Question

The two wires A and B of the same material have their lengths in the ratio 1 : 2 and their diameters in the ratio 2 : 1. If they are stretched with the same force, the ratio of the increase in the length of A to that of B will be:

• A. 1 : 2
• B. 4 : 1
• C. 1 : 8
• D. 1 : 4

Hint:

The extension in the wire is inversely proportional to the cross-sectional area and directly proportional to the length of the wire.

Answer 1

The Correct Option is B

The extension in the wire is given by Hooke's law: \[ \Delta L = \frac{F L}{A Y} \] where: - \( F \) is the force, - \( L \) is the length of the wire, - \( A \) is the cross-sectional area of the wire, - \( Y \) is the Young's modulus. The cross-sectional area \( A \) is proportional to the square of the diameter of the wire. The ratio of the extensions of A and B is: \[ \frac{\Delta L_A}{\Delta L_B} = \frac{L_A A_B}{L_B A_A} \] Substituting the given ratios \( L_A : L_B = 1 : 2 \) and \( A_A : A_B = 2^2 : 1^2 = 4 : 1 \): \[ \frac{\Delta L_A}{\Delta L_B} = \frac{1 \times 1}{2 \times 4} = \frac{1}{8} \] Thus, the ratio is 4 : 1.