A cube of aluminium of sides $0.1\, m$ 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$0.002$
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D$0.002$
d (d) Shearing strain $\varphi = \frac{x}{L} = \frac{{0.02cm}}{{10cm}}$
$\varphi = 0.002$
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