JEE Main 2020PhysicsWork Power EnergyEasyMCQ

JEE Main 2020 — Work Power Energy Question with Solution

JEE Main 2020 (02 Sep Shift 1)

Question

In a reactor, 2 kg of U23592 fuel is fully used up in 30 days. The energy released fission is 200 MeV. Given that the Avogadro number, N=6.023×1026 per kilo mole and 1 eV=1.6×10-19 J. The power output of the reactor is close to:

Choose an option

Show full solutionCorrect option: B
Correct answer
B60 MW

Step-by-step explanation

P=Et

=2235×6.023×1026×200×1.6×10-1930×24×60×60=60 MW

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About this question

This is a previous-year question from JEE Main 2020, covering the Work Power Energy chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.