1. Physics
  2. Properties of Solids and Liquids
  3. Elastic Moduli

Properties of Solids and Liquids · JEE & NEET Physics

Elastic Moduli: notes and previous year questions

How stiff a material is: Young's modulus, the bulk modulus, the shear modulus, Poisson's ratio and how they are linked.

Elastic Moduli in short

  • A modulus of elasticity is stress ÷ strain: the stiffness of a material.
  • Young's modulus is for length: ΔL = FL/(AY).
  • Longer wires stretch more; thicker wires stretch less (∝ 1/r²).
  • A rope under its own weight stretches half as much as with the weight at its end.

1How stiff?

Modulus of elasticity=stressstrain\text{Modulus of elasticity} = \frac{\text{stress}}{\text{strain}}Unit Pa. It depends on the material, not on the size or shape of the piece.
ModulusStressStrain
Young's YYtensile / compressiveΔL/L\Delta L/L
Bulk KKextra pressure all roundΔV/V\Delta V/V
Shear η\etasidewaysθ\theta

A bigger modulus means a stiffer material: twice the modulus, half the strain for the same stress.

2Young's modulus

Y=F/AΔL/L=FLA ΔLΔL=FLAYY = \frac{F/A}{\Delta L/L} = \frac{FL}{A\,\Delta L}\qquad \Delta L = \frac{FL}{AY}A wire acts like a spring with k = YA/L.
MaterialY (Pa)
Steel2×10112\times10^{11}
Copper1.2×10111.2\times10^{11}
Brass1.0×10111.0\times10^{11}
Aluminium, glass0.7×10110.7\times10^{11}
Wood, bone≈1.5×1010\approx 1.5\times10^{10}
Rubber≈5×106\approx 5\times10^{6}

3Longer and thicker wires

ΔL∝FL/r2\Delta L \propto FL/r^2: twice the length, twice the stretch; twice the radius, a quarter. Radii 1 : 2 with the same load and length give stretches 4 : 1. For length 2L2L and radius 2r2r, the same stretch needs 2F2F.

ΔL=ρgL22Y\Delta L = \frac{\rho g L^2}{2Y}A rope stretched by its own weight: half the stretch of the same weight hung at the end.

4Bulk modulus

K=−ΔPΔV/V=−VΔPΔVK = -\frac{\Delta P}{\Delta V/V} = -V\frac{\Delta P}{\Delta V}The minus sign makes K positive. Compressibility = 1/K.
MaterialK (Pa)
Steel1.6×10111.6\times10^{11}
Copper1.4×10111.4\times10^{11}
Glass0.4×10110.4\times10^{11}
Water2.2×1092.2\times10^{9}
Air≈105\approx 10^{5}

For a gas: K=PK = P at steady temperature, K=γPK = \gamma P when no heat flows.

5Shear modulus

η=F/Aθ,θ=ΔxL\eta = \frac{F/A}{\theta},\qquad \theta = \frac{\Delta x}{L}Also called the modulus of rigidity. Steel ≈ 8 × 10¹⁰ Pa; for metals η ≈ Y/3.

A 5 cm rubber cube, 100 N along the top, slides 0.2 cm: stress 4×1044\times10^4 Pa, θ=0.04\theta = 0.04, η=106\eta = 10^6 Pa. With 200 N it slides 0.4 cm.

6Poisson's ratio

σ=−Δd/dΔL/L\sigma = -\frac{\Delta d/d}{\Delta L/L}No unit; between 0 and 0.5 for real materials.

Stretch a bar and it gets thinner. Cork has σ≈0\sigma \approx 0 (it hardly bulges sideways, so it makes a good stopper); steel about 0.3; rubber about 0.5 (its volume stays the same).

Summary

Key ideas

  • A modulus of elasticity is stress ÷ strain: the stiffness of a material.
  • Young's modulus is for length: ΔL = FL/(AY).
  • Longer wires stretch more; thicker wires stretch less (∝ 1/r²).
  • A rope under its own weight stretches half as much as with the weight at its end.
  • The bulk modulus is for volume; its reciprocal is the compressibility.
  • The shear modulus is for shape; fluids have none.
  • Poisson's ratio: a stretched bar gets thinner, σ between 0 and 0.5.
  • The moduli are linked: Y = 2η(1 + σ) = 3K(1 − 2σ).

Every equation

Young's modulus
Y=FL/(AΔL)Y = FL/(A\Delta L)
Stretch
ΔL=FL/(AY)\Delta L = FL/(AY)
Wire as a spring
k=YA/Lk = YA/L
Own weight
ΔL=ρgL2/2Y\Delta L = \rho g L^2/2Y
Bulk modulus
K=−VΔP/ΔVK = -V\Delta P/\Delta V
Compressibility
κ=1/K\kappa = 1/K
Gas
K=P or γPK = P\ \text{or}\ \gamma P
Shear modulus
η=(F/A)/θ\eta = (F/A)/\theta
Poisson's ratio
σ=−(Δd/d)/(ΔL/L)\sigma = -(\Delta d/d)/(\Delta L/L)
Y and η
Y=2η(1+σ)Y = 2\eta(1+\sigma)
Y and K
Y=3K(1−2σ)Y = 3K(1-2\sigma)
Y, K and η
Y=9Kη/(3K+η)Y = 9K\eta/(3K+\eta)

Previous year questions with solutions

Real JEE and NEET questions on elastic moduli. Try each one before you open the solution.

