Question

If a copper wire is stretched to increase its length by \(20 \%\) The percentage increase in resistance of the wire is _______\(\%\).

Hint:

When a wire is stretched, the resistance increases as R ∝ L² (if volume remains constant).

Answer 1

Correct Answer: 44

To determine the percentage increase in resistance when a copper wire is stretched, we use the formula for resistance: \( R = \rho \frac{L}{A} \), where \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. When a wire is stretched, its volume remains constant, i.e., \( L \times A = \text{constant} \). Let's assume the original length is \( L_0 \) and the new length is \( L_1 = 1.2L_0 \) since the wire is stretched by 20%.

The area \( A \) will change inversely with \( L \) to keep the volume constant. Hence, \( A_1 = \frac{A_0L_0}{L_1} = \frac{A_0L_0}{1.2L_0} = \frac{A_0}{1.2} \).

Substituting in the resistance formula, the original resistance \( R_0 = \rho \frac{L_0}{A_0} \) and the new resistance \( R_1 = \rho \frac{L_1}{A_1} = \rho \frac{1.2L_0}{\frac{A_0}{1.2}} = \rho \frac{1.44L_0}{A_0} = 1.44R_0 \).

The percentage increase in resistance is calculated as:

\(\text{Percentage Increase} = \left(\frac{R_1 - R_0}{R_0}\right) \times 100\% = \left(\frac{1.44R_0 - R_0}{R_0}\right) \times 100\% = 0.44 \times 100\% = 44\%\).

This calculated percentage increase in resistance is \(44\%\) and falls within the provided range of 44 to 44.