Question

A wire is uniformly stretched to make its area of cross section $\frac{1^{th}}{n}$ times $(n>0)$. What will be its new resistance?

• A. $n^2$ times
• B. $\frac{1}{n}$ times
• C. $\frac{1}{n^2}$ times
• D. $n$ times

Answer 1

The correct option is A $n^2$ times

Volume remains the same.

$\therefore A\cdot l=A^{\prime }\cdot l^{\prime }⟹A\cdot l=\frac{A}{n}\cdot l^{\prime }\therefore l^{\prime }=nl{R}^{\prime }=\frac{\rho l^{\prime }}{A^{\prime }}=\frac{\rho \cdot nl}{\frac{A}{n}}=n^{2}R$