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Question

# Heater-1, takes 3 minutes to boil a given amount of water. Heater-2 takes 6 minutes. If t1 and t2 are the time taken to boil the same amount of water, when the two heaters are connected in series and parallel, respectively, then t1t2=? [Assume same cell is connected in all the cases]

A
92
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B
29
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C
49
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D
94
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Solution

## The correct option is A 92Here, the heat required (say H) to boil water is the same. Let P1 and P2 are the powers of the heaters. Then, H=P1t1=P2t2 So, P1=Ht1=H3 P2=Ht2=H6 In series, P=P1P2P1+P2−(H3×H6)(H3+H6)=H9 Now, t1=HP=H(H9)=9 minutes In parallel, P=P1+P2=H3+H6=H2 ∴t2=HP=HH2=2 minutes Further, t1t2=92

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