Two wires '$A$' and '$B$' of the same material have their lengths in the ratio $1 : 2$ and radii in the ratio $2 : 1$. The two wires are connected in parallel across a battery. The ratio of the heat produced in '$A$' to the heat produced in '$B$' for the same time is
Medium
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(d) ${R_1} = \rho \frac{{{l_1}}}{{{A_1}}}$ and ${R_2} = \rho \frac{{{l_2}}}{{{A_2}}}$$ \Rightarrow $$\,\frac{{{R_1}}}{{{R_2}}} = \frac{{{l_1}}}{{{l_2}}}.\frac{{{A_2}}}{{{A_1}}} = \frac{{{l_1}}}{{{l_2}}}{\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^2}$
Given $\frac{{{l_1}}}{{{l_2}}} = \frac{1}{2}$ and $\frac{{{r_1}}}{{{r_2}}} = \frac{2}{1}$ or $\frac{{{r_2}}}{{{r_1}}} = \frac{1}{2}$$ \Rightarrow $ $\frac{{{R_1}}}{{{R_2}}} = \frac{1}{8}$
$\therefore $ Ratio of heats $\frac{{{H_1}}}{{{H_2}}} = \frac{{{V^2}/{R_1}}}{{{V^2}/{R_2}}} = \frac{{{R_2}}}{{{R_1}}} = \frac{8}{1}$
Given $\frac{{{l_1}}}{{{l_2}}} = \frac{1}{2}$ and $\frac{{{r_1}}}{{{r_2}}} = \frac{2}{1}$ or $\frac{{{r_2}}}{{{r_1}}} = \frac{1}{2}$$ \Rightarrow $ $\frac{{{R_1}}}{{{R_2}}} = \frac{1}{8}$
$\therefore $ Ratio of heats $\frac{{{H_1}}}{{{H_2}}} = \frac{{{V^2}/{R_1}}}{{{V^2}/{R_2}}} = \frac{{{R_2}}}{{{R_1}}} = \frac{8}{1}$
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