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

Answer 1

Answer: (B)

$\frac{\pi \eta r^4{\alpha }^2}{4L}$

$\begin{array}{c}\text{Given, wire of length}=L\text{and}\\ \text{radius}=r\\ \text{medulus of rigidity}=\eta \end{array}$

Due to twisting there will shear stress and torque will be generated.

$\tau =\frac{\alpha \cdot \eta \cdot \gamma }{L}\begin{array}{c}\text{where}\tau =\text{torque}\\ n=\text{modulus of}\\ \text{rigidity}\end{array}$$\alpha =angletwrited$

$\begin{array}{c}\text{Now, elastic potential energy stored}\\ E=\frac{{\tau }^2}{4n}\times \text{volume}\end{array}$

$\begin{array}{c}=\frac{{\alpha }^2\cdot {\eta }^2\cdot r^2}{L^2\cdot 4\eta }\times \pi r^{2}L\\ E=\frac{\pi \eta r^4{\alpha }^2}{4L}\end{array}$