A cube has side length 5 cm and modulus of rigidity $$10^5$$ N/m$$^2$$. The displacement produced by a force of 10 N in the upper face of cube is _______ mm.

Correct Answer: 2

Use shear relation:

shear modulus $$G=\frac{\text{stress}}{\text{strain}}$$

$$G=\frac{F/A}{x/L}\Rightarrow x=\frac{FL}{AG}$$

given:

side = 5 cm = 0.05 m
area $$A=(0.05)2=2.5\times10^{-3}m^2$$
L=0.05m
F=10
$$G=10^5$$

$$x=\frac{10\times0.05}{2.5\times10^{-3}\times10^5}$$

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

=2 mm

Create a FREE account and get:

Predict your JEE Main percentile, rank & performance in seconds

Check Your Score Now

Ask our AI anything

AI can make mistakes. Please verify important information.

Download resources