Question

Young's moduli of the material of wires A and B are in the ratio of 1:4, while its area of cross sections are in the ratio of 1:3. If the same amount of load is applied to both the wires, the amount of elongation produced in the wires A and B will be in the ratio of (Assume length of wires A and B are same)

• A. 1:12
• B. 12:1
• C. 36:1
• D. 1:36

Hint:

When force and length are constant, elongation is inversely proportional to the product of area and Young's modulus ($ \Delta L \propto \frac{1}{AY} $). Just invert the given ratios and multiply them to get the answer quickly. \

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the ratio of elongation \((\Delta L)\) in two wires A and B given the ratios of their Young's moduli, cross-sectional areas, and assuming the same load and length for both.

Step 2: Key Formula or Approach:
Young's modulus \((Y)\) is defined as: \[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L} \] Rearranging for elongation \((\Delta L)\): \[ \Delta L = \frac{FL}{AY} \]

Step 3: Detailed Explanation:
Given: - Ratio of Young's moduli: \( \frac{Y_A}{Y_B} = \frac{1}{4} \)
- Ratio of areas: \( \frac{A_A}{A_B} = \frac{1}{3} \)
- Same load: \( F_A = F_B = F \)
- Same length: \( L_A = L_B = L \)
The ratio of elongations is: \[ \frac{\Delta L_A}{\Delta L_B} = \frac{\frac{FL}{A_A Y_A}}{\frac{FL}{A_B Y_B}} = \frac{A_B Y_B}{A_A Y_A} \] \[ \frac{\Delta L_A}{\Delta L_B} = \left( \frac{A_B}{A_A} \right) \times \left( \frac{Y_B}{Y_A} \right) = \left( \frac{3}{1} \right) \times \left( \frac{4}{1} \right) = 12 \]

Step 4: Final Answer:
The ratio of elongation produced in wires A and B is \(12:1\).