A steel wire and a rubber band both stretch when you pull them. But the steel wire snaps back to its original length; the rubber band may not. The difference — and the exam trap — lives in understanding what stress, strain, and Hooke's law actually guarantee.
Stress is force per unit cross-sectional area: σ = F/A. It is NOT force alone. Two wires under identical force but different diameters experience different stress. SI unit: Pa (N/m²). (NCERT Class 11 Physics Chapter 8, page 168.)
Strain is fractional deformation: ε = ΔL/L for longitudinal strain, ΔV/V for volumetric, and angular displacement for shear. Strain is dimensionless — no unit. (NCERT Class 11 Physics Chapter 8, page 168.)
Hooke's law states that stress is directly proportional to strain within the elastic limit: σ ∝ ε, or σ = Y·ε, where Y is Young's modulus. Beyond the elastic limit, Hooke's law fails — the material enters plastic deformation. (NCERT Class 11 Physics Chapter 8, page 169.)
The stress-strain curve (NCERT Class 11 Physics Chapter 8, page 170) is a high-frequency conceptual target. Key landmarks: proportional limit (Hooke's law valid), elastic limit (returns to original shape), yield point (permanent deformation begins), ultimate stress, fracture point. Ductile materials show large plastic region; brittle materials fracture shortly after elastic limit.
The trap that costs marks: confusing which modulus to use. Young's modulus Y = FL/(AΔL) applies to longitudinal stretching of a wire or rod. Bulk modulus K = −V(dP/dV) applies to uniform volumetric compression. Shear modulus G applies to angular deformation. NEET distractors routinely offer K-formula answers to a Y-formula question. Match the modulus to the deformation type before substituting.
Watch out: strain is ΔL/L, not ΔL alone. If a question gives extension without original length, you cannot compute strain.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The SI unit of stress is:
Answer: C. Stress = Force/Area = N/m² = Pa. (NCERT Class 11 Physics Chapter 8, page 168.)
Why A is wrong: A is wrong because N is the unit of force, not force per unit area. Stress requires dividing force by area.
Why B is wrong: B is wrong because N/m is the unit of surface tension, not stress. Stress has dimensions of N/m² (pressure).
Why D is wrong: D is wrong because strain is dimensionless, not stress. Stress has dimensions of [ML⁻¹T⁻²].
Strain is defined as:
Answer: A. Strain = ΔL/L (longitudinal), ΔV/V (volumetric), or angular displacement (shear) — always a ratio of change to original dimension. It is dimensionless. (NCERT Class 11 Physics Chapter 8, page 168.)
Why B is wrong: B is wrong because force per unit area is stress, not strain.
Why C is wrong: C is wrong because restoring force per unit extension defines spring constant (k), not strain.
Why D is wrong: D is wrong because energy stored per unit volume is strain energy density (½ × stress × strain), not strain itself.
Hooke's law is valid:
Answer: B. Hooke's law (stress ∝ strain) holds only within the elastic limit, specifically up to the proportional limit on the stress-strain curve. (NCERT Class 11 Physics Chapter 8, page 169.)
Why A is wrong: A is wrong because beyond the elastic limit, the stress-strain relationship becomes non-linear and Hooke's law breaks down.
Why C is wrong: C is wrong because Hooke's law applies to both ductile and brittle materials within their respective elastic limits.
Why D is wrong: D is wrong because beyond the yield point the material deforms plastically — Hooke's law has already ceased to apply.
A wire of original length 2.0 m and cross-sectional area 1.0 × 10⁻⁶ m² is stretched by a force of 100 N. If Young's modulus Y = 2.0 × 10¹¹ Pa, the extension ΔL is:
Answer: D. Y = FL/(AΔL), so ΔL = FL/(AY) = (100 × 2.0)/(1.0 × 10⁻⁶ × 2.0 × 10¹¹) = 200/200000 = 1.0 × 10⁻³ m. (NCERT Class 11 Physics Chapter 8, page 171.)
Why A is wrong: A is wrong because 2.5 × 10⁻⁴ m comes from dividing by L instead of multiplying: F/(LAY) = 100/(2.0 × 2.0 × 10⁵).
