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(A) The rate of heat developed (power) in a wire is given by $P = \frac{V^2}{R}$, where $V$ is the constant voltage and $R$ is the resistance.Since $R

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Will be doubled

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(A) The rate of heat developed (power) in a wire is given by $P = \frac{V^2}{R}$, where $V$ is the constant voltage and $R$ is the resistance.Since $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$, we have $P \propto \frac{1}{R} \propto \frac{r^2}{l}$.Let the initial length be $l_1 = l$ and radius be $r_1 = r$. Then $P_1 \propto \frac{r^2}{l}$.When both length and radius are doubled, $l_2 = 2l$ and $r_2 = 2r$.The new power is $P_2 \propto \frac{r_2^2}{l_2} = \frac{(2r)^2}{2l} = \frac{4r^2}{2l} = 2 \left( \frac{r^2}{l} \right)$.Therefore, $P_2 = 2P_1$. The rate of heat developed will be doubled.

$A$ constant voltage is applied between the two ends of a metallic wire. If both the length and the radius of the wire are doubled, the rate of heat developed in the wire

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