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(B) The heat developed in a wire with constant voltage $V$ is given by $H = \frac{V^2}{R} t$. For the heat to be doubled,the resistance $R$ must be ha

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both the length and radius of the wire are doubled

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(B) The heat developed in a wire with constant voltage $V$ is given by $H = \frac{V^2}{R} t$. For the heat to be doubled,the resistance $R$ must be halved $(R' = R/2)$.Resistance is given by $R = \frac{\rho \ell}{A} = \frac{\rho \ell}{\pi r^2}$.If both length $\ell$ and radius $r$ are halved,$\ell' = \ell/2$ and $r' = r/2$.Then $R' = \frac{\rho (\ell/2)}{\pi (r/2)^2} = \frac{\rho \ell / 2}{\pi r^2 / 4} = 2 \left( \frac{\rho \ell}{\pi r^2} \right) = 2R$.If both length $\ell$ and radius $r$ are doubled,$\ell' = 2\ell$ and $r' = 2r$.Then $R' = \frac{\rho (2\ell)}{\pi (2r)^2} = \frac{2\rho \ell}{4\pi r^2} = \frac{1}{2} \left( \frac{\rho \ell}{\pi r^2} \right) = \frac{R}{2}$.Since $R' = R/2$,the heat $H' = \frac{V^2}{R'} t = \frac{V^2}{R/2} t = 2 \left( \frac{V^2}{R} t \right) = 2H$. Thus,the heat is doubled.

$A$ constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if

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