Water boils in an electric kettle in $15\,\min$ after switching on. If the length of the heating wire is decreased to $2/3$ of its initial value, then the same amount of water will boil with the same supply voltage in ............. $min$
Diffcult
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$H = \frac{{{V^2}}}{R}t$
Since supply voltage is same and equal amount of heat will produce, therefore
$\frac{{{R_1}}}{{{t_1}}} = \frac{{{R_2}}}{{{t_2}}}$ or $\frac{{{R_1}}}{{{R_2}}} = \frac{{{t_1}}}{{{t_2}}}$ ..... $(i)$
But $R \propto l$ $ \Rightarrow $ $\frac{{{R_1}}}{{{R_2}}} = \frac{{{l_1}}}{{{l_2}}}$..... $(ii)$
By $(i)$ and $(ii)$, $\frac{{{l_1}}}{{{l_2}}} = \frac{{{t_1}}}{{{t_2}}}$..... $(iii)$
Now ${l_2} = \frac{2}{3}\,\,{l_1}$ $ \Rightarrow $ $\frac{{{l_1}}}{{{l_2}}} = \frac{3}{2}$
By equation $(iii)$, $\frac{3}{2} = \frac{{15}}{{{t_2}}}$ $ \Rightarrow $ ${t_2} = 10\,\min$
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