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 2.)
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 2.)
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 3.)
The stress-strain curve (NCERT Class 11 Physics Chapter 8, page 4) 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:
Strain is defined as:
Hooke's law is valid:
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:
Two wires of the same material and length are pulled 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:
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:
A steel wire of length 1.0 m, cross-sectional area 2.0 × 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:
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:
Pattern: Wire stretching — given Y, length, area, find elongation under applied force. (PYQ pattern: Young's modulus wire problems, observed 2024.)
Given
A copper wire has length L = 3.0 m, cross-sectional area A = 2.0 × 10⁻⁶ m², and Young's modulus Y = 1.2 × 10¹¹ Pa. A load of F = 4.8 × 10³ N is applied along its length.
Required
Find the extension ΔL of the wire.
Concept
Within the elastic limit, Hooke's law applies. The deformation is longitudinal (wire stretching along one axis), so we use Young's modulus — NOT bulk modulus (which is for volumetric compression) and NOT shear modulus (which is for angular deformation).
Formula
Y = FL/(AΔL) Rearranging: ΔL = FL/(AY)
Substitution
ΔL = (4.8 × 10³ × 3.0) / (2.0 × 10⁻⁶ × 1.2 × 10¹¹)
Calculation
Numerator: 4.8 × 10³ × 3.0 = 14.4 × 10³ = 1.44 × 10⁴ N·m Denominator: 2.0 × 10⁻⁶ × 1.2 × 10¹¹ = 2.4 × 10⁵ N ΔL = 1.44 × 10⁴ / 2.4 × 10⁵ = 0.060 × 10⁻¹ = 6.0 × 10⁻² m Wait — let me recheck: 1.44 × 10⁴ / 2.4 × 10⁵ = (1.44/2.4) × 10⁴⁻⁵ = 0.60 × 10⁻¹ = 6.0 × 10⁻² m. That's 6 cm — too large for a metal wire. Let me verify the numbers are reasonable by checking stress: σ = F/A = 4.8 × 10³ / 2.0 × 10⁻⁶ = 2.4 × 10⁹ Pa. This exceeds the elastic limit of copper (~70 MPa). Let me adjust to a realistic problem. **Revised 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). ### Step 5 (revised) — Substitution F = mg = 20 × 10 = 200 N ΔL = FL/(AY) = (200 × 2.0) / (1.0 × 10⁻⁶ × 2.0 × 10¹¹) ### Step 6 (revised) — Calculation Numerator: 200 × 2.0 = 400 = 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 Note on exact constants: g = 10 m/s² is a problem-defined exact value, and the mass 20 kg is a counting/given exact integer. These do not limit significant figures in 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, p. 2Within 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, p. 3Typical 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, p. 4Energy stored per unit volume = ½ × stress × strain = ½ Y ε² (for longitudinal). Total energy = ½ × (F·ΔL) for a stretched wire.
-- NCERT, p. 8Conservation of energy along a streamline of incompressible non-viscous flow.
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | pressure | Pa |
| rho | density | kg/m^3 |
| v | speed | m/s |
| g | gravity | m/s^2 |
| h | height | m |
Resistance of a material to uniform compression. Inverse: compressibility.
| Symbol | Quantity | SI Unit |
|---|---|---|
| K | bulk modulus | Pa |
| V | volume | m^3 |
| P | pressure | Pa |
Height a liquid rises (or falls) in a capillary tube. cos(theta) > 0: rises (wetting); < 0: depresses.
| Symbol | Quantity | SI Unit |
|---|---|---|
| h | capillary height | m |
| T | surface tension | N/m |
| theta | contact angle | rad |
| rho | density | kg/m^3 |
| r | tube radius | m |
Pressure at depth h below free surface of fluid of density rho.
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | total pressure | Pa |
| P0 | atmospheric/surface pressure | Pa |
| rho | density | kg/m^3 |
| h | depth | m |
Heat absorbed/released during phase change at constant T. L_fusion or L_vaporisation.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Q | heat | J |
| m | mass | kg |
| L | latent heat | J/kg |
Heat required to raise mass m by temperature Delta_T. Specific heat c is material property.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Q | heat | J |
| m | mass | kg |
| c | specific heat | J/kg/K |
| Delta_T | temp change | K |
Radiation power from a body. Black body epsilon=1; net to surroundings P = sigma*epsilon*A*(T^4 - T_s^4).
| Symbol | Quantity | SI Unit |
|---|---|---|
| sigma | Stefan-Boltzmann = 5.67e-8 | W/m^2/K^4 |
| epsilon | emissivity (0-1) | - |
| A | surface area | m^2 |
| T | absolute temp | K |
Drag force on a sphere of radius r moving with velocity v through viscous fluid (low Reynolds number).
| Symbol | Quantity | SI Unit |
|---|---|---|
| F | drag force | N |
| eta | viscosity | Pa*s |
| r | sphere radius | m |
| v | velocity | m/s |
Excess internal pressure due to surface tension. Bubble has 2 surfaces, hence factor 4.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Delta_P | excess pressure | Pa |
| T | surface tension | N/m |
| r | radius | m |
Constant velocity reached when net force is zero (gravity balanced by buoyancy + viscous drag).
| Symbol | Quantity | SI Unit |
|---|---|---|
| v_t | terminal velocity | m/s |
| r | sphere radius | m |
| rho_s | sphere density | kg/m^3 |
| rho_f | fluid density | kg/m^3 |
| eta | viscosity | Pa*s |
Fractional change in length, area, volume per degree temperature change.
| Symbol | Quantity | SI Unit |
|---|---|---|
| alpha | linear coefficient | 1/K |
| beta | volume coefficient | 1/K |
| Delta_T | temperature change | K |
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 |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Student uses 2T/r for soap bubble (drop formula). Bubble has 2 surfaces → 4T/r.
Question mentions soap bubble OR liquid drop OR air bubble in liquid.
Drop in air: 1 surface → 2T/r. Soap bubble in air: 2 surfaces → 4T/r. Air bubble in liquid: 1 surface → 2T/r.
Category: Similar Terms
Student uses Y formula when problem is about volumetric compression (use K) or vice versa.
Problem describes longitudinal stretching (use Y), volumetric pressure (use K), or shear (use G).
Y: longitudinal stress/strain. K: volumetric. G: shear. Match modulus to deformation type.
Root cause: concept gap
Drop in air: 1 surface → ΔP = 2T/r. Soap bubble: 2 surfaces (inner + outer) → ΔP = 4T/r. Air bubble inside liquid: 1 surface → 2T/r.
Root cause: formula misuse
v_t ∝ r² (because Stokes drag ∝ r v, gravity ∝ r³ - r³ = r³_diff). Doubling radius quadruples terminal velocity, not doubles.
Root cause: formula misuse
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.
If a soap bubble expands, the pressure inside the bubble
forgets height term
Drops rho*g*h term
equates pressures incorrectly
Picks wrong reference points
ignores mass difference
Compares only specific heats, not masses
uses 2T r instead of 4T r for bubble
Treats soap bubble like a drop
expects linear acceleration
Default to constant acceleration without recognising drag-induced terminal v
uses bulk modulus formula
Confuses Y with K
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