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In a nucleus reactor 0.96 grams of .(92)...

In a nucleus reactor `0.96` grams of `._(92)U^(235)` is consumed in one day. If `0.1%` of mass of `._(92)U^(235)` is available as energy fine the power of reactor (in `MW`)

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The correct Answer is:
A
Power`=("energy")/("time") , P=(E)/(T),P=(mc^(2))/(t)`
`P=(((0.1)/(100)M)C^(2))/(t), M=0.96` grams `=96xx10^(5)`
kg, t= 1 day = 86400 seconds
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