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(B) The heat energy $H$ required to boil a given quantity of water is constant.Using the formula for heat energy $H = \frac{V^2}{R} t$,where $V$ is th

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$20$

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(B) The heat energy $H$ required to boil a given quantity of water is constant.Using the formula for heat energy $H = \frac{V^2}{R} t$,where $V$ is the voltage,$R$ is the resistance,and $t$ is the time.Since $H$ and $R$ are constant,we have $H_1 = H_2$.$\frac{V_1^2}{R} t_1 = \frac{V_2^2}{R} t_2$Given $V_1 = V$,$t_1 = 5 	ext{ min}$,and $V_2 = \frac{V}{2}$.Substituting these values: $\frac{V^2}{R} (5) = \frac{(\frac{V}{2})^2}{R} t_2$$5 = \frac{V^2}{4R} \cdot \frac{R}{V^2} \cdot t_2$$5 = \frac{t_2}{4}$$t_2 = 20 	ext{ min}$.

$A$ given quantity of water is boiled by an electric heater in $5 \text{ min}$. If the supply voltage of the heater is reduced to half,then the time taken to boil the same quantity of water will be . . . . . . $\text{min}$. (Assume the resistance of the heater remains constant.)

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