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(C) The rate of heat dissipation (power) is given by $P = \frac{V^2}{R}$. Since the voltage $V$ is the same, $P \propto \frac{1}{R}$.Resistance $R$ is

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$20$

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(C) The rate of heat dissipation (power) is given by $P = \frac{V^2}{R}$. Since the voltage $V$ is the same, $P \propto \frac{1}{R}$.Resistance $R$ is given by $R = ho \frac{l}{A}$. Since the mass $m = 	ext{density} 	imes 	ext{volume} = ho_d 	imes A 	imes l$, we have $A = \frac{m}{ho_d 	imes l}$.Substituting $A$ in the resistance formula: $R = ho \frac{l^2 ho_d}{m}$.For wires of the same material and mass, $R \propto l^2$.Therefore, $\frac{P_A}{P_B} = \frac{R_B}{R_A} = \frac{l_B^2}{l_A^2}$.Given $\frac{l_A}{l_B} = \frac{1}{2}$, we have $\frac{P_A}{P_B} = (\frac{2}{1})^2 = 4$.Given $P_B = 5\,W$, then $P_A = 4 	imes 5\,W = 20\,W$.

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

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