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Work Energy Power

JEE Main level, single correct, medium

160 of 183

Question 160 of 183

In a reactor, $2kg$ of $U23592$ fuel is fully used up in $30$ days. The energy released fission is $200MeV$ . Given that the Avogadro number, $N=6.023\times {10}^{26}$ per kilo mole and $1eV=1.6\times {10}^{-19}J$ . The power output of the reactor is close to:

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Answer: (B) $60MW$

P=\dfrac{E}{t}

=\dfrac{2}{235}\times \dfrac{6.023\times {10}^{26}\times 200\times 1.6\times {10}^{-19}}{30\times 24\times 60\times 60}=60MW

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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