Question

Young's moduli of the material of wires A and B are in the ratio of 1:4, while its area of cross section 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 in the ratio of
[Assume length of wires A and B are same]

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

Answer 1

The Correct Option is D

To solve this problem, we need to understand the relation between the elongation produced in a wire, its material properties, and its dimensions.

For a wire, the elongation (\Delta L) produced by applying a force can be calculated using Hooke's Law, described by the formula:

\Delta L = \frac{FL}{AE}

• F = Force applied
• L = Original length of the wire
• A = Area of cross-section of the wire
• E = Young's modulus of the material

According to the question:

• Young's moduli ratio of wires A and B = 1:4 (E_A:E_B = 1:4)
• Area of cross-section ratio of wires A and B = 1:3 (A_A:A_B = 1:3)

The formula for the elongation of wire A can be written as:

\Delta L_A = \frac{FL}{A_A E_A}

Similarly, for wire B:

\Delta L_B = \frac{FL}{A_B E_B}

To find the ratio of elongations, we divide the expression for \Delta L_A by \Delta L_B:

\frac{\Delta L_A}{\Delta L_B} = \frac{\frac{FL}{A_A E_A}}{\frac{FL}{A_B E_B}}

This simplifies to:

\frac{\Delta L_A}{\Delta L_B} = \frac{A_B E_B}{A_A E_A}

Substitute the given ratios:

\frac{\Delta L_A}{\Delta L_B} = \frac{3 \times 4}{1 \times 1} = \frac{12}{1}

Thus, the ratio of the elongation of wire A to wire B is 12:1.

Therefore, the correct answer is: 12:1