Law
Hooke's law
Within 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. 169Elastic Behaviour
8 MCQs4 revision cards9-step worked example
Source: NCERT Properties of Solids and LiquidsOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Elastic Behaviour, explained for NEET
A common trap in elastic behaviour questions: confusing which modulus applies. A wire stretched by a hanging weight requires Young's modulus (longitudinal stress ÷ longitudinal strain). A solid cube submerged and compressed uniformly by fluid pressure requires bulk modulus (volume stress ÷ volume strain). Using the wrong formula gives an answer that looks plausible but is physically wrong — and NEET distractors are built on exactly this confusion.
Hooke's law states that within the elastic limit, stress is directly proportional to strain (NCERT Class 11 Physics, Chapter 8 — Mechanical Properties of Solids, page 169). The proportionality constant is the elastic modulus. The stress-strain curve for a ductile material shows a linear (Hookean) region, a yield point, a plastic region, and fracture. NEET does not typically ask you to draw the curve, but it does test whether you know that Hooke's law fails beyond the elastic limit — this is a core evaluable fact (NCERT Chapter 9, page 170).
Young's modulus Y = FL/(AΔL) quantifies resistance to longitudinal deformation. It applies when a rod or wire is stretched or compressed along its length with uniform cross-section.
Bulk modulus K = −V(dP/dV) quantifies resistance to uniform volumetric compression. It applies when pressure acts equally from all sides.
The watch-out: when a problem describes "a wire under tension," use Y. When it describes "a sphere subjected to uniform pressure increase," use K. If the problem describes a shape change at constant volume (shearing), neither Y nor K applies — that's the shear modulus G, though NEET rarely tests G computationally.
Can you answer these Elastic Behaviour 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
Hooke's law is valid
Show answer and why every option is right or wrong
Answer: C. Hooke's law (stress ∝ strain) holds only within the elastic limit, as stated in NCERT Class 11 Physics, Chapter 8, page 169. Beyond this point the material deforms plastically and the linear relationship breaks down.
Why A is wrong: A is wrong because beyond the elastic limit the material enters the plastic region where stress is no longer proportional to strain. The breaking point is far past the Hookean region.
Why B is wrong: B is wrong because Hooke's law is explicitly limited to the elastic region. At large strains the material yields and eventually fractures — the linear law fails.
Why D is wrong: D is wrong because Hooke's law applies to all types of elastic deformation (longitudinal, volumetric, shear) within the elastic limit, not exclusively to shear.
MCQ 2Easy RecallPractice
A body that regains its original size and shape completely once the deforming force is removed is described as:
Show answer and why every option is right or wrong
Answer: A. A is correct. Elasticity is the property of recovering from deformation. A perfectly elastic body recovers completely; real materials come close to this only within their elastic limit. Quartz fibre is the usual near-example.
Why B is wrong: B is wrong because a perfectly plastic body is the opposite case: it does not recover at all, and stays in its deformed shape. Putty is the usual example.
Why C is wrong: C is wrong because ductile describes a material that can be drawn into wires because it undergoes a large PLASTIC deformation before fracture. That is permanent deformation, not recovery.
Why D is wrong: D is wrong because brittle describes a material that fractures soon after its elastic limit, with little plastic deformation. It says how the material fails, not that it recovers.
MCQ 3Easy RecallPractice
Which elastic modulus is used to describe the resistance of a material to uniform compression from all sides?
Show answer and why every option is right or wrong
Answer: B. Bulk modulus K = −V(dP/dV) quantifies resistance to volumetric compression. NCERT Class 11 Physics, Chapter 8, page 173.
Why A is wrong: A is wrong because Young's modulus measures resistance to longitudinal stretching or compression along one axis, not uniform compression from all directions (trap: confusing Y with K).
Why C is wrong: C is wrong because shear modulus measures resistance to shape change at constant volume — not volumetric compression.
Why D is wrong: D is wrong because Poisson's ratio is a dimensionless ratio (lateral strain to longitudinal strain), not a modulus measuring resistance to compression.
MCQ 4Concept TrapPractice
On the stress–strain curve of a metal wire under increasing load, the point beyond which the wire no longer returns to its original length when the load is removed is the:
Show answer and why every option is right or wrong
Answer: A. A is correct. Up to the elastic limit the deformation is reversible. Beyond it the wire yields: removing the load leaves a permanent set, so the wire does not return to its original length.
Why B is wrong: B is wrong because the proportional limit is where stress stops being proportional to strain, where Hooke's law ends. The wire can still be elastic slightly beyond it, so it is not the point where recovery stops.
Why C is wrong: C is wrong because the fracture point is where the wire breaks. Permanent deformation begins much earlier, at the elastic limit.
Why D is wrong: D is wrong because the ultimate tensile strength is the MAXIMUM stress the wire can bear, reached well into the plastic region.
MCQ 5Direct ApplicationPractice
A copper wire of length 1.0 m and diameter 2.0 mm is stretched by a load. If the same load is applied to a copper wire of length 2.0 m and diameter 4.0 mm, the ratio of their elongations (ΔL₁/ΔL₂) is
Show answer and why every option is right or wrong
Answer: D. ΔL = FL/(AY). Same material (Y) and same force (F). A = π d²/4. ΔL₁/ΔL₂ = (L₁/A₁)/(L₂/A₂) = (L₁ × A₂)/(L₂ × A₁) = (1.0 × π(4.0)²/4)/(2.0 × π(2.0)²/4) = (16)/(2 × 4) = 16/8 = 2. Ratio is 2 : 1.
Why A is wrong: A is wrong — this assumes elongation depends only on length ratio, ignoring the change in cross-sectional area. Since diameter doubles, area quadruples, which matters.
