An ideal heat engine working between temperature T1 and T2 has an efficiency η , the new efficiency if both the source and sink temperature are 4 times, will be
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a
η2
b
η
c
2η
d
3η
answer is B.
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Detailed Solution
As η=1−T2T1 When T1&T2 are 4 times, ‘η ’ remains the same.