If a force F displace 0.25cm with respect to bottom.What is F if modulus of rigidity is $5\times 10^5$.

$\begin{array}{1 1}(A)\;50N\\(B)\;94N\\(C)\;107N\\(D)\;25N\end{array}$

Given $\Delta x=.25cm$
$l=2cm$
$\theta=\large\frac{.25}{2}=\frac{1}{8}$
$\eta=5\times 10^5$
$\eta=(\large\frac{F}{a})\times \frac{l}{\Delta x}$
$5\times 10^5=\large\frac{F}{3\times 5\times 10^{-4}}\times \frac{2}{.25}$
$F=\large\frac{5\times 10^5\times 3\times 5\times 10^{-4}}{8}$
$\;\;\;=93.75N$
$\;\;\;\approx 94N$
Hence (B) is the correct answer.