A 2.0cm cube of some substance has its upper face displaced by 0.15 cm, by a tangential force of 0.30N fixing its lower face. Calculate the rigidity modulus of the substance.

Given,

$$\begin{gathered} \mathrm{Force} \mathrm{F}=0.30\mathrm{N} \\ \mathrm{Area}=\mathrm{A}=4\times {10}^{-2}{\mathrm{m}}^2 \end{gathered}$$

we know that the shearing strain,

$\theta =\frac{∆L}{L}$

$\theta =\frac{0.15}{2.0}; \theta =0.075$

Now the shearing stress is given by,

$\frac{\mathrm{F}}{\mathrm{A}}=\frac{0.30}{4\times {10}^{-4}} ;\frac{\mathrm{F}}{\mathrm{A}}=0.075\times {10}^4$

$$\begin{gathered} \mathrm{Rigidity} \mathrm{modulus}, \\ \mathrm{\eta }=\frac{\mathrm{F}}{\mathrm{A}\times \mathrm{\theta }} \end{gathered}$$

$\mathrm{\eta }=\frac{0.075\times {10}^4}{0.075}; \mathrm{\eta }=1\times {10}^4\mathrm{N}/{\mathrm{m}}^2$

Hence the correct answer is $1\times {10}^{4}N/m^2.$