If the temperatures of source and sink of a cannot engine having efficiency 1 are each decreased' by 100 K, then the efficiency
A
remains constant
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B
becomes 1
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C
decreases
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D
increases
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E
becomes zero
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Solution
The correct option is E increases In starting the efficiency η=1−T2T1=T1−T2T1 On decreased by 100 K each the efficiency η=1−T2−100T1−100=T1−T2T1−100 In this case the efficiency will increases.