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(B) The heat produced $H$ in a wire with resistance $R$ under constant voltage $V$ is given by $H = \frac{V^2}{R} t$. For a fixed time $t$,$H \propto

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Both the length and radius of the wire are doubled.

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(B) The heat produced $H$ in a wire with resistance $R$ under constant voltage $V$ is given by $H = \frac{V^2}{R} t$. For a fixed time $t$,$H \propto \frac{1}{R}$.To double the heat $(H' = 2H)$,the resistance must be halved $(R' = R/2)$.The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$.If the length $l$ is halved $(l' = l/2)$ and the radius $r$ is kept constant,$R' = \rho \frac{l/2}{\pi r^2} = R/2$. This satisfies the condition.However,checking the options provided: If the length $l$ is halved $(l' = l/2)$ and radius $r$ is halved $(r' = r/2)$,then $R' = \rho \frac{l/2}{\pi (r/2)^2} = \rho \frac{l/2}{\pi r^2 / 4} = 2R$. This does not work.If the length $l$ is doubled $(l' = 2l)$ and radius $r$ is doubled $(r' = 2r)$,then $R' = \rho \frac{2l}{\pi (2r)^2} = \rho \frac{2l}{4\pi r^2} = R/2$. This satisfies the condition.Therefore,the correct option is $B$.

When a constant voltage is applied across the ends of a metallic wire,the heat produced in the wire will be doubled if:

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