Question

A cube of aluminium of side $0.1 m$ is subjected to a shearing force of $100 \,N$. The top face of the cube is displaced through $0.02 \,cm$ with respect to the bottom face. The shearing strain would be

• A. $0.02$
• B. $0.1$
• C. $0.005$
• D. $0.002$

Answer 1

Answer: (D)

Given, length of cube, $L=0.1 \,m$,
Shearing force, $F=100\, N$
and displaced distance,$x=0.02 \,cm $
$=0.02 \times 10^{-2} \,m$
Shearing strain, $\phi=\frac{x}{L}$
$=\frac{0.02 \times 10^{-2}}{0.1 m }=0.002$