Question

Two copper wires A and B of lengths in the ratio \( 1:2 \) and diameters in the ratio \( 3:2 \) are stretched by forces in the ratio \( 3:1 \). The ratio of the elastic potential energies stored per unit volume in the wires A and B is

• A. \( 2:1 \)
• B. \( 4:9 \)
• C. \( 16:9 \)
• D. \( 4:3 \)

Hint:

Energy density depends on both force per unit area and strain. Consider all given ratios before applying the formula.

Answer 1

The Correct Option is C

Let \( L_A \) and \( L_B \) be the lengths of wires A and B, respectively.
Let \( d_A \) and \( d_B \) be the diameters of wires A and B, respectively.
Let \( F_A \) and \( F_B \) be the forces applied to wires A and B, respectively.
Given: \begin{itemize} \item \( \frac{L_A}{L_B} = \frac{1}{2} \) \item \( \frac{d_A}{d_B} = \frac{3}{2} \) \item \( \frac{F_A}{F_B} = \frac{3}{1} \) \end{itemize} The elastic potential energy stored per unit volume (energy density) is given by:
\[ U = \frac{1}{2} \times \text{stress} \times \text{strain} \] Also, stress \( \sigma = \frac{F}{A} \) and strain \( \epsilon = \frac{\sigma}{Y} \), where \( A \) is the cross-sectional area and \( Y \) is Young's modulus.
Therefore, \( U = \frac{1}{2} \frac{\sigma^2}{Y} = \frac{1}{2} \frac{F^2}{A^2 Y} \). Since both wires are copper, Young's modulus \( Y \) is the same for both wires. The cross-sectional area \( A = \pi (d/2)^2 = \frac{\pi d^2}{4} \).
Thus, \( U \propto \frac{F^2}{d^4} \).
We need to find the ratio \( \frac{U_A}{U_B} \).
\[ \frac{U_A}{U_B} = \frac{F_A^2/d_A^4}{F_B^2/d_B^4} = \left( \frac{F_A}{F_B} \right)^2 \times \left( \frac{d_B}{d_A} \right)^4 \]
Substituting the given ratios: \[ \frac{U_A}{U_B} = \left( \frac{3}{1} \right)^2 \times \left( \frac{2}{3} \right)^4 = 9 \times \frac{16}{81} = \frac{16}{9} \]
Therefore, the ratio of the elastic potential energies stored per unit volume in the wires A and B is \( 16:9 \). Final Answer: The correct answer is (3) \( 16:9 \).