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}
\[ U = \frac{1}{2} K_1 (\Delta l_1)^2 + \frac{1}{2} K_2 (\Delta l_2)^2 \] \[ = \frac{1}{2} \times \frac{Y(4 \pi R^2)}{L} \times \left( \frac{WL}{4 \pi R^2 Y} \right)^2 + \frac{1}{2} \times \frac{Y(\pi R^2)}{L} \times \left( \frac{WL}{\pi R^2 Y} \right)^2 \] \[ = \frac{5W^2 L}{8 \pi R^2 Y} \]

Answer 2

Step 1: The formula for elastic potential energy stored in a stretched wire is:
\( U = \frac{1}{2} \cdot \frac{F^2 L}{A Y} \), where:
- \( F \) is the force (here, equal to weight \( W \))
- \( L \) is the length of the wire
- \( A \) is the cross-sectional area
- \( Y \) is Young's modulus

Step 2: There are two wires in series:
- Wire 1: radius \( R \), length \( L \), area \( A_1 = \pi R^2 \)
- Wire 2: radius \( 2R \), length \( L \), area \( A_2 = \pi (2R)^2 = 4\pi R^2 \)

Step 3: Since they are in series, both wires experience the same force \( F = W \).
Compute the energy in each wire using the formula:
\( U = \frac{1}{2} \cdot \frac{W^2 L}{A Y} \)

Step 4: Energy stored in wire 1 (thin wire):
\( U_1 = \frac{1}{2} \cdot \frac{W^2 L}{\pi R^2 Y} \)

Step 5: Energy stored in wire 2 (thicker wire):
\( U_2 = \frac{1}{2} \cdot \frac{W^2 L}{4\pi R^2 Y} \)

Step 6: Total energy in the system:
\( U = U_1 + U_2 = \frac{1}{2} \cdot \frac{W^2 L}{\pi R^2 Y} + \frac{1}{2} \cdot \frac{W^2 L}{4\pi R^2 Y} \)
Take common factors:
\( U = \frac{W^2 L}{2\pi R^2 Y} \left(1 + \frac{1}{4}\right) = \frac{W^2 L}{2\pi R^2 Y} \cdot \frac{5}{4} = \frac{5W^2 L}{8\pi R^2 Y} \)

Final Answer: \( \frac{5W^2 L}{8\pi R^2 Y} \)