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 Concept:
Young's modulus ($Y$) measures the resistance of a material to elastic deformation under load. The elongation ($\Delta L$) produced in a wire depends on the applied force, the original length, the cross-sectional area, and the material's Young's modulus.
Step 2: Key Formula or Approach:
The formula for Young's modulus is $Y = \frac{F L}{A \Delta L}$.
Rearranging for elongation: $\Delta L = \frac{F L}{A Y}$.
Step 3: Detailed Explanation:
1. Let the parameters for wire A be $F, L, A_A, Y_A$ and for wire B be $F, L, A_B, Y_B$.
2. Given ratios: $\frac{Y_A}{Y_B} = \frac{1}{4}$ and $\frac{A_A}{A_B} = \frac{1}{3}$.
3. Since load ($F$) and length ($L$) are same for both: \[ \Delta L \propto \frac{1}{A Y} \]
4. Taking the ratio of elongations: \[ \frac{\Delta L_A}{\Delta L_B} = \frac{A_B Y_B}{A_A Y_A} = \left(\frac{A_B}{A_A}\right) \times \left(\frac{Y_B}{Y_A}\right) \]
5. Substitute the values: \[ \frac{\Delta L_A}{\Delta L_B} = \left(\frac{3}{1}\right) \times \left(\frac{4}{1}\right) = \frac{12}{1} \]
Step 4: Final Answer
The ratio of elongation produced in wires A and B is 12:1.