1. Physics
  2. Properties of Solids and Liquids
  3. Stress and Strain

Properties of Solids and Liquids · JEE & NEET Physics

Stress and Strain: notes and previous year questions

Elasticity, stress F/A and its kinds, strain, Hooke's law, the stress–strain curve and the energy stored in a stretch.

Stress and Strain in short

  • Elastic bodies spring back; plastic ones keep the change.
  • Stress is the internal restoring force per area, measured in pascals.
  • Stress can be tensile, compressive, shearing or bulk.
  • Strain is change ÷ original size and has no unit.

1Elastic or plastic?

Elasticity is the property of a body to regain its shape and size when the deforming force is removed. A spring springs back; clay stays squashed.

TypeBehaviourExamples
Nearly perfectly elasticreturns fullyquartz fibre, steel (not overloaded)
Plasticthe change staysclay, putty, wet mud
Partly elasticpartly recoversmost real materials

2Stress

Stress is the internal restoring force per unit area inside a deformed body:

Stress=FA\text{Stress} = \frac{F}{A}Unit N/m² = pascal (Pa); 1 MPa = 10⁶ Pa, 1 GPa = 10⁹ Pa. Dimensions [M L⁻¹ T⁻²], the same as pressure.

The same load on a thinner wire gives a bigger stress: 100 N on 1 mm² is 100 MPa, on 4 mm² only 25 MPa. Areas in the ratio 1 : 2 give stresses 2 : 1.

  • Tensile: stretching forces along the length.
  • Compressive: squashing forces along the length (together, longitudinal stress).
  • Shearing: a force along the surface, Ftangential/AF_{tangential}/A (scissors, rivets).
  • Bulk (hydraulic): an extra pressure on every face at once (a submarine hull).

3Strain

Strain=change in sizeoriginal size\text{Strain} = \frac{\text{change in size}}{\text{original size}}A ratio: it has no unit.
  • Longitudinal: ΔL/L\Delta L/L.
  • Shearing: the angle θ=Δx/L\theta = \Delta x/L (in radians).
  • Volume: ΔV/V\Delta V/V.

2.0 m stretched to 2.002 m: strain 0.001=0.1%0.001 = 0.1\%. A 1000 cm³ ball squeezed to 998 cm³: volume strain −0.2%-0.2\%. A 2 m wire of 2 mm² stretched 1 mm by 100 N: stress 50 MPa, strain 5×10−45\times10^{-4}.

4Hooke's law

Stress=E×Strain\text{Stress} = E \times \text{Strain}Within the elastic limit. E is a modulus of elasticity, a property of the material.
F=kxF = kxFor a spring. k = 200 N/m stretched 5 cm: F = 10 N, the weight of 1 kg.

It holds only for small deformations below the elastic limit.

5The stress–strain curve

  1. A, proportional limit: Hooke's law holds up to here (a straight line).
  2. B, elastic limit: below it the wire springs back; beyond it a permanent stretch remains.
  3. C, yield point: the metal starts to flow, stretching a lot for little extra stress.
  4. D, ultimate tensile strength: the greatest stress it can bear; then it necks.
  5. E, fracture: it breaks, at a stress a little below D because the neck is thinner.

Ductile materials (copper, iron) have a long plastic region and can be drawn into wires. Brittle materials (glass) break soon after the straight part.

6Energy in a stretch

U=12F ΔL=12kx2U = \tfrac12 F\,\Delta L = \tfrac12 kx^2The area under the force–extension line.
u=UV=12×stress×strainu = \frac{U}{V} = \tfrac12 \times \text{stress} \times \text{strain}

A spring of 500 N/m compressed 10 cm stores 2.5 J; one of 200 N/m compressed 5 cm stores 0.25 J.

Summary

Key ideas

  • Elastic bodies spring back; plastic ones keep the change.
  • Stress is the internal restoring force per area, measured in pascals.
  • Stress can be tensile, compressive, shearing or bulk.
  • Strain is change ÷ original size and has no unit.
  • Hooke's law: stress ∝ strain below the elastic limit.
  • The stress–strain curve has proportional, elastic, yield, ultimate and fracture points.
  • Ductile materials stretch a lot before breaking; brittle ones do not.
  • Stretching stores energy ½FΔL; hysteresis turns part of it into heat.

Every equation

Stress
σ=F/A\sigma = F/A
Longitudinal strain
ΔL/L\Delta L/L
Shear strain
θ=Δx/L\theta = \Delta x/L
Volume strain
ΔV/V\Delta V/V
Hooke's law
stress=E×strain\text{stress} = E \times \text{strain}
Spring
F=kxF = kx
Stored energy
U=12FΔL=12kx2U = \tfrac12 F\Delta L = \tfrac12 kx^2
Energy per volume
u=12 stress×strainu = \tfrac12\,\text{stress}\times\text{strain}

Previous year questions with solutions

Real JEE and NEET questions on stress and strain. 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 stress and strain 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.