Two wires A and B of same material and mass h | vedclass

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Question

(c) $\frac{H}{t} = P = \frac{{{V^2}}}{R} \Rightarrow P \propto \frac{1}{R}$ also $R \propto \frac{l}{A} \propto \frac{{{l^2}\rho }}{{A.l\rho }}$
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Vedclass · Accepted Answer

$20$

Vedclass · Answer

(c) $\frac{H}{t} = P = \frac{{{V^2}}}{R} \Rightarrow P \propto \frac{1}{R}$ also $R \propto \frac{l}{A} \propto \frac{{{l^2}\rho }}{{A.l\rho }}$
$ \Rightarrow $ $R \propto \frac{{{l^2}}}{m}$ $ \Rightarrow $ $R \propto {l^2}$ (for same mass)
So $\frac{{{P_A}}}{{{P_B}}} = \frac{{l_B^2}}{{l_A^2}} = \frac{4}{1} \Rightarrow {P_A} = 20\,W$

Two wires $A$ and $B$ of 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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