Question

Two wires of the same material having radius in ratio 2 : 1 and lengths in ratio 1 : 2. If the same force is applied on them, then the ratio of their change in length will be

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

Hint:

The change in length of a wire is inversely proportional to its cross-sectional area. For wires of the same material, the ratio of the change in length depends on the ratio of their areas and lengths.

Answer 1

The Correct Option is D

The change in length of a wire due to an applied 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, and \( Y \) is the Young's modulus of the material. 
Since both wires are made of the same material, \( Y \) is constant for both. 
The cross-sectional area \( A \) of a wire is proportional to the square of the radius: \[ A = \pi r^2 \] 
Let the radius of the first wire be \( r_1 \) and the radius of the second wire be \( r_2 \), with \( r_1 : r_2 = 2 : 1 \), so: \[ A_1 : A_2 = (2^2) : (1^2) = 4 : 1 \] Let the lengths of the wires be \( L_1 \) and \( L_2 \), with \( L_1 : L_2 = 1 : 2 \). Now, using the formula for change in length: \[ \Delta L_1 : \Delta L_2 = \frac{F L_1}{A_1} : \frac{F L_2}{A_2} = \frac{L_1}{A_1} : \frac{L_2}{A_2} \] Substitute the ratios: \[ \Delta L_1 : \Delta L_2 = \frac{1}{4} : \frac{2}{1} = 1 : 8 \] 
Thus, the ratio of their change in length is \( 1 : 8 \).