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
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(d) Shearing strain $\varphi = \frac{x}{L} = \frac{{0.02cm}}{{10cm}}$
$\varphi = 0.002$
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