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A cube of aluminium of sides $0.1 \mathrm{~m}$ is subjected to a shearing force of $100 \mathrm{~N}$. The top face of the cube is displaced through $0.02 \mathrm{~cm}$ with respect to the bottom face. The shearing strain would be
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0.002
Shearing strain $\phi=\frac{x}{L}=\frac{0.02 \mathrm{~cm}}{10 \mathrm{~cm}} \quad \therefore \phi=0.002$
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