Question

The length of a metallic wire is increased by 20% and its area of cross section is reduced by 4%. The percentage change in resistance of the metallic wire is ______

Answer 1

Correct Answer: 25

Resistance Change in a Wire Problem

Step 1: Resistance Formula 

The resistance (\( R \)) of a wire is given by:

\( R = \frac{\rho l}{A} \)

where \( \rho \) is the resistivity, \( l \) is the length, and \( A \) is the cross-sectional area.

Step 2: New Resistance

The new length is \( l' = l + 0.20l = 1.2l \). The new area is \( A' = A - 0.04A = 0.96A \). The new resistance (\( R' \)) is:

\( R' = \frac{\rho l'}{A'} = \frac{\rho (1.2l)}{0.96A} = \frac{1.2}{0.96} \frac{\rho l}{A} = \frac{1.2}{0.96} R = 1.25R \)

Step 3: Percentage Change in Resistance

The percentage change in resistance is:

\( \frac{R' - R}{R} \times 100 = \frac{1.25R - R}{R} \times 100 = 0.25 \times 100 = 25\% \)

Conclusion:

The percentage change in resistance is a \( \mathbf{25\%} \) increase.