Why B is wrong: B is wrong — this results from using ΔL ∝ L/d (linear in diameter) instead of ΔL ∝ L/d². The area scales as d², not d (trap: confusing Y with bulk modulus or forgetting the squared dependence on diameter).
Why C is wrong: C is wrong — this results from considering only the area ratio (A₂/A₁ = 4) while ignoring the length ratio (L₂/L₁ = 2). Both L and A affect ΔL.
MCQ 6Direct ApplicationPractice
A rod of cross-sectional area 2.0 × 10⁻⁴ m² is subjected to a tensile force. The stress in the rod is 1.5 × 10⁸ Pa. The tensile force applied is
Show answer and why every option is right or wrong
Answer: B. Stress = F/A, so F = Stress × A = 1.5 × 10⁸ × 2.0 × 10⁻⁴ = 3.0 × 10⁴ N. Direct application of the stress definition.
Why A is wrong: A is wrong — this results from dividing stress by area (F = σ/A) instead of multiplying, giving an absurdly large force.
Why C is wrong: C is wrong — this results from a power-of-ten error, obtaining 10³ instead of 10⁴.
Why D is wrong: D is wrong — this results from halving the answer, perhaps by mistakenly using A = 1.0 × 10⁻⁴ m² instead of 2.0 × 10⁻⁴ m².
MCQ 7CalculationPractice
Two wires of the same material have lengths in the ratio 1 : 2 and diameters in the ratio 2 : 1. If the same force is applied to both, the ratio of their elongations (ΔL₁ : ΔL₂) is
Show answer and why every option is right or wrong
Answer: C. ΔL = FL/(AY). Same material (Y) and same force (F). A ∝ d². So ΔL ∝ L/d². ΔL₁/ΔL₂ = (L₁/d₁²) / (L₂/d₂²) = (1/4) / (2/1) = 1/8. Ratio is 1 : 8. Two steps: compute area ratio from diameter ratio, then combine with length ratio.
Why A is wrong: A is wrong — this results from considering only the length ratio (1/2) and ignoring the diameter difference entirely.
Why B is wrong: B is wrong — this results from using ΔL ∝ L/d (linear in d) instead of ΔL ∝ L/d², giving (1/2)/(2/1) = 1/4.
Why D is wrong: D is wrong — 2 : 1 is just the diameter ratio, as if elongation grew with diameter; a thicker wire stretches less, since ΔL ∝ L/d².
MCQ 8Direct ApplicationPractice
A solid sphere is placed inside a fluid whose pressure is uniformly increased. To calculate the change in volume of the sphere, which modulus should be used?
Show answer and why every option is right or wrong
Answer: D. Uniform pressure from all sides causes volumetric strain. The appropriate modulus is bulk modulus K = −V(dP/dV). NCERT Class 11 Physics, Chapter 8, page 173. This tests the concept of matching deformation type to the correct modulus.
Why A is wrong: A is wrong because Young's modulus applies to longitudinal stress/strain along one axis, not uniform volumetric compression. Using Y here is the classic Y-vs-K confusion trap.
Why B is wrong: B is wrong because while Y = 3K(1 − 2σ) relates the two moduli, the problem asks which modulus to use directly for volumetric compression — that is K, not Y/3.
Why C is wrong: C is wrong because shear modulus applies to tangential (shape-changing) deformation at constant volume, not to uniform compression.
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Elastic Behaviour: quick recall before you leave
How do you solve a Elastic Behaviour question? A worked example
- 1
Given
A steel wire has original length L = 1.5 m, cross-sectional area A = 2.0 × 10⁻⁶ m², and is subjected to a tensile force F = 600 N. Young's modulus of steel Y = 2.0 × 10¹¹ Pa.
- 2
Required
Find the elongation ΔL of the wire.
- 3
Concept
This is longitudinal stretching along one axis → use Young's modulus (not bulk modulus). Y = stress/strain = FL/(AΔL).
- 4
Formula
Y = FL/(AΔL), rearranged: ΔL = FL/(AY)
- 5
Substitution
ΔL = (600 × 1.5) / (2.0 × 10⁻⁶ × 2.0 × 10¹¹)
- 6
Calculation
Numerator: 600 × 1.5 = 900
Denominator: 2.0 × 10⁻⁶ × 2.0 × 10¹¹ = 4.0 × 10⁵
ΔL = 900 / (4.0 × 10⁵) = 2.25 × 10⁻³ m
Note on exact constants: the numerical coefficients in the formula (the "1" in FL/AΔL) are exact and do not affect significant-figure counting. The given values each have 2 significant figures, so the answer is reported to 2 significant figures. - 7
Final answer
ΔL = 2.3 × 10⁻³ m (to 2 significant figures), equivalently 2.3 mm.
- 8
Common trap
If you see "sphere under uniform pressure" instead of "wire under tension," you must switch from Y = FL/(AΔL) to K = −V(dP/dV). The numbers might look identical, but the physics is different — and NEET distractors exploit this Y-vs-K confusion.
- 9
Similar NEET-style question
A copper wire of length 2.0 m and cross-sectional area 5.0 × 10⁻⁷ m² stretches by 4.0 × 10⁻³ m under a certain load. Find Young's modulus of copper if the applied force is 400 N. (Answer: Y = FL/(AΔL) = (400 × 2.0)/(5.0 × 10⁻⁷ × 4.0 × 10⁻³) = 4.0 × 10¹¹ Pa.)
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What to remember before solving Elastic Behaviour questions
Key Fact
Stress-strain curve regions
Typical 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. 169More in Properties of Bulk Matter: 5 exam traps and mistakes · 12 formulas · 5 question patterns from its other lessons.
Elastic Behaviour questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
All 17 past-paper questions from Properties of Bulk Matter →
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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