Question

The resistance of a 10 m long wire is 10 $ \Omega $. Its length is increased by 25\% by stretching the wire uniformly. The new resistance is

• A. 18.6 \( \Omega \)
• B. 15.6 \( \Omega \)
• C. 12.8 \( \Omega \)
• D. 14.9 \( \Omega \)

Hint:

When a wire is stretched uniformly, its length increases, and its cross-sectional area decreases. The resistance increases because both length and area influence the resistance.

Answer 1

The Correct Option is B

The resistance of a wire is given by the formula: \[ R = \rho \frac{L}{A} \] Where: 
- \( R \) is the resistance, - \( \rho \) is the resistivity of the material (which remains constant), 
- \( L \) is the length of the wire, 
- \( A \) is the cross-sectional area of the wire. 
When the wire is stretched, the length increases by 25%, meaning the new length \( L' \) is: \[ L' = L(1 + 0.25) = 1.25L \] Since the volume of the wire is conserved (it’s just being stretched), the volume \( V = A \times L \) remains constant. The new area \( A' \) after stretching is related to the initial area by: \[ A' = \frac{A}{1.25} \] 
Thus, the new resistance \( R' \) is given by: \[ R' = \rho \frac{L'}{A'} = \rho \frac{1.25L}{A/1.25} = R \times \left( \frac{1.25}{\frac{1}{1.25}} \right) = R \times 1.25^2 = 10 \times 1.25^2 = 10 \times 1.5625 = 15.6 \, \Omega \] Therefore, the new resistance is \( 15.6 \, \Omega \).