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

Answer: 9

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 $=\frac{M}{V}=\frac{M}{4\pi r_1^2{l}_1}=\frac{M}{4\pi r_2^2{l}_2}$
$\Rightarrow r_2^2=\frac{l_1{r}_1^2}{l_2}=\frac{r_1^2}{3}\dots ..(i)\left(\because l_2=3l_1\right)$
Now, $R_2\propto \frac{l_2}{r_2^2}$
$\therefore R_2\propto \frac{9l_1}{r_1^2}....\left(ii\right)$
From $(i)$ and $(ii)$
$\therefore R_2:R_1=9:1$