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

wire \(=P=\frac{V^{2}}{R}\)

Resistance of the wire of length, \(L\)

\(R_{1}=\frac{\rho L}{A}=\frac{\rho L}{\pi r^{2}}\)

\(\therefore\) Power, \(P_{1}=\frac{V^{2}}{R_{1}}\)

Resistance of the wire when length is halved

i.e., \(L / 2\)

\(R_{2}=\frac{\rho \frac{L}{2}}{\pi(2 r)^{2}}=\frac{\rho L}{\pi 8 r^{2}}=\frac{R_{1}}{8}\)

\(\therefore\) Power, \(P_{2}=\frac{V}{\frac{R_{1}}{8}}=\frac{8 V}{R_{1}}\)

or, \(\mathrm{P}_{2}=8 \mathrm{P}_{1}\) i.e., power increased \(8\) times of previous or original wire.