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 is given by the formula: \[ \Delta L = \frac{F L}{A Y} \] where: \( F \) is the applied force, \( L \) is the length of the wire, \( A \) is the cross-sectional area, \( Y \) is the Young's modulus (which is constant for both wires as they are made of the same material). 
Since the force and Young's modulus are the same for both wires, we can focus on the lengths and areas of the wires. 
Step 1: Lengths and areas of the wires
The length of wire A and wire B are related by: \[ \frac{L_A}{L_B} = \frac{1}{3} \] So, \( L_A = \frac{L_B}{3} \). The cross-sectional area \( A \) of a wire is related to its diameter \( d \) by the formula: \[ A = \frac{\pi d^2}{4} \] The ratio of the areas of wires A and B is: \[ \frac{A_A}{A_B} = \left( \frac{d_A}{d_B} \right)^2 = 2^2 = 4 \] 
Thus, \( A_A = 4 A_B \). 
Step 2: Ratio of elongations
The ratio of the elongations \( \Delta L_A \) and \( \Delta L_B \) is given by: \[ \frac{\Delta L_A}{\Delta L_B} = \frac{F L_A / A_A Y}{F L_B / A_B Y} \] Simplifying: \[ \frac{\Delta L_A}{\Delta L_B} = \frac{L_A}{L_B} \times \frac{A_B}{A_A} \] Substituting the known values: \[ \frac{\Delta L_A}{\Delta L_B} = \frac{1/3}{1} \times \frac{1}{4} = \frac{1}{12} \] 
Thus, the ratio of the elongations is: \[ 1 : 12 \] So, the correct answer is option (2): \( 1 : 12 \).