Question

A $0.05\text{m}$ cube has its upper face displaced by $0.2\text{cm}$ by a tangential force of $8\text{N}$. Calculate the modulus of rigidity of the material of the cube.

• A. $8\times {10}^4{\text{N/m}}^2$
• B. $16\times {10}^4{\text{N/m}}^2$
• C. $24\times {10}^4{\text{N/m}}^2$
• D. $32\times {10}^4{\text{N/m}}^2$

Answer 1

The correct option is A $8\times {10}^4{\text{N/m}}^2$

$l=5\times {10}^{-2}\text{m},\Delta l=0.2\text{cm}=0.2\times {10}^{-2}\text{m}$
$F=8\text{N}$

Shearing strain$=\frac{\Delta l}{l}=\frac{0.2}{5}=0.04$

Shearing stress$=\frac{F}{l\times l}=\frac{8}{(5\times {10}^{-2}{)}^2}=3200{\text{N/m}}^2$

Modulus of rigidity is given by,

$\eta =\frac{\text{Shearing stress}}{\text{Shearing strain}}=\frac{3200}{0.04}=8\times {10}^4{\text{N/m}}^2$

Hence, $\left(A\right)$ is the correct answer.

Why this question? Modulus of Rigidity is, $\eta =\frac{\text{Shearing stress}}{\text{Shearing strain}}=\frac{F}{A\varphi }$