Question

Wire A and B have their Young’s moduli in the ratio 1:3, area of cross section in the ratio of 1:2 and lengths in ratio of 3:4. If same force is applied on the two wires to elongate then ratio of elongation is equal to

• A. 8:1
• B. 1:12
• C. 1:8
• D. 9:2

Answer 1

The Correct Option is D

To find the ratio of elongation of wires A and B, let's begin by using the formula for elongation due to a force applied on a wire:

\(\Delta L = \frac{F \cdot L}{A \cdot Y}\)

where:

• \(\Delta L\) is the elongation,
• \(F\) is the force applied,
• \(L\) is the original length of the wire,
• \(A\) is the cross-sectional area,
• \(Y\) is Young's modulus of the material.

Given the ratios:

• \(Y_A:Y_B = 1:3\)
• \(A_A:A_B = 1:2\)
• \(L_A:L_B = 3:4\)

Assuming the same force \(F\) is applied to both wires, the elongation ratio for A and B can be calculated as follows:

1. Calculate the expression for elongation for wire A: \(\Delta L_A = \frac{F \cdot L_A}{A_A \cdot Y_A}\)
2. Calculate the expression for elongation for wire B: \(\Delta L_B = \frac{F \cdot L_B}{A_B \cdot Y_B}\)
3. Find the ratio: \(\frac{\Delta L_A}{\Delta L_B} = \frac{F \cdot L_A}{A_A \cdot Y_A} \cdot \frac{A_B \cdot Y_B}{F \cdot L_B}\)
4. Simplifying the ratio: \(\frac{\Delta L_A}{\Delta L_B} = \frac{L_A \cdot A_B \cdot Y_B}{A_A \cdot L_B \cdot Y_A}\)
5. Substitute the given ratios: \(\frac{\Delta L_A}{\Delta L_B} = \frac{3 \cdot 2 \cdot 3}{1 \cdot 4 \cdot 1} = \frac{18}{4} = \frac{9}{2}\)

Therefore, the ratio of elongation of wire A to wire B is \(9:2\), which corresponds to the correct answer option.