JEE PYQ: Nuclear Physics Question 10
Question 10 - 2020 (02 Sep Shift 1)
In a reactor, 2 kg of $_{92}\text{U}^{235}$ fuel is fully used up in 30 days. The energy released per fission is 200 MeV. Given that the Avogadro number, $N = 6.023 \times 10^{26}$ per kilo mole and 1 eV $= 1.6 \times 10^{-19}$ J. The power output of the reactor is close to:
(1) 35 MW
(2) 60 MW
(3) 125 MW
(4) 54 MW
Show Answer
Answer: (2)
Solution
Power output of the reactor, $P = \frac{\text{energy}}{\text{time}}$ $= \frac{2}{235} \times \frac{6.023 \times 10^{26} \times 200 \times 1.6 \times 10^{-19}}{30 \times 24 \times 60 \times 60} = 60$ MW
BETA
