An electric heater, operating at $220 V$ , boils $5 l$ of water in $5 minutes$ . If it is used on a $110 V$ line, it will boil the same amount of water in:

10 minutes

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20 minutes

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5 minutes

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1 minute

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Solution

The correct option is B $20 minutes$
Heat required to boil the water will be same in the two cases.
Heat produced by heater is:
$H = \frac{V^{2}}{R} t$
$V^{2} = \frac{H R}{t}$
Hence, other things remaining constant, $V^{2} \propto \frac{1}{t}$

Let the first case be labeled by subscript 1 and the second by subscript 2. Then,

$\frac{t_{2}}{t_{1}} = \frac{V_{1}^{2}}{V_{2}^{2}}$
$t_{2} = \frac{220^{2} \times 5}{110^{2}}$
$t_{2} = 20 m i n$

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An electric heater, operating at 220 V, boils 5 l of water in 5 minutes. If it is used on a 110 V line, it will boil the same amount of water in:

The correct option is B 20 minutesHeat required to boil the water will be same in the two cases.Heat produced by heater is:H=V2RtV2=HRtHence, other things remaining constant, V2∝1tLet the first case be labeled by subscript 1 and the second by subscript 2. Then,t2t1=V21V22t2=2202×51102t2=20 min

An electric heater, operating at $220 V$ , boils $5 l$ of water in $5 minutes$ . If it is used on a $110 V$ line, it will boil the same amount of water in: