JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Elasticity & SHM

A cube has side length 5 cm and modulus of rigidity

1 0^{5}

N/m

^{2}

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

Answer: 2

Solution

Under a tangential force the shear strain is

ϕ = \frac{x}{h}

, and the modulus of rigidity is

η = \frac{F / A}{x / h} = \frac{F h}{A x}

, so

x = \frac{F h}{A η}

. With

F = 10

h = 5 cm = 0.05

A = (0.05)^{2} = 25 \times 1 0^{- 4} m^{2}

and

η = 1 0^{5} N/m^{2}

x = \frac{10 \times 0.05}{25 \times 1 0 ^{- 4} \times 1 0 ^{5}} = \frac{0.5}{250} = 2 \times 1 0^{- 3}

m. The upper face is displaced by

2

mm.

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Integer answer