Question

The ratio of the lengths of two wires A and B made of same material is 2:1. The diameter of wire A is twice the diameter of wire B. If both the wires are stretched by same tension, the ratio of the energies stored in wires A and B is:

• A. \( 1:4 \)
• B. \( 1:1 \)
• C. \( 1:2 \)
• D. \( 1:8 \)

Hint:

The energy stored in a stretched wire depends on the length, cross-sectional area, and the square of the diameter.

Answer 1

The Correct Option is C

Step 1:
The energy stored in a stretched wire is given by: \[ E = \frac{1}{2} \cdot \frac{T L}{A} \cdot \Delta L \] where: - \( E \) is the energy stored, - \( T \) is the tension in the wire, - \( L \) is the length of the wire, - \( A \) is the cross-sectional area of the wire, - \( \Delta L \) is the elongation of the wire.
Step 2:
The elongation \( \Delta L \) is proportional to \( \frac{L}{A} \) for a constant tension \( T \). Hence, we focus on the expression for the energy stored.
Step 3:
For wires A and B, the energy stored is proportional to: \[ E \propto \frac{L}{A} \] Since the length of wire A is twice that of wire B (\( L_A = 2L_B \)), and the diameter of wire A is twice that of wire B (\( d_A = 2d_B \)), the cross-sectional area \( A \) is proportional to \( d^2 \), so: \[ A_A = \pi \left( \frac{d_A}{2} \right)^2 = 4 \pi \left( \frac{d_B}{2} \right)^2 = 4A_B \]
Step 4:
The ratio of the energies stored in wires A and B is: \[ \frac{E_A}{E_B} = \frac{L_A / A_A}{L_B / A_B} = \frac{2L_B / 4A_B}{L_B / A_B} = \frac{2}{4} = \frac{1}{2} \]
Step 5:
Thus, the ratio of the energies stored in wires A and B is \( 1:2 \), so Option (3) is correct.