Question

The metal cube of side $10\, cm$ 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 $0.05 \,cm$ with respect to its bottom is

• A. $2\times10^{6} N$ $m^{-2}$
• B. $10^{5}N$ $m^{-2}$
• C. $1\times10^{7} N$ $m^{-2}$
• D. $4\times10^{5} N$ $m^{-2}$

Answer 1

Answer: (A)

Here, $l=10 \,cm$, $\Delta l=0.05 \,cm$
Shearing stress$=10^{4} N$ $m^{-2}$
Shearing strain$=\frac{\Delta l}{l}$ $=\frac{0.05}{10}$
$\therefore \quad$ $\eta$ $=\frac{Shearing \,stress}{Shearing \,strain}$ $=\frac{10^{4}}{0.005}$ $=2\times10^{6} N m^{-2}$