JEE Mains Physics · Mechanical Properties of Solids

Bulk Modulus, Shear Modulus and Poisson's Ratio

The bulk modulus B = ΔP/(ΔV/V) measures resistance to squeezing from all sides, the shear modulus η = (F/A)/θ resistance to a change of shape, and Poisson's ratio the thinning of a stretched wire; Y = 2η(1 + σ) = 3B(1 − 2σ) ties them together.

All Mechanical Properties of Solids notes JEE Mains Physics strategy

Why this matters

Nineteen PYQs, nine of them asking for a number, and two from 2026. Seven use the bulk modulus directly, five take a body down into the sea, and seven deal with shear, Poisson's ratio or the relations between the moduli. Percentages and the face a shear force acts on are where the marks go.

Concept 1 of 3: Bulk modulus and the change in volume

Squeeze a body equally from all sides and its volume shrinks. The bulk modulus is the extra pressure needed for each unit of fractional loss of volume, so a large B means the body is hard to compress. Its reciprocal is the compressibility. The mass does not change, so a smaller volume means a larger density.

Definition

$B = -\dfrac{\Delta P}{\Delta V/V}$ . The minus sign makes B positive, since the volume falls as the pressure rises.
$\dfrac{\Delta V}{V} = \dfrac{\Delta P}{B}$ and $\Delta V = \dfrac{\Delta P\,V}{B}$ . Compressibility $= 1/B$ .
Rise in density: $\Delta\rho = \dfrac{\rho\,\Delta P}{B}$ .
A gas with $P = aV^{-n}$ : $B = -V\dfrac{dP}{dV} = nP$ . An ideal gas at constant temperature $(n = 1)$ has $B = P$ .
Two materials under the same pressure: $B \propto \dfrac{1}{\Delta V/V}$ .

Bulk modulus

B = -\frac{\Delta P}{\Delta V / V} \qquad \Delta V = \frac{\Delta P\,V}{B} \qquad \Delta\rho = \frac{\rho\,\Delta P}{B}

Worked example

A steel sphere of volume 2 litres is put under an extra pressure of

8 \times 10^{7}\ \text{Pa}

. The bulk modulus of steel is

1.6 \times 10^{11}\ \text{N/m}^{2}

. By how much does its volume fall?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 28 January 2025 · Q98Moderate

Example 1 · Mechanical Properties of Solids · Bulk Modulus, Shear Modulus and Poisson's Ratio

The volume contraction of a solid copper cube of edge length 10 cm , when subjected to a hydraulic pressure of

7 \times10^{6}\text{ }Pa

, would be

\_\_\_\_

mm^{3}

. (Given bulk modulus of copper

= 1.4 \times10^{11}Nm^{- 2}

)

A percentage is not a fraction

A 0.2% fall in volume is ΔV/V = 2 × 10⁻³, not 0.2. Divide by 100 before using the formula.

Volume units

1 litre is 10⁻³ m³, and 1 m³ is 10⁶ cm³ or 10⁹ mm³. An answer asked in mm³ needs the cube of the length conversion.

Drill 6 more on bulk modulus and the change in volume

Concept 2 of 3: Compression at a depth in water

Under water the extra pressure on a body is ρgh, the weight of the water column above it. Put that pressure into the bulk modulus and the fractional loss of volume follows; turn it round to find the depth that causes a given loss.

Definition

$\Delta P = \rho g h$ . Atmospheric pressure acts at the surface as well, so it adds nothing to the change.
$\dfrac{\Delta V}{V} = \dfrac{\rho g h}{B}$ and $h = \dfrac{B}{\rho g}\cdot\dfrac{\Delta V}{V}$ .
Hydraulic stress divided by hydraulic strain is the bulk modulus: $B = \dfrac{\rho g h}{\Delta V/V}$ .
Use the B of the body being squeezed: the ball's B for a ball taken down, water's B for the water at the bottom of a sea.

At a depth h

\frac{\Delta V}{V} = \frac{\rho g h}{B} \qquad h = \frac{B}{\rho g}\cdot\frac{\Delta V}{V}

Worked example

To what depth in water must a rubber ball

(B = 5 \times 10^{8}\ \text{N/m}^{2})

be taken so that it loses 0.1% of its volume?

