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(B) Let $V$ be the voltage applied. The power consumed by the filaments are $P_1$ and $P_2$ respectively.The heat required to boil the water is $H$. W

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$6$

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(B) Let $V$ be the voltage applied. The power consumed by the filaments are $P_1$ and $P_2$ respectively.The heat required to boil the water is $H$. We know that $H = P_1 t_1 = P_2 t_2$.When connected in parallel,the total power is $P_{eq} = P_1 + P_2$.The time $t$ taken to boil the same amount of water $H$ is given by $H = (P_1 + P_2)t$.Thus,$t = \frac{H}{P_1 + P_2}$.Substituting $P_1 = \frac{H}{t_1}$ and $P_2 = \frac{H}{t_2}$,we get:$t = \frac{H}{\frac{H}{t_1} + \frac{H}{t_2}} = \frac{1}{\frac{1}{t_1} + \frac{1}{t_2}} = \frac{t_1 t_2}{t_1 + t_2}$.Given $t_1 = 10 	ext{ min}$ and $t_2 = 15 	ext{ min}$:$t = \frac{10 	imes 15}{10 + 15} = \frac{150}{25} = 6 	ext{ minutes}$.

An electric kettle has two filaments. The first filament boils a certain amount of water in $10 \text{ minutes}$ and the second filament boils it in $15 \text{ minutes}$. If both filaments are connected in parallel,the water will boil in .............. minutes.

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