Why B is wrong: B is wrong because 2.0 × 10⁻³ m counts the length twice: FL²/(AY) = 100 × 4.0/(2.0 × 10⁵) = 2.0 × 10⁻³ m, double the correct extension.
Why C is wrong: C is wrong because this results from using L = 1 m instead of the given L = 2.0 m, halving the correct extension.
Which statement about strain is correct?
Answer: B. B is correct. Longitudinal strain is ΔL/L and volume strain ΔV/V, each a change divided by an original value of the same kind, and shearing strain is an angle. So strain has no units and no dimensions, which is why Young's modulus has the same unit as stress.
Why A is wrong: A is wrong because N/m² (the pascal) is the unit of STRESS, force per unit area. Stress and strain are the two quantities Hooke's law relates.
Why C is wrong: C is wrong because strain is a change in length divided by a length, so the lengths cancel. The change in length ΔL on its own has the dimensions of length, but that is the extension, not the strain.
Why D is wrong: D is wrong because N/m is the unit of a spring constant, force per unit extension. That depends on the size of the object, while strain does not.
A uniform rod is subjected to a longitudinal tensile force. The stress in the rod is 3.0 × 10⁸ Pa and the strain produced is 1.5 × 10⁻³. Young's modulus of the material is:
Answer: C. Y = stress/strain = 3.0 × 10⁸ / 1.5 × 10⁻³ = 2.0 × 10¹¹ Pa. This is a direct application of Hooke's law in modulus form. (NCERT Class 11 Physics Chapter 8, page 169.)
Why A is wrong: A is wrong because this value arises from multiplying stress × strain (3.0 × 10⁸ × 1.5 × 10⁻³ = 4.5 × 10⁵) instead of dividing.
Why B is wrong: B is wrong because this likely comes from an incorrect power-of-ten calculation during the division — possibly treating 10⁸/10⁻³ as 10⁵ instead of 10¹¹.
Why D is wrong: D is wrong because 4.5 comes from multiplying the coefficients (3.0 × 1.5) while dividing the powers of ten correctly — a half-finished division. 3.0/1.5 = 2.0.
A steel wire of length 1.0 m, cross-sectional area 1.0 × 10⁻⁶ m² and Young's modulus 2.0 × 10¹¹ Pa is stretched by a force. If the elastic potential energy stored in the wire is 0.10 J, the extension of the wire is:
Answer: D. D is correct. A stretched wire behaves as a spring of stiffness k = YA/L, so the energy it stores is U = ½kΔL² = ½(YA/L)ΔL². Rearranging, ΔL = √(2UL/(YA)) = √(2 × 0.10 × 1.0 / (2.0 × 10¹¹ × 1.0 × 10⁻⁶)) = √(0.20 / 2.0 × 10⁵) = √(1.0 × 10⁻⁶) = 1.0 × 10⁻³ m (NCERT Class 11 Physics Chapter 8, page 174).
Why A is wrong: A is wrong because 1.0 × 10⁻² m comes from putting the length in centimetres, L = 100, while every other quantity is in SI units: √(2 × 0.10 × 100 / 2.0 × 10⁵) = √(10⁻⁴). Mixed units inflate the answer tenfold.
Why B is wrong: B is wrong because 7.1 × 10⁻⁴ m comes from dropping the ½ in U = ½(YA/L)ΔL², so that ΔL² = UL/(YA) = 5.0 × 10⁻⁷ m². The ½ is there because the force rises from zero to its final value during the stretch; the work done is the average force times the extension.
Why C is wrong: C is wrong because 1.0 × 10⁻⁶ m is ΔL², not ΔL — the square root has been left off. The units give it away: 2UL/(YA) comes out in m².
A metal cube is placed deep under the ocean where it is compressed uniformly from all sides. The appropriate elastic modulus to describe this deformation is:
Answer: A. Uniform compression from all sides is volumetric deformation. The modulus that relates volumetric stress (pressure change) to volumetric strain (ΔV/V) is bulk modulus K. (NCERT Class 11 Physics Chapter 8, page 172.) This is the exact scenario where students confuse Y with K — match the modulus to the deformation type.
