b
(b)
$\text { Modulus of rigidity }(G)=\frac{\text { Force } \times \text { Length }}{\text { Area } \times \text { Lateral displacement }}=\frac{F L}{A \times \Delta x}$
$F=10 \,kN =10 \times 10^3 \,N$
$L=4 \,cm =0.04 \,m$
$A=1 \,cm ^2=1 \times 10^{-4} \,m ^2$
$G=8 \times 10^{11}\,N / m ^2$
Substituting values
$8 \times 10^{11}=\frac{10 \times 10^3 \times 0.04}{1 \times 10^{-4} \times \Delta x}$
$\Delta x=\frac{10 \times 10^3 \times 0.04}{1 \times 10^{-4} \times 8 \times 10^{11}}=5 \times 10^{-6} \,m$
