Question

Under the same load, wire A having length 5.0 m and cross-section \( 2.5 \times 10^{-5} \, \text{m}^2 \) stretches uniformly by the same amount as another wire B of length 6.0 m and a cross-section \( 3.0 \times 10^{-5} \, \text{m}^2 \) stretches. The ratio of the Young's modulus of wire A to that of wire B will be:

• A. 1:4
• B. 1:1
• C. 1:10
• D. 1:2

Hint:

Young’s modulus is a material property and remains the same for identical materials, even if length and cross-sectional area differ.

Answer 1

The Correct Option is B

Step 1: Young’s Modulus Formula
Young’s modulus \( Y \) is defined as: \[ Y = \frac{F L}{A \Delta L} \] where:
\( F \) denotes the applied force,
\( L \) is the wire's length,
\( A \) represents the cross-sectional area,
\( \Delta L \) is the elongation.
Step 2: Ratio of Young’s Modulus for Both Wires
Given that both wires exhibit the same elongation under identical forces, the relationship is established as: \[ \frac{Y_A}{Y_B} = \frac{F_A \cdot L_A}{A_A} \times \frac{A_B}{F_B \cdot L_B} \] Substituting the provided values: \[ \frac{Y_A}{Y_B} = \frac{5.0}{6.0} \times \frac{3.0 \times 10^{-5}}{2.5 \times 10^{-5}} \] The simplified result is: \[ \frac{Y_A}{Y_B} = \frac{5}{6} \times \frac{6}{5} = 1 \]Final Answer: The ratio of Young's modulus for wire A to wire B is 1:1 .