The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. Then is the vertical deflection of this face

The vertical deflection of the face can be calculated as follows:$\mathrm{Shear} \mathrm{stress}= \frac{\mathrm{Shear} \mathrm{force}}{ \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{face} \mathrm{Shear} \mathrm{force}}$

$\mathrm{Shear} \mathrm{stress} =100\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2(\mathrm{mass}\times \mathrm{acceleration} \mathrm{due} \mathrm{to} \mathrm{gravity})$

Area of the face = ${\left(10\mathrm{cm}\right)}^2=0.01{\mathrm{m}}^2$

$\mathrm{Shear} \mathrm{stress}=\frac{100\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2}{0.01{\mathrm{m}}^2}=98000\mathrm{Pa}$

$\mathrm{Shear} \mathrm{strain} =\frac{ \mathrm{Shear} \mathrm{stress}}{\mathrm{Shear} \mathrm{modulus}}$

 $\mathrm{Shear} \mathrm{strain} = \frac{98,000 \mathrm{Pa}}{25 \mathrm{GPa}}= 3.92\times {10}^{-7}\mathrm{m}$

.$\mathrm{Shear} \mathrm{strain} \approx 4\times {10}^{-7}\mathrm{m}$

Hence the correct answer is $4\times {10}^{-7}\mathrm{m}$.