(\rho = 1000\ \text{kg/m}^{3},\ g = 10\ \text{m/s}^{2})

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 31 January 2024 · Q28Moderate

Example 2 · Mechanical Properties of Solids · Bulk Modulus, Shear Modulus and Poisson's Ratio

The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by

0.02\%

\ \_\_\_\_\_\_ m

. (Take density of sea water

=10^{3}kgm^{- 3}

, Bulk modulus of rubber

= 9 \times10^{8}Nm^{- 2}

, and

g = 10{\text{ }ms}^{- 2}

)

The wrong body's bulk modulus

A rubber ball taken down is squeezed by the water, but it is the ball that shrinks, so its own B goes in the formula.

Adding atmospheric pressure

The ball already felt atmospheric pressure at the surface. Only the extra pressure ρgh changes its volume.

Drill 4 more on compression at a depth in water

Concept 3 of 3: Shear modulus, Poisson's ratio and the relations between the moduli

A shearing force acts along a face, not across it, and changes the body's shape rather than its volume. The top slides by x over the height h, so the shear strain is the angle θ = x/h. Separately, a stretched wire also gets thinner: Poisson's ratio compares that sideways strain with the lengthwise one.

Definition

Shear: $\eta = \dfrac{F/A}{\theta}$ with $\theta = \dfrac{x}{h}$ , so $x = \dfrac{Fh}{A\eta}$ . A is the face the force acts on; h is the distance from that face to the fixed one.
A square slab of side l and thickness d sheared on its narrow face: $A = ld$ , $h = l$ , so $\theta = \dfrac{F}{\eta l d}$ .
A cylinder of radius r sheared at its top: $\theta = \dfrac{F}{\pi r^{2}\eta}$ .
Poisson's ratio $\sigma = \dfrac{\text{lateral strain}}{\text{longitudinal strain}}$ , taken as a positive number.
$Y = 2\eta(1 + \sigma) = 3B(1 - 2\sigma)$ . Removing σ gives $\dfrac{9}{Y} = \dfrac{1}{B} + \dfrac{3}{\eta}$ ; removing Y gives $\sigma = \dfrac{3B - 2\eta}{6B + 2\eta}$ .
Kinds of stress: a normal force on one pair of faces (tensile or compressive) goes with Y, a tangential force with η, and an equal pressure on every face with B. The restoring force per unit area is the stress.

Shear and the relations

\eta = \frac{F/A}{x/h} \qquad Y = 2\eta(1 + \sigma) = 3B(1 - 2\sigma)

Worked example

A cube of jelly of side 10 cm has a modulus of rigidity of

2 \times 10^{4}\ \text{N/m}^{2}

. Its bottom is held fixed and a force of 6 N acts along its top face. How far does the top move?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 1 · Q21Moderate

Example 3 · Mechanical Properties of Solids · Bulk Modulus, Shear Modulus and Poisson's Ratio

A cube has side length 5 cm and modulus of rigidity

10^{5}\text{ }N/m^{2}

. The displacement produced by a force of 10 N in the upper face of cube is

\_\_\_\_

mm .

The area of the wrong face

A is the face the force acts on. A slab pushed on its narrow face has A = side × thickness, and the height is the side, not the thickness.

Shear strain is an angle

The strain is θ = x/h in radians, not the displacement x. Find θ first, then multiply by h if the question asks how far the top moves.

Drill 6 more on shear modulus, poisson's ratio and the relations between the moduli

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Bulk modulus and the change in volume
Bulk modulus
$B = -\frac{\Delta P}{\Delta V / V} \qquad \Delta V = \frac{\Delta P\,V}{B} \qquad \Delta\rho = \frac{\rho\,\Delta P}{B}$
Compression at a depth in water
At a depth h
$\frac{\Delta V}{V} = \frac{\rho g h}{B} \qquad h = \frac{B}{\rho g}\cdot\frac{\Delta V}{V}$
Shear modulus, Poisson's ratio and the relations between the moduli
Shear and the relations
$\eta = \frac{F/A}{x/h} \qquad Y = 2\eta(1 + \sigma) = 3B(1 - 2\sigma)$

Watch out for (6)

A percentage is not a fraction
→ Bulk modulus and the change in volume
Volume units
→ Bulk modulus and the change in volume
The wrong body's bulk modulus
→ Compression at a depth in water
Adding atmospheric pressure
→ Compression at a depth in water
The area of the wrong face
→ Shear modulus, Poisson's ratio and the relations between the moduli
Shear strain is an angle
→ Shear modulus, Poisson's ratio and the relations between the moduli

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