Question

A wire of resistance R is stretched till its radius is half of the original value. Then, the resistance of the stretched wire is

• A. 2 R
• B. 4 R
• C. 8 R
• D. 16 R

Answer 1

Answer: (D)

16 R

Answer 2

R = $\frac{\rho\mathcal{l}}{A}$ $\left\{ \begin{matrix} massm = A\mathcal{l}d \\ \therefore\mathcal{l} = \frac{m}{Ad} \end{matrix} \right.\$

R = $\frac{\rho m}{dA^{2}}$ ̃ R µ $\frac{1}{A^{2}}$ ̃ $\begin{matrix} R \propto \frac{1}{r^{4}} \end{matrix}$

\ $\frac{R'}{R}$= $\left( \frac{r}{r'} \right)^{4}$ \ R¢ = R$\left( \frac{r}{r/2} \right)^{4}$ = 16 R