Q1JEE Main 2026One correct option

The Young's modulus of steel wire of radius rr and length LL is YY.

If the radius rr and length LL of the wire are doubled then the value of YY

  1. Aincreases by two times
  2. Breduces by half
  3. Cremains unchanged
  4. Dbecomes one fourth
Show answer and solution

Answer: Option C

YY is a property of steel, and the wire is still steel. Doubling rr makes the area 44 times larger and doubling LL doubles the length, so under a given load the new wire stretches 24=12\dfrac{2}{4} = \dfrac{1}{2} as much. The stress and the strain both change, but their ratio, YY, does not.

The traps come from reading the formula Y=FLA ΔLY = \dfrac{FL}{A\,\Delta L} as a recipe for changing YY: holding FF and ΔL\Delta L fixed while LL doubles and AA quadruples gives 24\dfrac{2}{4}, B, and using the area alone gives D. But ΔL\Delta L does not stay fixed; it halves, and YY comes out unchanged.

Q2NEET 2026One correct option

Match List I with List II

List I | List I I

A. Young's Modulus | I. ΔdΔL(Ld)\frac{\Delta d}{\Delta L}(\frac{L}{d})

B. Compressibility | II. FLA(ΔL)\frac{FL}{A(\Delta L)}

C. Bulk Modulus | III. −1ΔP(ΔVV)-\frac{1}{\Delta P}(\frac{\Delta V}{V})

D. Poisson's Ratio | IV. −P(VΔV)-P(\frac{V}{\Delta V})

Choose the correct answer from the options given below:

  1. AA-IV, B-I, C-II, D-III
  2. BA-III, B-II, C-I, D-IV
  3. CA-I, B-IV, C-III, D-II
  4. DA-II, B-III, C-IV, D-I
Show answer and solution

Answer: Option D

Young's modulus is F/AΔL/L=FLA ΔL\dfrac{F/A}{\Delta L/L} = \dfrac{FL}{A\,\Delta L}: A–II. Compressibility is 1B=−1ΔP ΔVV\dfrac{1}{B} = -\dfrac{1}{\Delta P}\,\dfrac{\Delta V}{V}: B–III. The bulk modulus is −PΔV/V=−P VΔV-\dfrac{P}{\Delta V/V} = -P\,\dfrac{V}{\Delta V}: C–IV. Poisson's ratio is the lateral strain over the longitudinal strain, Δd/dΔL/L=ΔdΔL(Ld)\dfrac{\Delta d/d}{\Delta L/L} = \dfrac{\Delta d}{\Delta L}\left(\dfrac{L}{d}\right): D–I. That is option D.

List II leaves the minus sign off Poisson's ratio, but the matching is still unique. The trap is swapping III and IV: they are reciprocals, and the one with ΔP\Delta P in the denominator is the compressibility.

Q3JEE Advanced 2013One correct option

One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another

horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces

at two ends, the ratio of the elongation in the thin wire to that in the thick wire is

  1. A0.25
  2. B0.50
  3. C2.00
  4. D4.00
Show answer and solution

Answer: Option C

Welded end to end and pulled at the two ends, both wires carry the same force FF, and both are copper, so ΔL∝Lr2\Delta L \propto \dfrac{L}{r^{2}}. Thin wire: LR2\dfrac{L}{R^{2}}. Thick wire: 2L4R2=L2R2\dfrac{2L}{4R^{2}} = \dfrac{L}{2R^{2}}. So thin : thick =2= 2.

The thick wire is twice as long, but its area is four times as big, and the area wins. The trap is B, 0.500.50, the ratio turned upside down: the thin wire, not the thick one, is the one that stretches more. D, 4.004.00, forgets that the thick wire is also twice as long.

Practice questions, easy to hard

Three questions from the elastic moduli practice ladder: one easy, one medium, one hard.

Q4One correct option

A wire's cross-section is a circle, so its area is A=πr2A = \pi r^{2}. A screw gauge measures the diameter d=2rd = 2r, and in terms of that the same area is A=πd24A = \dfrac{\pi d^{2}}{4}.

A wire of diameter 2 mm2\ \mathrm{mm} carries a tension of 314 N314\ \mathrm{N}. Taking π=3.14\pi = 3.14, what is the stress in it?

