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 :

The heat produced in the metallic wire is given using

H = ​$\frac{V^{2}t}{R}$

As R =  $\frac{\rho l}{A}$

or

H = ​$\frac{V^{2}t{\mathrm{\pi r}}^2}{\rho l}$

Therefore, the heat produced will be 8 times if length is halved and the radius of the wire are doubled.