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Q.

200 MeV of energy may be obtained per fission of U235 . A reactor is generating 1000 KW of power. The rate of nuclear fission in the reactor is

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a

1000 per sec   

b

3.125×1013  per sec

c

2×108  per sec

d

3.125×1016   per sec

answer is C.

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Detailed Solution

Rate of nuclear fission 

=106200×1.6×10−13=3.125×1016

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