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(B) The heat required to raise the temperature of a given mass of water is $H = \frac{V^2}{R} t$. Since $H$ and $V$ are the same for both heaters,we h

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$\frac{20}{3}$ minutes

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(B) The heat required to raise the temperature of a given mass of water is $H = \frac{V^2}{R} t$. Since $H$ and $V$ are the same for both heaters,we have $t \propto R$. Let $R_1$ and $R_2$ be the resistances of the two heaters. Then $R_1 \propto t_1 = 10$ and $R_2 \propto t_2 = 20$. When connected in parallel,the equivalent resistance $R_{	ext{eq}}$ is given by $\frac{1}{R_{	ext{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{R_1 + R_2}{R_1 R_2}$,so $R_{	ext{eq}} = \frac{R_1 R_2}{R_1 + R_2}$. Since $t \propto R$,the equivalent time $t_{	ext{eq}}$ is $t_{	ext{eq}} = \frac{t_1 t_2}{t_1 + t_2}$. Substituting the given values: $t_{	ext{eq}} = \frac{10 	imes 20}{10 + 20} = \frac{200}{30} = \frac{20}{3}$ minutes.

Two heaters take $10$ minutes and $20$ minutes respectively to heat the same amount of water. If the heaters are connected in parallel,what would be the time required to heat the same amount of water?

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