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. The elastic potential energy 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. \( \frac{W^2 L}{\pi R^2 Y} \)

Hint:

Elastic potential energy is stored in the stretching of both wires, with energy distributed based on their cross-sectional areas.

Answer 1

The Correct Option is C

Step 1: {Calculate elongations}
\[ \Delta l_1 = \frac{WL}{(4 \pi R^2)Y}, \quad \Delta l_2 = \frac{WL}{\pi R^2 Y} \] Step 2: {Find elastic potential energy}
\[ U = \frac{1}{2} K_1 (\Delta l_1)^2 + \frac{1}{2} K_2 (\Delta l_2)^2 \] \[ = \frac{1}{2} \times \frac{Y(4 \pi R^2)}{L} \times \left( \frac{WL}{4 \pi R^2 Y} \right)^2 + \frac{1}{2} \times \frac{Y(\pi R^2)}{L} \times \left( \frac{WL}{\pi R^2 Y} \right)^2 \] \[ = \frac{5W^2 L}{8 \pi R^2 Y} \]