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, $R=\frac{\rho V}{{\pi }^2{r}^4}\dots$ (i)
When radius is halved $R^{\prime }=\frac{\rho V}{{\pi }^2{\left(\frac{r}{2}\right)}^4}\dots$ (ii)
$\therefore \frac{R^{\prime }}{R}=\frac{16\rho V}{{\pi }^2{r}^4}\times \frac{{\pi }^2{r}^4}{\rho V}=16$
$\Rightarrow R^{\prime }=16R$
Hence, new resistance increases to sixteen times its original value.