Q 12-03-283JEE MainJEE Main 2018 (15 Apr, Shift 2)Easy
A constant voltage is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be:
Answer: (A) Increased 8 times
$R = \dfrac{\rho l}{\pi r^2}$. Halving $l$ and doubling $r$:
$$R' = \frac{\rho(l/2)}{\pi(2r)^2} = \frac{R}{8}$$
At constant voltage, $P = \dfrac{V^2}{R}$, so the rate of heating increases 8 times.
Solution by Sreeraj P, M.Sc Physics