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Q.

1016 fissions per second are taking place in a nuclear reactor having efficiency 25%. The energy released per fission is 200 MeV. The power output of the nuclear reactor is 

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a

20 KW

b

40 KW

c

60 KW

d

80 KW

answer is D.

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Detailed Solution

P=25100×nEt
Energy released in one sec
=10165200MeV=1016×200×106×1.6×1019J
Given that efficiency of the reactor is 25%. 
Power output of the nuclear reactor
=1016×200×106×1.6×1019×25100=80 kW

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1016 fissions per second are taking place in a nuclear reactor having efficiency 25%. The energy released per fission is 200 MeV. The power output of the nuclear reactor is