Question

When a wire is stretched and its radius becomes $\frac {r}{2}$. then its resistance will be

• A. 16 R
• B. 4 R
• C. 2 R
• D. zero

Answer 1

Answer: (A)

16 R

Answer 2

Resistance of wire initially is $R =\rho \frac{l}{A}$ or $R =\rho \frac{l}{\pi r^{2}}$ Volume of wire remains same even after stretching it. Therefore, $\pi r^{2} l=\pi r^{' 2} l'$ where, $r$ and $l=$ initial radius and length of wire $r'$ and $l'=$ final radius and length of wire Since, $r'=\frac{r}{2}$, therefore $\pi r^{2} l=\pi\left(\frac{r}{2}\right)^{2} l'$ $\Rightarrow l'=4 l$ Now, new resistance of wire is given by $R'=\rho \frac{l'}{-A'}=\rho \frac{l'}{-\pi r^{2}}$ $=\rho \cdot \frac{4 l}{\pi\left(\frac{r}{2}\right)^{2}}$ $=16 \times \rho \cdot \frac{l}{\pi r^{2}}$ $\Rightarrow R'=16 R$