Question

Two wires A and B of same cross-section are connected end to end. When same tension is created in both wires, the elongation in B is twice that of A. If $ L_A $ and $ L_B $ are initial lengths and the Young's modulus are: $$ Y_A = 2 \times 10^{11} \text{ Nm}^{-2}, \quad Y_B = 1.1 \times 10^{11} \text{ Nm}^{-2} $$ Then the ratio $ \frac{L_A}{L_B} $ is:

• A. \( \dfrac{10}{11} \)
• B. \( \dfrac{4}{5} \)
• C. \( \dfrac{9}{11} \)
• D. \( \dfrac{3}{7} \)

Hint:

When same force is applied, elongation is proportional to \( \frac{L}{Y} \) for identical cross-sectional area.

Answer 1

The Correct Option is A

Elongation is given by: \[ \Delta L = \frac{FL}{AY} \] Given: \( \Delta L_B = 2 \Delta L_A \), and \( A \) is same for both. \[ \frac{F L_B}{A Y_B} = 2 \cdot \frac{F L_A}{A Y_A} \Rightarrow \frac{L_B}{Y_B} = 2 \cdot \frac{L_A}{Y_A} \Rightarrow \frac{L_A}{L_B} = \frac{Y_A}{2Y_B} = \frac{2 \times 10^{11}}{2 \cdot 1.1 \times 10^{11}} = \frac{10}{11} \]