A 0.05 m cube has its upper face displaced by 0.2 cm by a tangential force of 8 N. Calculate the modulus of rigidity $\left(\mathrm{in}\times {10}^4{\mathrm{Nm}}^{-2}\right)$ of the material of the cube.

$\mathrm{l}=5\times {10}^{-2}\mathrm{m},\mathrm{\Delta l}=0.2\mathrm{cm}=0.2\times {10}^{-2}\mathrm{m},\mathrm{F}=8\mathrm{N}$

$\text{ Shearing strain }=\frac{\mathrm{\Delta l}}{\mathrm{l}}=\frac{0.2}{5}=0.04$

$\text{ Shearing stress }=\frac{\mathrm{F}}{\mathrm{l}\times \mathrm{l}}=\frac{8}{{\left(5\times {10}^{-2}\right)}^2}=3200{\mathrm{Nm}}^{-2}$

$\text{η= }\frac{\text{Shearing stress }}{\mathrm{Shearing} strain}=\frac{3200}{0.04}$$=80000{\mathrm{Nm}}^{-2}=8\times {10}^4{\mathrm{Nm}}^{-2}$