Question

The resistance of the wire is $R_{1}$ . It is now stretched (keeping the volume constant) without breaking so that its length becomes three times. Now the resistance becomes $R_{2}$ , then what is the value of $R_{2}:R_{1}?$

Answer 1

For a given wire, its resistance
$R \propto \frac{l}{A}$
$\therefore R_{1} \propto \frac{l_{1}}{r_{1}^{2}}$
The wire is stretched such that $l_{2}=3l_{1}$
But density is unchanged.
$\therefore $ Density $=\frac{M}{V}=\frac{M}{4 \pi r_{1}^{2} l_{1}}=\frac{M}{4 \pi r_{2}^{2} l_{2}}$
$\Rightarrow r_{2}^{2}=\frac{l_{1} r_{1}^{2}}{l_{2}}=\frac{r_{1}^{2}}{3}\ldots ..\left(\right.i\left.\right)\left(\because l_{2} = 3 l_{1}\right)$
Now, $R_{2} \propto \frac{l_{2}}{r_{2}^{2}}$
$\therefore R_{2} \propto \frac{9 l_{1}}{r_{1}^{2}}....\left(ii\right)$
From $\left(\right.i\left.\right)$ and $\left(\right.ii\left.\right)$
$\therefore R_{2}:R_{1}=9:1$