Q74 / 100 · BITSAT 2024 (Memory Based Paper 3) · Physics
Two wires of the same material (Young's modulus $=Y$ ) and same length $L$ but radii $R$ and $2 R$ 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$\frac{3 w^2 L}{4 \pi R^2 Y}$
- B$\frac{3 w^2 L}{8 \pi R^2 Y}$
- Correct answerC$\frac{5 w^2 L}{8 \pi R^2 Y}$
- D$\frac{w^2 L}{\pi R^2 Y}$
Solution
$\begin{aligned} & \Delta l_1=\frac{w L}{\left(4 \pi R^2\right) Y}, \Delta l_2=\frac{w L}{\pi R^2 Y} \\ & \therefore U=\frac{1}{2} K_1\left(\Delta l_1\right)^2+\frac{1}{2} K_2\left(\Delta l_2\right)^2 \\ &=\frac{1}{2} \times \frac{Y\left(4 \pi R^2\right)}{L} \times\left[\frac{w L}{4 \pi R^2 Y}\right]^2+\frac{1}{2} \times \frac{Y\left(\pi R^2\right)}{L} \\ & \times\left[\frac{w L}{\pi R^2 Y}\right]^2\left(\because K=\frac{Y A}{L}\right) \\ &=\frac{5 w^2 L}{8 \pi R^2 Y}\end{aligned}$
Hence, the correct answer is option: 'C'.