Q.
Two wires of the same material (Young’s modulus Y) and same length $L\ $but radii R and 2R respectively are joined end to end and a weight $w$ is suspended from the combination as shown in the figure. The elastic potential energy in the system is
A) $\frac{3w^{2}L}{4\pi R^{2}Y}\ $
B) $\frac{3w^{2}L}{8\pi R^{2}Y}$
C) $\frac{5w^{2}L}{8\pi R^{2}Y}$
D) $\frac{w^{2}L}{\pi R^{2}Y}\ $
Explanation
(c) $k_{1} = \ \ \frac{Y\pi(2R)^{2}}{L},\ k_{2} = \frac{Y\pi(R)^{2}}{L}$
Equivalent $\frac{1}{k_{1}} + \frac{1}{k_{2}} = \frac{L}{4Y\pi R^{2}} + \frac{L}{Y\pi R^{2}}\ $
Since, $k_{1}x_{1} = k_{2}x_{2} = w\ $
Elastic potential energy of the system
$U = \frac{1}{2}k_{1}x_{1}^{2} + \frac{1}{2}\ k_{2}x_{2}^{2}$
$U = \frac{1}{2}\ k_{1}\left( \frac{w}{k_{1}} \right)^{2} + \frac{1}{2}k_{2}\left( \frac{w}{k_{2}} \right)^{2}$
= $\frac{1}{2}w^{2}\ \left\{ \frac{1}{k_{1}} + \frac{1}{k_{2}} \right\} = \frac{1}{2}w^{2}\left( \frac{5L}{4Y\pi R^{2}} \right)$
$U$ = $\frac{5w^{2}L}{8\pi YR^{2}}\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.