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, p. 5A 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 9 — Mechanical Properties of Solids, page 5).
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:
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.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Young's modulus is defined as the ratio of:
The SI unit of Young's modulus is the same as that of:
Young's modulus is applicable when the deformation is:
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:
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:
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:
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:
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:
Pattern: Wire stretching — given Y, dimensions, find elongation or compare wires (pattern: NEET 2024, wire-stretching type).
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).
Required
Find the extension ΔL of the wire.
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 9, page 5).
Formula
Y = FL / (AΔL), rearranged: ΔL = FL / (AY)
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)
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).
Final answer
ΔL ≈ 6.4 × 10⁻⁴ m (0.64 mm)
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.
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.) ---
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, p. 5Energy 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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