Answer: Option D
:
D
We know that $Volume=\frac{Mass}{density}$
Let m be the mass of the given wire, d its density and $\rho$its resistivity.
These quantities remain unchanged when the wire is drawn out.
Let $l_1$ be the length of the smaller wire and $r_1$ its corresponding radius of cross section. If $l_2$ and $r_2$ represent the length and radius of cross section of the elongated wire,
$⟹m=\pi r_1^2{l}_{2}d$ for smaller wire and for elongated wire, $m=\pi r_2^2{l}_{2}d$.
Since mass and density are constant
$⟹r_1^2{l}_1=r_2^2{l}_2$
If $R_1$ is the resistance of the smaller wire and $R_2$ is the resistance of theelongated wire, we have,
$R_1=\frac{\rho l_1}{\pi r_1^2}$
$R_2=\frac{\rho l_2}{\pi r_2^2}$
Now, $\frac{R_2}{R_1}=\frac{l_2}{l_1}\times \frac{r_1^2}{r_2^2}$
It is given that, $l_2=3l_1.$
Therefore $r_2^2=\frac{r_1^2}{3}$
Substituting the values,
$\frac{R_2}{R_1}=9$
Since, $R_1=5\Omega$
$R_2=45\Omega$.