If the radius and length of a copper rod are both doubled, the rate of flow of heat along the rod increases ....... times
Medium
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(b) $Q \propto \frac{A}{l} \propto \frac{{{r^2}}}{l}$
==> $\frac{{{Q_2}}}{{{Q_1}}} = \frac{{r_2^2}}{{r_1^2}} \times \frac{{{l_1}}}{{{l_2}}}$
==> $\frac{{{Q_2}}}{{{Q_1}}} = \frac{4}{1} \times \frac{1}{2}$
==> ${Q_2} = 2{Q_1}$
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