JEE Main 2026 — Mechanical Properties of Solids Question with Solution

Given:
Side length of the cube, $L=5$ cm $=0.05$ m
Modulus of rigidity, $\eta =10^5$ N/m${}^2$
Force applied on the upper face, $F=10$ N

Area of the upper face, $A=L^2=(0.05{)}^2=2.5\times 10^{-3}$ m${}^2$

Shear stress is given by:
$\text{Stress}=\frac{F}{A}=\frac{10}{2.5\times 10^{-3}}=4000$ N/m${}^2$

Shear strain is given by:
$\text{Strain}=\frac{\Delta x}{L}$

Modulus of rigidity is defined as the ratio of shear stress to shear strain:
$\eta =\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta x/L}$

Rearranging for displacement $\Delta x$:
$\Delta x=\frac{F\cdot L}{A\cdot \eta }$

Substituting the values:
$\Delta x=\frac{10\times 0.05}{2.5\times 10^{-3}\times 10^5}$

$\Delta x=\frac{0.5}{250}=0.002$ m

Converting to millimeters:
$\Delta x=0.002\times 1000=2$ mm

Answer: $2$