QJEEM18Ph3019: JEE Main 2018 Physics - Class 12 JEEM18Ph3019 Solution: A constant voltages is applied between two ends of a metallic…

Since rate of heat $$=\frac{V^2}{R}\Rightarrow$$ rate of heat $$\propto \frac{1}{R}$$, where $$R=\frac{\rho L}{A}=\frac{\rho L}{\pi r^2}$$; here r is radius of wire and L is length of wire. Therefore,

$$R\propto \frac{L}{r^2}$$.

Thus, rate of heat $$\propto \frac{r^2}{L}$$.

When length in halved and radius is doubled then, rate of heat $$\propto \frac{{(2r)}^2}{L/2}=\frac{8r^2}{L}$$

Therefore, rate of heat developed in wire will be increased 8 times.