Question

When length of wire is increased by 10%, then percentage increase in resistance is

• A. 10%
• B. 21%
• C. 25%
• D. 35%

Answer 1

Answer: (B)

21%

Answer 2

A wire’s resistance is given by

$R=\frac{\rho L}{A}$,

where $\rho$ is the resistivity, $L$ the length, and $A$ the cross‐sectional area. When a wire is stretched (i.e. increased in length while keeping its volume constant), we have

$\text{Volume} = A\,L = \text{constant}$.

If the length increases by 10%, then

$L' = 1.10\,L \quad \Rightarrow \quad A' = \frac{A}{1.10}$.

Thus the new resistance is

$R'=\frac{\rho L'}{A'}=\frac{\rho (1.10\,L)}{A/1.10} = \rho\,L \cdot \frac{1.10^2}{A}=1.21\,R$.

This means a 21% increase in resistance.