JEE Main 2026 — Apr 5, Shift 1 — Physics solution

JEE Main2026Apr 5, Shift 1Physics

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.

Your answer

Solution

Modulus of rigidity relates shear stress to shear strain: $η = \frac{F / A}{x / h}$ , so $x = \frac{F h}{A η}$ .

With $F = 10$ N, $h = 5$ cm $= 5 \times 1 0^{- 2}$ m, $A = (5 \times 1 0^{- 2})^{2} = 25 \times 1 0^{- 4}$ m $^{2}$ , $η = 1 0^{5}$ N/m $^{2}$ :

$x = \frac{10 \times 5 \times 1 0 ^{- 2}}{25 \times 1 0 ^{- 4} \times 1 0 ^{5}} = \frac{0.5}{250} = 2 \times 1 0^{- 3}$ m $= 2$ mm.

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →