Question

If a wire is stretched to n times of its original length, its new resistance will be

• A. 1/n times
• B. $n^2$ times
• C. n times
• D. None

Answer 1

The correct option is B $n^2$ times

$\text{Step 1: Volume of the wire}$

Volume of the wire remains same before and after stretching.

$V_i=V_f=V$

$\text{Step 2: Resistance of the wire}$

Initial resistance: $R=\frac{\rho l}{A}$ where $A=V/l$

$\Rightarrow R=\frac{\rho l^2}{V}$ $....\left(1\right)$

Final resistance : $R^{\prime }=\frac{\rho l^{{}^{\prime }2}}{V}$ $....\left(2\right)$

$\text{Step 3: Dividing equation (2) by (1)}$

$\Rightarrow \frac{R^{\prime }}{R}=\frac{l^{{}^{\prime }2}}{l^2}$

Given $l^{\prime }=nl$

$\Rightarrow R^{\prime }=\frac{(nl{)}^{2}R}{l^2}=n^{2}R$

Hence option B is correct.