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A Carnot engine, having an efficiency of `eta=1//10` as heat engine, is used as a refrigerator. If the work done on the system is 10J, the amount of energy absorbed from the reservoir at lower temperature is
A. 1 J
B. 90 J
C. 99 J
D. 100 J

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Correct Answer - B
`beta=(1-eta)/(eta)`
`=(1-(1)/(10))/((1)/(10))=((9)/(10))/((1)/(10))`
`beta=9`
`beta=(Q_(2))/(W)`
`Q_(2)=9xx10=90J`

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