Question

Two wires of the young's modulii $Y$ and $2Y$ having lengths $2L,L$ and radii $2R,R$ respectively, are joined end to end as shown in the image. The elastic potential energy stored in the system in equilibrium, is


[Assume the wires are massless and $w$ is the weight hung at the bottom]

• A. $\frac{4w^{2}L}{\pi R^{2}Y}$
• B. $\frac{w^{2}L}{4\pi R^{2}Y}$
• C. $\frac{2w^{2}L}{\pi R^{2}Y}$
• D. $\frac{w^{2}L}{2\pi R^{2}Y}$

Answer 1

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

Potential energy stored in the system is given by $E=\frac{F^{2}L}{2AY}$
where $F=$ Force acting on wire
$L=$ Length of wire
$A=$ Area of cross section
$Y=$ Young's Modulus

Total potential energy of system
$E=\frac{w^{2}2L}{2AY}+\frac{w^{2}L}{2A\left(2Y\right)}$
[Here, $w$ is the force acting on each wire]
$\Rightarrow E=\frac{w^2\times 2L}{2\left(\pi \right(2R{)}^2)Y}+\frac{w^{2}L}{2\left(\pi R^2\right)\times 2Y}$
$\Rightarrow E=\frac{2w^{2}L}{8\pi R^{2}Y}+\frac{w^{2}L}{4\pi R^{2}Y}$
$\Rightarrow E=\frac{w^{2}L}{4\pi R^{2}Y}+\frac{w^{2}L}{4\pi R^{2}Y}$
$\Rightarrow E=\frac{2w^{2}L}{4\pi R^{2}Y}$
$\Rightarrow E=\frac{w^{2}L}{2\pi R^{2}Y}$
Hence, the correct option is (d).