Why B is wrong: B is wrong because Young's modulus applies to longitudinal stretching or compression along one axis (like a wire being pulled), not uniform compression from all directions (trap: confusing Y with K).
Why C is wrong: C is wrong because shear modulus applies to angular deformation where layers slide over each other (tangential force on a face), not uniform isotropic compression.
Why D is wrong: D is wrong because Poisson's ratio is a dimensionless ratio of lateral strain to longitudinal strain — it is not a modulus and does not describe volumetric compression behaviour.
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Pattern: Wire stretching — given Y, length, area, find elongation under applied force. (PYQ pattern: Young's modulus wire problems, observed 2024.)
Given
A steel wire has length L = 2.0 m, cross-sectional area A = 1.0 × 10⁻⁶ m² and Young's modulus Y = 2.0 × 10¹¹ Pa. A load of mass 20 kg is hung from it. Take g = 10 m/s² (exact, problem-defined).
Required
Find the extension ΔL of the wire.
Concept
Within the elastic limit, Hooke's law applies. The deformation is longitudinal (the wire stretches along one axis), so the modulus is Young's modulus — NOT bulk modulus (volumetric compression) and NOT shear modulus (angular deformation). Before trusting the answer, check the stress is below the elastic limit; if it is not, Hooke's law does not apply and the formula gives a meaningless number.
Formula
Y = FL/(AΔL)
Rearranging: ΔL = FL/(AY)
Substitution
F = mg = 20 × 10 = 200 N
ΔL = (200 × 2.0) / (1.0 × 10⁻⁶ × 2.0 × 10¹¹)
Calculation
Numerator: 200 × 2.0 = 4.0 × 10² N·m
Denominator: 1.0 × 10⁻⁶ × 2.0 × 10¹¹ = 2.0 × 10⁵ N
ΔL = 4.0 × 10² / 2.0 × 10⁵ = 2.0 × 10⁻³ m = 2.0 mm
Elastic-limit check: stress = F/A = 200 / 1.0 × 10⁻⁶ = 2.0 × 10⁸ Pa = 200 MPa, which is within the elastic range of steel, so Hooke's law holds.
Note on exact constants: g = 10 m/s² is a problem-defined exact value, and the 20 kg mass is a given exact value. Neither limits the significant figures of the answer.
Final answer
ΔL = 2.0 × 10⁻³ m (2.0 mm)
The answer has 2 significant figures, matching the precision of the given data (L = 2.0 m, A = 1.0 × 10⁻⁶ m², Y = 2.0 × 10¹¹ Pa — all 2 sig figs).
Common trap
Using bulk modulus K instead of Young's modulus Y. This problem involves a wire stretched along its length — longitudinal deformation — so Y is correct. If the problem described a cube compressed uniformly from all sides in a fluid, then K would apply. Always match the modulus to the deformation geometry.
Similar NEET-style question
Two steel wires of equal length but radii r and 2r are stretched by the same force. What is the ratio of their extensions? (Answer: 4:1 — area scales as r², so extension is inversely proportional to r².)
---
Stress is the restoring force per unit area in a deformed body: σ = F/A (units: Pa = N/m²). Strain is the fractional deformation: ε = ΔL/L (longitudinal) or ΔV/V (volumetric) or shear-angle (shear). Strain is dimensionless.
-- NCERT Class 11 Physics, Ch. 8, p. 168Within the elastic limit, stress is directly proportional to strain. The constant of proportionality is the modulus of elasticity. Beyond the elastic limit, the relationship is non-linear.
-- NCERT Class 11 Physics, Ch. 8, p. 169Typical curve: (1) proportional region (Hooke's law holds), (2) elastic region (recoverable), (3) yield point, (4) plastic region (permanent deformation), (5) ultimate strength, (6) breaking point.
-- NCERT Class 11 Physics, Ch. 8, p. 169Energy 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. 174More in Properties of Bulk Matter: 5 exam traps and mistakes · 12 formulas · 5 question patterns from its other lessons.
All 17 past-paper questions from Properties of Bulk Matter →
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.
Test yourself on this topic with real past-paper questions:
Practice this topic →