Question

A metal cube of side $10cm$ is subjected to a shearing stress of ${10}^{4}N{m}^{-2}$. The modulus of rigidity if the top of the cube is displaced by $0.05cm$ with respect to its bottom is

• A. $2\times {10}^{6}N{m}^{-2}$
• B. ${10}^{5}N{m}^{-2}$
• C. $1\times {10}^{7}N{m}^{-2}$
• D. $4\times {10}^{5}N{m}^{-2}$

Answer 1

The correct option is A $2\times {10}^{6}N{m}^{-2}$

We know that shearing stress$=\frac{F}{A}={10}^{4}N{m}^{-2}$

Length of side of cube$=10cm=\frac{10}{100}m=0.1m$

Shearing displacement$=△n=0.05cm$

$=\frac{0.05}{100}m=0.0005m$

We know that modulus of rigidity is $\eta =\frac{F}{A\theta }$

$=\frac{FL}{A△x}[\tan \theta \approx \theta =\frac{△x}{L}]$

$=\frac{{10}^4\times 0.1}{5\times {10}^{-4}}\Rightarrow \frac{{10}^7}{5^1}\Rightarrow 2\times {10}^{6}N{m}^{-2}$

Hence, modulus of rigidity is equal to $2\times {10}^{6}N{m}^{-2}$

(Image)