Here, side of cube, L = 10 cm = 0.1 m $A=$ Area of the face $(1)=L \times L = L^{2} = (0.1)^{2} = 0.01 \text{m}^{2}$ Mass attached to face $(1), M=100 \text{kg}$ If F be the tangential force on face (1) due to this mass, then $F=M g=100 \times 9.8 \text{N}$ ∴ Shear stress on the face $= \frac{F}{A}= \frac{100 \times 9.8}{0.01} \text{N} \text{m}^{-2}$ Shear modulus of aluminium, $\eta = 25 \text{GPa} = 25 \times 10^{9} \text{N} \text{m}^{-2}$ We know that $\eta = \frac{\text{ Shearing stress }}{\text{ Shearing strain }}$Let $\Delta y$ = vertical displacement of the face = ?$\therefore$ Shearing strain $= \frac{\Delta y}{L}= \frac{\text{ Shearing stress}}{\eta}$$\Delta y = \frac{\text{Shearing stress }}{\eta} \times L$$= \frac{9.8 \times 10^{4} \times 0.1}{25 \times 10^{9}}$$=0.0392 \times 10^{-5} \text{m}$$=0.0392 \times 10^{-3} \text{cm}$$=4 \times 10^{-7} \text{m}$