Question

A given piece of wire of length I and radius $r$ is having a resistance $R$. This wire is stretched uniformly to a wire of radius $ \frac{r}{2} $ . What is the new resistance?

• A. 4 R
• B. 8 R
• C. 16 R
• D. 2 R

Answer 1

Answer: (C)

Resistance of a wire of length $I$, area $A$ and specific resistance $\rho$ is given by $R=\rho \frac{l}{A}$
Also, Volume $(V)$ Length $(l)$ Area where $A=\pi r^{2}$ (r is radius)
When the wire is stretched its volume $(V)$ remains constant.
Hence, $R=\frac{\rho V}{\pi^{2} r^{4}} \ldots$ (i)
When radius is halved $R'=\frac{\rho V}{\pi^{2}\left(\frac{r}{2}\right)^{4}} \ldots$ (ii)
$\therefore \frac{R'}{R}=\frac{16 \rho V}{\pi^{2} r^{4}} \times \frac{\pi^{2} r^{4}}{\rho V}=16$
$\Rightarrow R'=16 R$
Hence, new resistance increases to sixteen times its original value.