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. $\frac{3WL}{8\pi R^{2}Y}$
• 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 $\frac{5W^{2}L}{8\pi R^{2}Y}$

Force constant of wires can be considered as
$k_1=\frac{Y\pi (2R{)}^2}{L},k_2=\frac{Y\pi (R{)}^2}{L}$
Hence, equivalent force constant $k$
$\frac{1}{k}=\frac{1}{k_1}+\frac{1}{k_2}$ [As two wires are joined in series]
$\frac{1}{k}=\frac{L}{4Y\pi R^2}+\frac{L}{Y\pi R^2}=\frac{5L}{4Y\pi R^2}$
Since, $k_1{x}_1=k_2{x}_2=W$
Elastic potential energy of the system is
$U=\frac{1}{2}{k}_1{x}_1^2+\frac{1}{2}{k}_2{x}_2^2$
$=\frac{1}{2}{k}_1{\left(\frac{W}{k_1}\right)}^2+\frac{1}{2}{k}_2{\left(\frac{W}{k_2}\right)}^2$
$=\frac{1}{2}{W}^2[\frac{1}{k_1}+\frac{1}{k_2}]$
$=\frac{1}{2}{W}^2\left[\frac{5L}{4Y\pi R^2}\right]=\frac{5W^{2}L}{8\pi Y{R}^2}$
Hence, the correct answer is option (c),