  1. A2.5×107 Pa2.5 \times 10^{7}\ \mathrm{Pa}
  2. B1×102 Pa1 \times 10^{2}\ \mathrm{Pa}
  3. C3.14×108 Pa3.14 \times 10^{8}\ \mathrm{Pa}
  4. D1×108 Pa1 \times 10^{8}\ \mathrm{Pa}
Show answer and solution

Answer: Option D

The radius is 1 mm=10−3 m1\ \mathrm{mm} = 10^{-3}\ \mathrm{m}, so A=πr2=3.14×10−6 m2A = \pi r^{2} = 3.14 \times 10^{-6}\ \mathrm{m^{2}} and σ=3143.14×10−6=108 Pa\sigma = \dfrac{314}{3.14 \times 10^{-6}} = 10^{8}\ \mathrm{Pa}. From the diameter it is the same: πd24=3.14×4×10−64\dfrac{\pi d^{2}}{4} = \dfrac{3.14 \times 4 \times 10^{-6}}{4}.

The trap is A: putting the diameter into πr2\pi r^{2} makes the area four times too big and the stress four times too small. C leaves out π\pi, and B forgets that the area came out in square millimetres.

Q5One correct option

Squeeze a body equally from every side, by raising the pressure all round it by ΔP\Delta P, and it shrinks in volume without changing its shape. The stress is now the pressure change, and the strain is the fractional change in volume, the volume strain ΔVV\dfrac{\Delta V}{V}. Their ratio is the bulk modulus:

B=−ΔPΔV/VB = -\dfrac{\Delta P}{\Delta V/V}

A rise in pressure (ΔP>0\Delta P > 0) makes the volume fall (ΔV<0\Delta V < 0); the minus sign keeps BB positive. Its unit is Pa\mathrm{Pa}.

A 1.0×10−3 m31.0 \times 10^{-3}\ \mathrm{m^{3}} block of metal shrinks by 2.0×10−7 m32.0 \times 10^{-7}\ \mathrm{m^{3}} when the pressure on it rises by 8.0×106 Pa8.0 \times 10^{6}\ \mathrm{Pa}. What is its bulk modulus?

  1. A4×1013 Pa4 \times 10^{13}\ \mathrm{Pa}
  2. B4×1010 Pa4 \times 10^{10}\ \mathrm{Pa}
  3. C1.6×103 Pa1.6 \times 10^{3}\ \mathrm{Pa}
  4. D2.5×10−11 Pa2.5 \times 10^{-11}\ \mathrm{Pa}
Show answer and solution

Answer: Option B

The volume strain is 2.0×10−71.0×10−3=2×10−4\dfrac{2.0 \times 10^{-7}}{1.0 \times 10^{-3}} = 2 \times 10^{-4}, so B=8.0×1062×10−4=4×1010 PaB = \dfrac{8.0 \times 10^{6}}{2 \times 10^{-4}} = 4 \times 10^{10}\ \mathrm{Pa}.

The trap is A, which divides the pressure by the change in volume itself, 2.0×10−7 m32.0 \times 10^{-7}\ \mathrm{m^{3}}, instead of by the fractional change. C multiplies where it should divide, and D is the right calculation upside down, ΔV/VΔP\dfrac{\Delta V/V}{\Delta P}.

Q6One or more correct options

When a solid warms, every length in it grows by the same fraction α ΔT\alpha\,\Delta T, so its volume grows by three times that fraction (just as ΔVV=3 ΔRR\dfrac{\Delta V}{V} = 3\,\dfrac{\Delta R}{R} for a sphere). The coefficient of volume expansion of a solid is therefore

γ=3α\gamma = 3\alpha, with ΔV=Vγ ΔT\Delta V = V\gamma\,\Delta T

The mass does not change, so the density falls by the same small fraction. A warmer metal is also a softer one: the Young's modulus of a metal falls as its temperature rises.

Take α=1.2×10−5 K−1\alpha = 1.2 \times 10^{-5}\ \mathrm{K^{-1}} for steel. Which statements are correct?

  1. AA steel ball warmed by 100 K100\ \mathrm{K} grows in volume by 0.36%0.36\%
  2. BWarming a steel ball makes it denser
  3. CA steel ball warmed by 100 K100\ \mathrm{K} grows in radius by 0.36%0.36\%
  4. DThe same load stretches a steel wire further at 300 ∘C300\ ^{\circ}\mathrm{C} than at 20 ∘C20\ ^{\circ}\mathrm{C}
Show answer and solution

Answer: Options A, D

A: γ ΔT=3×1.2×10−5×100=3.6×10−3=0.36%\gamma\,\Delta T = 3 \times 1.2 \times 10^{-5} \times 100 = 3.6 \times 10^{-3} = 0.36\%. D: the hot wire has the smaller YY, and ΔL=FLAY\Delta L = \dfrac{FL}{AY}, so the same load stretches it further.

B has it backwards: the same mass in a larger volume is less dense. C gives the radius the volume's fraction; the radius grows by α ΔT=0.12%\alpha\,\Delta T = 0.12\%, a third of it. The trap there is forgetting that γ=3α\gamma = 3\alpha adds up three directions, each of which grows only by α ΔT\alpha\,\Delta T.