where Hooke's law holds.
Y = FL / (AΔL)
where F is the applied force along the length, L is the original length, A is the cross-sectional area, and ΔL is the extension (NCERT Class 11 Physics, Chapter 8 — Mechanical Properties of Solids, page 171).
The modulus-type trap. Three elastic moduli appear in this chapter: Young's modulus (Y) for longitudinal deformation, bulk modulus (K) for volumetric compression, and shear modulus (G) for tangential deformation. NEET questions on wire stretching require Y. If the problem describes uniform pressure compressing a volume, that is K. Mixing them up costs marks and is a documented confusion pattern.
Key checks before substituting:
- Is the deformation along the length of a rod or wire? → Use Y.
- Is a diameter given? Convert to area: A = πd²/4. A common arithmetic slip is forgetting the factor of 4.
- Are units consistent? Force in newtons, lengths in metres, area in m². Y comes out in pascals (Pa).
- Is the material within its elastic limit? Y is undefined beyond yield point.
Watch out: When two wires of different materials or dimensions are compared, set up the ratio Y₁/Y₂ = (F₁L₁A₂ΔL₂)/(F₂L₂A₁ΔL₁). Cancel what is common before computing — NEET rewards clean ratio work over brute-force substitution.
Can you answer these 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
Young's modulus is defined as the ratio of:
Show answer and why every option is right or wrong
Answer: A. Young's modulus is the ratio of longitudinal (tensile or compressive) stress to longitudinal strain, as defined in NCERT Class 11 Physics, Chapter 8, page 171.
Why B is wrong: B describes bulk modulus (K), not Young's modulus. Bulk modulus applies to volumetric compression under uniform pressure.
Why C is wrong: C describes shear modulus (G, also called modulus of rigidity). Shear involves tangential forces, not longitudinal stretching.
Why D is wrong: D describes Poisson's ratio, a dimensionless ratio of lateral contraction to longitudinal extension — not a modulus at all.
MCQ 2Easy RecallPractice
The SI unit of Young's modulus is the same as that of:
Show answer and why every option is right or wrong
Answer: A. Young's modulus = stress / strain. Since strain is dimensionless, Y has the same unit as stress: pascal (Pa) or N/m². NCERT Class 11 Physics, Chapter 8, page 171.
Why B is wrong: Force has units of newtons (N). Young's modulus has units of N/m² (Pa), which is stress, not force.
Why C is wrong: Strain is dimensionless (length/length) and has no unit. Young's modulus has units of Pa.
Why D is wrong: Energy has units of joules (J = N·m). Young's modulus has units of Pa (N/m²) — different dimensions entirely.
MCQ 3Easy RecallPractice
Young's modulus is applicable when the deformation is:
Show answer and why every option is right or wrong
Answer: B. Young's modulus specifically describes the response to longitudinal (tensile or compressive) stress — deformation along the direction of the applied force. NCERT Class 11 Physics, Chapter 8, page 171.
Why A is wrong: Volume change under uniform pressure is described by the bulk modulus (K), not Young's modulus. (Trap: Young's vs bulk modulus confusion.)
Why C is wrong: Tangential displacement of parallel faces is shear deformation, described by the shear modulus (G).
Why D is wrong: Torsion (twist) involves shear stresses in the material, governed by the shear modulus, not Young's modulus.
MCQ 4Direct ApplicationPractice
A steel wire of length 2.0 m and cross-sectional area 1.0 × 10⁻⁶ m² is stretched by a force of 200 N. If Young's modulus of steel is 2.0 × 10¹¹ Pa, the extension of the wire is:
Show answer and why every option is right or wrong
Answer: B. ΔL = FL/(AY) = (200 × 2.0)/(1.0 × 10⁻⁶ × 2.0 × 10¹¹) = 400/(2.0 × 10⁵) = 2.0 × 10⁻³ m. Direct substitution into Y = FL/(AΔL), rearranged.
Why A is wrong: A (1.0 × 10⁻³ m) results from dropping the factor of L = 2.0 m in the numerator — computing F/(AY) instead of FL/(AY).
Why C is wrong: C (4.0 × 10⁻³ m) results from doubling the correct answer, likely by using A = 0.5 × 10⁻⁶ m² (halving the area by mistake).
Why D is wrong: D (2.0 × 10⁻⁴ m) is off by a factor of 10 — a power-of-ten arithmetic error during exponent manipulation.
