A rubber cube of side 5 ~cm has one face fixed, while a tangential...

Shear stress =$\frac{\text { Tangential force }}{\text { Area }}=\frac{1800 \mathrm{~N}}{(0.05)^2}$Shear strain $=\tan \theta$$\begin{aligned}& =\frac{x}{h}=\frac{\text { Lateral displacement }}{\text { Height of cube }}=\frac{x}{5} \mathrm{~cm}\\ & =\frac{x \mathrm{~m}}{0.05 \mathrm{~m}} \end{aligned}$Modulus of rigidity, \eta=$\frac{\text { Shear stress }}{\text { Shear strain }}$2.4 $\times$10^6 $\frac{\mathrm{N}}{\mathrm{m}^2}=\frac{1800}{25 \times 10^{-4}} \times \frac{0.05}{x} \mathrm{~N}$/ $\mathrm{m}^2 or\begin{aligned}x \text{ and } =\frac{1800 \times 0.05}{25 \times 10^{-4} \times 2.4 \times 10^6}\\ \text{ and } =15 \times10^{-3} \mathrm{~m}\\ \text{ and } =15 \mathrm{~mm} \end{aligned}$$