JEE Main 2020 · PhysicsMultiple choiceSingle correctMediumCalculation

JEE Main 2 September 2020, Shift 1, Physics Q18

Question 18 of 75 in this shift, Physics question 18 of 25.

In a reactor, 2 kg of

_{92}\mathrm{U}^{235}

fuel is fully used up in 30 days. The energy released per fission is 200 MeV. Given that the Avogadro number,

N=6.023\times10^{26}

per kilo mole and

1\ \mathrm{eV}=1.6\times10^{-19}

J. The power output of the reactor is close to :

(1)60 MWOfficial answer
(2)125 MW
(3)54 MW
(4)35 MW

Official answer

Option 1

NTA final key (Sep 2020).

Chapter: Nuclei
Topic: Fission and Fusion
Idea tested: Reactor Power Calculation

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.