Question

Two wires A and B are made of the same material. Their diameters are in the ratio 1:2 and lengths are in the ratio 1:3. If they are stretched by the same force, then increase in their lengths will be in the ratio of

• A. 3 : 2
• B. 4 : 3
• C. 3 : 4
• D. 2 : 3

Hint:

When dealing with the stretching of wires under the same force, the increase in length 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 increase in the length of a wire under an applied force is given by the formula: \[ \Delta L = \frac{F L}{A Y} \] where: - \( F \) is the applied force, - \( L \) is the original length of the wire, - \( A \) is the cross-sectional area of the wire, and - \( Y \) is Young's modulus of the material. Since the wires are made of the same material, the value of \( Y \) is the same for both wires. Also, the applied force \( F \) is the same for both wires. Therefore, the increase in length is proportional to the ratio \( \frac{L}{A} \). The cross-sectional area \( A \) of a wire with diameter \( d \) is given by: \[ A = \frac{\pi d^2}{4} \] Let the diameter of wire A be \( d_1 \) and the diameter of wire B be \( d_2 \). Since \( d_1 : d_2 = 1 : 2 \), we can write: \[ A_1 = \frac{\pi d_1^2}{4}, \quad A_2 = \frac{\pi d_2^2}{4} = \frac{\pi (2 d_1)^2}{4} = 4 A_1 \] Now, the increase in length for wire A and wire B will be: \[ \frac{\Delta L_A}{\Delta L_B} = \frac{L_A / A_A}{L_B / A_B} = \frac{L_A A_B}{L_B A_A} \] Given that \( L_A : L_B = 1 : 3 \) and \( A_A : A_B = 1 : 4 \), we get: \[ \frac{\Delta L_A}{\Delta L_B} = \frac{1 \times 4}{3 \times 1} = \frac{4}{3} \] Thus, the ratio of the increase in lengths of wires A and B is \( 4 : 3 \).