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(C) The supply voltage $V$ and the heat required $H$ to boil the same amount of water are constant.The heat produced is given by $H = \frac{V^2}{R} t$

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$10$

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(C) The supply voltage $V$ and the heat required $H$ to boil the same amount of water are constant.The heat produced is given by $H = \frac{V^2}{R} t$,where $R$ is the resistance of the wire and $t$ is the time.Since $H$ and $V$ are constant,we have $t \propto R$.Therefore,$\frac{t_1}{t_2} = \frac{R_1}{R_2}$.Resistance $R$ is proportional to the length $l$ of the wire $(R = ho \frac{l}{A})$,so $\frac{R_1}{R_2} = \frac{l_1}{l_2}$.Substituting this into the time equation: $\frac{t_1}{t_2} = \frac{l_1}{l_2}$.Given $l_2 = \frac{2}{3} l_1$,we have $\frac{l_1}{l_2} = \frac{3}{2}$.Thus,$\frac{15}{t_2} = \frac{3}{2}$.Solving for $t_2$: $t_2 = 15 	imes \frac{2}{3} = 10 \ min$.

Water in an electric kettle starts boiling in $15 \ min$ after being switched on. If the length of the heating wire is reduced to $2/3$ of its original length,the time taken to boil the same amount of water will be ........... $min$.

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