MCQ 5Direct ApplicationPractice
A copper wire of diameter 2.0 mm and original length 1.0 m is stretched by a load. If the extension is 0.50 mm and Y for copper is 1.2 × 10¹¹ Pa, the applied force is closest to:
Show answer and why every option is right or wrong
Answer: C. A = π(d/2)² = π(1.0 × 10⁻³)² = π × 10⁻⁶ m². F = YAδL/L = 1.2 × 10¹¹ × π × 10⁻⁶ × 5.0 × 10⁻⁴ / 1.0 = 1.2π × 10¹ ≈ 60π/10 ≈ 188 N.
Why A is wrong: A (60 N) results from leaving π out of the area, A = r² = 1.0 × 10⁻⁶ m², which gives F = 60 N; the cross-section is πr².
Why B is wrong: B (754 N) results from using the diameter as the radius, A = πd² = 4π × 10⁻⁶ m², making the area and the force four times too large (trap: diameter vs radius).
Why D is wrong: D (47 N) results from halving the diameter twice — taking r = 0.50 mm — which makes the area, and the force, four times too small.
MCQ 6Direct ApplicationPractice
Two wires of the same material and length are stretched by the same force. Wire P has diameter d and wire Q has diameter 2d. The ratio of their extensions ΔL_P / ΔL_Q is:
Show answer and why every option is right or wrong
Answer: D. Same material → same Y. Same F and L. ΔL = FL/(AY), so ΔL ∝ 1/A ∝ 1/d². Ratio = (2d)²/d² = 4. So ΔL_P/ΔL_Q = 4 : 1.
Why A is wrong: A (1:1) wrongly assumes extension depends only on material and length, ignoring the cross-sectional area entirely.
Why B is wrong: B (2:1) comes from assuming extension is inversely proportional to diameter (1/d) instead of inversely proportional to area (1/d²). Since A = πd²/4, doubling d quadruples A.
Why C is wrong: C (1:4) inverts the ratio — this would mean the thicker wire extends more, which contradicts ΔL ∝ 1/A.
MCQ 7CalculationPYQ Pattern
A steel wire (Y = 2.0 × 10¹¹ Pa) of length 1.0 m and radius 1.0 mm supports a mass of 10 kg. If the wire is now replaced by another steel wire of the same length but radius 2.0 mm, carrying a mass of 40 kg, the ratio of the extension of the first wire to the second wire is:
Show answer and why every option is right or wrong
Answer: D. ΔL = FL/(AY). For wire 1: F₁ = 10g, A₁ = π(10⁻³)². For wire 2: F₂ = 40g, A₂ = π(2×10⁻³)² = 4π×10⁻⁶. Ratio ΔL₁/ΔL₂ = (F₁/A₁)/(F₂/A₂) = (10g / π×10⁻⁶) / (40g / 4π×10⁻⁶) = (10/π×10⁻⁶) × (4π×10⁻⁶/40) = 40/40 = 1. Extensions are equal.
Why A is wrong: A (1:4) results from considering only the force ratio (10:40 = 1:4) while ignoring the area difference entirely.
Why B is wrong: B (1:2) results from only accounting for the force ratio (10:40 = 1:4) and the radius ratio (1:2 → area ratio 1:4), but combining them as 1:(4/2) = 1:2 instead of correctly computing (F₁/A₁)/(F₂/A₂).
Why C is wrong: C (2:1) comes from inverting the area ratio without properly handling the force ratio — getting the algebra backwards.
MCQ 8CalculationPYQ Pattern
A uniform wire of length 3.0 m is stretched by 3.0 mm when a force of 150 N is applied. If the same wire is cut into three equal pieces and one piece is stretched by the same force, the extension of that piece is:
Show answer and why every option is right or wrong
Answer: C. ΔL = FL/(AY). Cutting into three equal pieces gives each piece length L/3. Same F, A, Y. So ΔL_new = F(L/3)/(AY) = ΔL_original/3 = 3.0 mm / 3 = 1.0 mm. Extension is proportional to original length.
Why A is wrong: A (3.0 mm) assumes extension is independent of the wire's length — but ΔL = FL/(AY) shows ΔL is directly proportional to L.
Why B is wrong: B (9.0 mm) results from multiplying instead of dividing by 3, perhaps reasoning that a shorter wire is 'stiffer per unit length' and incorrectly inverting the relationship.
Why D is wrong: D (0.33 mm) divides by 9 instead of 3 — possibly squaring the factor of 3, confusing this with an area-scaling problem.
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Young's Modulus: quick recall before you leave
How do you solve a Young's Modulus question? A worked example
Pattern: Wire stretching — given Y, dimensions, find elongation or compare wires (pattern: NEET 2024, wire-stretching type).
