Question

A uniform wire of resistance R is uniformely compressed along its length, until its radius becomes n times the original radius. Now, the resistance of the wire becomes :

• A. $R/n$
• B. $R/n^4$
• C. $R/n^2$
• D. $nR$

Answer 1

The correct option is B $R/n^4$

Answer is C.
The resistance of the conductor is given as $R=\rho \frac{l}{A}$.
Here, the new length can be derived as follows.
Old volume of wire = $\pi r^2\times l$ -- eqn 1. After compression, the volume changes as $\pi (nr{)}^2\times l^{{}^{\prime }}$ -- eqn 2.
Equating eqns 1 and 2 we get the new length.
So, the new length is given as $l^{{}^{\prime }}=\frac{l}{n^2}$
In this case, the radius of the wire is taken into consideration. So, we will have the area of cross section as $\pi r^2.$
So, $R=\rho \frac{l}{\pi r^2}$.
Now the wire of resistance R is uniformly compressed along its length, until its radius becomes n times the original radius. So the new length is $l^{{}^{\prime }}$ and the area of cross section is $\pi ({nr)}^2$.
Therefore, the resistance of the new wire is given as $R^{{}^{\prime }}=\rho \frac{l^{{}^{\prime }}}{\pi ({nr)}^2}$.
That is, $\rho l^{{}^{\prime }}\times \frac{1}{\pi n^2{r}^2}$.
Substituting the value of $l^{{}^{\prime }}$, we get,
$\rho \times \frac{l}{n^2}\times \frac{1}{\pi n^2{r}^2}$.
$=\frac{1}{n^4}\times \rho \frac{l}{\pi r^2}$.
So, $R^{{}^{\prime }}=\frac{1}{n^4}R$.
Hence, the new resistance is $\frac{R}{n^4}$.