Question

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. $\frac{w^{2}L}{\pi R^{2}Y}$

Answer 1

The correct option is C $\frac{5w^{2}L}{8\pi R^{2}Y}$

As the wires are joined end to end, tension in both the wires is same and equal to weight $w$.

From Hookes law,

Elongation in the first wire is
$e_1=\frac{\text{wL}}{\pi R^{2}Y}$
Energy stored in the first wire is, $U_1=\frac{1}{2}w\times e_1=\frac{w^{2}L}{2\pi R^{2}Y}$

Elongation in the second wire is $e_2=\frac{wL}{\pi {\left(2R\right)}^{2}Y}$
Energy stored in the second wire is,
$U_2=\frac{1}{2}w\times e_2=\frac{w^{2}L}{8\pi R^{2}Y}$

$\therefore$ Total potential energy stored is =$U_1+U_2=\frac{5w^{2}L}{8\pi R^{2}Y}$