- 1
Given
A steel wire of length L = 2.0 m and diameter d = 1.0 mm hangs vertically and supports a mass m = 5.0 kg at its lower end. Young's modulus for steel Y = 2.0 × 10¹¹ Pa. Take g = 10 m/s² (exact, problem-defined).
- 2
Required
Find the extension ΔL of the wire.
- 3
Concept
Young's modulus relates longitudinal stress (F/A) to longitudinal strain (ΔL/L). The wire is under tensile stress from the hanging mass. This is a longitudinal deformation → use Y, not K (NCERT Class 11 Physics, Chapter 8, page 171).
- 4
Formula
Y = FL / (AΔL), rearranged: ΔL = FL / (AY)
- 5
Substitution
F = mg = 5.0 × 10 = 50 N
r = d/2 = 0.50 mm = 5.0 × 10⁻⁴ m
A = πr² = π × (5.0 × 10⁻⁴)² = π × 2.5 × 10⁻⁷ m²
ΔL = (50 × 2.0) / (π × 2.5 × 10⁻⁷ × 2.0 × 10¹¹) - 6
Calculation
Numerator: 50 × 2.0 = 100
Denominator: π × 2.5 × 10⁻⁷ × 2.0 × 10¹¹ = π × 5.0 × 10⁴ = 1.571 × 10⁵
ΔL = 100 / (1.571 × 10⁵) = 6.37 × 10⁻⁴ m ≈ 0.64 mm
Note on exact constants: g = 10 m/s² is stated as an exact problem-defined value and the integer 2 in the length are exact counting/defined quantities. They do not limit significant figures. The answer is reported to 2 significant figures, matching the least precise given quantity (each datum has 2 sig figs). - 7
Final answer
ΔL ≈ 6.4 × 10⁻⁴ m (0.64 mm)
- 8
Common trap
Using bulk modulus K instead of Young's modulus Y. This wire is being stretched longitudinally → Y is correct. If the problem described uniform pressure compressing the steel from all sides, only then would K apply.
- 9
Similar NEET-style question
A copper wire (Y = 1.2 × 10¹¹ Pa) of length 1.5 m and cross-sectional area 2.0 × 10⁻⁶ m² is stretched by a force until the extension is 0.75 mm. Find the applied force. (Answer: F = YAδL/L = 1.2 × 10¹¹ × 2.0 × 10⁻⁶ × 7.5 × 10⁻⁴ / 1.5 = 120 N.)
---
What to remember before solving Young's Modulus questions
Energy stored per unit volume = ½ × stress × strain = ½ Y ε² (for longitudinal). Total energy = ½ × (F·ΔL) for a stretched wire.
-- NCERT Class 11 Physics, Ch. 8, p. 174Which Young's Modulus formulas do you need for NEET?
1 formula — click to collapse
Young's modulus
Ratio of longitudinal stress to longitudinal strain in a stretched wire/rod within elastic limit.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Y | Young's modulus | Pa |
| F | applied force | N |
| A | cross-section area | m^2 |
| L | original length | m |
| Delta_L | extension | m |
Valid when
- Within elastic limit (Hooke's law region)
- Uniform cross-section
- Force along length
Where do students lose marks on 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: Similar Terms
Student uses Y formula when problem is about volumetric compression (use K) or vice versa.
When it triggers
Problem describes longitudinal stretching (use Y), volumetric pressure (use K), or shear (use G).
How to avoid
Y: longitudinal stress/strain. K: volumetric. G: shear. Match modulus to deformation type.
Root cause: formula misuse
Correction
Y for longitudinal stretch (FL/A·ΔL); K for volumetric compression (-V·dP/dV); G for shear. Match modulus type to deformation type before computing.
More in Properties of Bulk Matter: 3 exam traps and mistakes · 11 formulas · 4 question patterns from its other lessons.
Young's Modulus questions from past NEET papers
2 questions from NEET 2024. Answers verified against NTA official keys. — click to collapse
NEET 2024Revised key
14 mm
20.4 mm
340 mm
48 mm
NTA Answer: Option 1(revised_final)
NEET 2024Revised key
15 × 10³ N
250 × 10³ N
3100 × 10³ N
42 × 10³ N
NTA Answer: Option 2(revised_final)
All 17 past-paper questions from Properties of Bulk Matter →
How does NEET ask about Young's Modulus?
1 recurring pattern from past papers — click to collapse
Wire stretching: given Y, length, area, force or stress, find elongation or stress at limit. Y = FL/(A delta_L).
Multi StepMedium
Common distractors
uses bulk modulus formula
Confuses Y with K
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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