Q 12-03-185JEE MainJEE Main 2022 (28 Jul, Shift 1)Easy
A wire of resistance $R_1$ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:
Answer: (A) $9 : 1$
Increasing the length by twice the original makes the new length $3L$. The volume stays the same, so $R = \rho\dfrac{L^2}{V} \propto L^2$:
$$\frac{R_2}{R_1} = 3^2 = 9$$
Solution by Sreeraj P, M.Sc Physics