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

As we know, using spring-rod analogy,
$k=\frac{YA}{L}$
where, $k=$ Spring constant
$Y=$ Young's modulus
So, $k_1=\frac{Y\pi (2R{)}^2}{L},k_2=\frac{Y\pi (R{)}^2}{L}$
Equivalent spring constant $=\frac{1}{k_1}+\frac{1}{k_2}=\frac{L}{4Y\pi R^2}+\frac{L}{Y\pi R^2}=\frac{5L}{4Y\pi R^2}$

Since block is in equilibrium, hence
$k_1{x}_1=k_2{x}_2=w$
[for springs in series. force acting is the same]
Elastic potential energy of the system
$U=\frac{1}{2}{k}_1{x}_1^2+\frac{1}{2}{k}_2{x}_2^2$
$U=\frac{1}{2}{k}_1{\left(\frac{w}{k_1}\right)}^2+\frac{1}{2}{k}_2{\left(\frac{w}{k_2}\right)}^2$
$U=\frac{1}{2}{w}^2\{\frac{1}{k_1}+\frac{1}{k_2}\}=\frac{1}{2}{w}^2\left(\frac{5L}{4Y\pi R^2}\right)$
$\therefore U=\frac{5w^{2}L}{8\pi Y{R}^2}$