Question

Two wires A and B are made of the same material, having the ratio of lengths $ \frac{L_A}{L_B} = \frac{1}{3} $ and their diameters ratio $ \frac{d_A}{d_B} = 2 $. If both the wires are stretched using the same force, what would be the ratio of their respective elongations?

• A. \( 1 : 6 \)
• B. \( 1 : 12 \)
• C. \( 3 : 4 \)
• D. \( 1 : 3 \)

Hint:

When dealing with elongation problems, use the formula \( \Delta L = \frac{F L}{A Y} \) and focus on the relationship between the lengths and cross-sectional areas of the wires.

Answer 1

The Correct Option is B

The elongation \( \Delta L \) of a wire under a stretching force \( F \) is described by \( \Delta L = \frac{F L}{A Y} \), where \( L \) is the wire's length, \( A \) is its cross-sectional area, and \( Y \) is the Young's modulus. Given that \( Y \) is constant for both wires as they are of the same material, and the applied force \( F \) is also the same for both, the elongations are determined by the wire lengths and areas.

Step 1: Wire Dimensions

The lengths of wire A and wire B are related by \( \frac{L_A}{L_B} = \frac{1}{3} \), implying \( L_A = \frac{L_B}{3} \). The cross-sectional area \( A \) of a wire is given by \( A = \frac{\pi d^2}{4} \), where \( d \) is its diameter. The ratio of the areas of wires A and B is \( \frac{A_A}{A_B} = \left( \frac{d_A}{d_B} \right)^2 \). Since \( \frac{d_A}{d_B} = 2 \), it follows that \( \frac{A_A}{A_B} = 2^2 = 4 \), or \( A_A = 4 A_B \).

Step 2: Elongation Ratio Calculation

The ratio of the elongations \( \Delta L_A \) to \( \Delta L_B \) is computed as follows:

\[ \frac{\Delta L_A}{\Delta L_B} = \frac{F L_A / A_A Y}{F L_B / A_B Y} \]

This simplifies to:

\[ \frac{\Delta L_A}{\Delta L_B} = \frac{L_A}{L_B} \times \frac{A_B}{A_A} \]

Substituting the derived ratios:

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

Therefore, the ratio of the elongations is \( 1 : 12 \), corresponding to option (2).