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 stored 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 $\frac{5W^{2}L}{8\pi R^{2}Y}$

The elongation produced in a wire is,

$\Delta l=\frac{FL}{YA}$

$\therefore \Delta l_1=\frac{WL}{\left(4\pi R^2\right)Y}$ and $\Delta l_2=\frac{WL}{\pi R^{2}Y}$

$U=\frac{1}{2}{K}_1(\Delta l_1{)}^2+\frac{1}{2}{K}_2(\Delta l_2{)}^2$

The equivalent spring constant of an elastic wire is given by, $K=\frac{YA}{L}$

$\therefore U=\frac{1}{2}\times \frac{Y\left(4\pi R^2\right)}{L}\times {\left[\frac{WL}{4\pi R^{2}Y}\right]}^2$
$+\frac{1}{2}\times \frac{Y\left(\pi R^2\right)}{L}\times {\left[\frac{WL}{\pi R^{2}Y}\right]}^2$

$=\frac{5W^{2}L}{8\pi R^{2}Y}$

Hence, $\left(C\right)$ is the correct answer.

Why this Q? Elastic Rods can be compared with springs for easy calculation of force and elastic potential energy.