Question

A uniform wire of length $L$ and radius $r$ is twisted by an angle $\alpha$. If modulus of rigidity of the wire is $\eta$, then the elastic potential energy stored in wire, is

• A. $\frac{\pi \eta r^{4} \alpha}{2 L^{2}}$
• B. $\frac{\pi \eta r^{4} \alpha^{2}}{4 L}$
• C. $\frac{\pi \eta r^{4} \alpha}{4 L^{2}}$
• D. $\frac{\pi \eta r^{4} \alpha^{2}}{2 L}$

Answer 1

Answer: (B)

$U=$ work done
We know
Work done $=\frac{\pi S r^{4} \phi^{2}}{4 L}$
Where,
$\phi=$ Angle of twist $=\alpha$
$S=$ Modulus of rigidity $=4$
Substituting values
$U=\frac{\pi \eta r^{4} \alpha^{2}}{4 L}$