An electric kettle consists of two filaments $F_{1}$ and $F_{2}$ of different resistances $R_{1}$ an

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An electric kettle consists of two filaments $F_{1}$ and $F_{2}$ of different resistances $R_{1}$ and $R_{2}$ respectively, connected in parallel with separate switches. With a certain fixed amount of water in the kettle pot to prepare the tea, when only $F_{1}$ is switched on, it takes 4 minute to boil the water. When $F_{1}$ and $F_{2}$ both are switched on simultaneously, it takes 3 minute to boil the same amount of water up to the same temperature. The time taken to boil the same amount of water when only $F_{2}$ is switched on (assume that entire heat radiated by filaments is used in boiling the water starting from the same room temperature in each case) is

Accepted Answer

Sol. Key concept: Heating times are inversely proportional to heating power.Let Q be the heat needed to boil the fixed quantity of water. F₁ supplies Q/4 per minute, while the two filaments together supply Q/3 per minute. Since the filaments operate in parallel at the same supply voltage, their powers add.$$P_{2} = \frac{Q}{3} - \frac{Q}{4} = \frac{Q}{12}$$$$t_{2} = \frac{Q}{P_{2}} = 12\ min$$This assumes identical starting temperatures and negligible heat loss, as stated. Hence (d).

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