Water boils in an electric kettle in 15,text{ | vedclass

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(C) The heat required to boil the water is given by $H = \frac{V^2}{R} t$,where $V$ is the supply voltage,$R$ is the resistance of the heating wire,an

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(C) The heat required to boil the water is given by $H = \frac{V^2}{R} t$,where $V$ is the supply voltage,$R$ is the resistance of the heating wire,and $t$ is the time taken.Since the supply voltage $V$ and the amount of heat $H$ required are the same in both cases,we have $H = \frac{V^2}{R_1} t_1 = \frac{V^2}{R_2} t_2$.This implies $\frac{t_1}{R_1} = \frac{t_2}{R_2}$,or $\frac{t_2}{t_1} = \frac{R_2}{R_1}$ ..... $(i)$We know that the resistance of a wire is proportional to its length,$R \propto l$,so $\frac{R_2}{R_1} = \frac{l_2}{l_1}$ ..... $(ii)$Substituting $(ii)$ into $(i)$,we get $\frac{t_2}{t_1} = \frac{l_2}{l_1}$.Given that the new length $l_2 = \frac{2}{3} l_1$,we have $\frac{l_2}{l_1} = \frac{2}{3}$.Substituting the values,$\frac{t_2}{15} = \frac{2}{3}$,which gives $t_2 = 15 	imes \frac{2}{3} = 10\,	ext{min}$.

Water boils in an electric kettle in $15\,\text{min}$ 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 ............. $\text{min}$.

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