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Two wires A and B of same material and mass have their lengths in the ratio 1 : 2. On connecting them separately to the same source, the rate of heat dissipation in B is found to be 5W. The rate of heat dissipation in A is

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detailed solution

Correct option is T

$\frac{Q}{t} = \frac{V^{2}}{R}$

$\Rightarrow P = \frac{V^{2}}{R} \Rightarrow P \propto \frac{1}{R} \Rightarrow \frac{P_{1}}{P_{2}} = \frac{R_{2}}{R_{1}}$

Also, $R = ρ \frac{l}{A} = ρ \frac{l^{2}}{v}$

$\Rightarrow R \propto l^{2}$

$\Rightarrow \frac{P_{A}}{5} = \frac{{l_{B}}^{2}}{{l_{A}}^{2}}$

$\Rightarrow \frac{P_{A}}{5} = 4 \Rightarrow P_{A} = 20 W$

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detailed solution

Correct answer is 20

$\frac{Q}{t} = \frac{V^{2}}{R}$

$\Rightarrow P = \frac{V^{2}}{R} \Rightarrow P \propto \frac{1}{R} \Rightarrow \frac{P_{1}}{P_{2}} = \frac{R_{2}}{R_{1}}$

Also, $R = ρ \frac{l}{A} = ρ \frac{l^{2}}{v}$

$\Rightarrow R \propto l^{2}$

$\Rightarrow \frac{P_{A}}{5} = \frac{{l_{B}}^{2}}{{l_{A}}^{2}}$

$\Rightarrow \frac{P_{A}}{5} = 4 \Rightarrow P_{A} = 20 W$

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