In a reactor, 2 kg of _{}\textrm{92}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  (in MW) of the reactor is close to:
Option: 1 60
Option: 2 35
Option: 3 125
Option: 4 54

Answers (1)

Number of uranium atoms in 2 kg=\frac{2 \times 6.023 \times 10^{26}}{235}

energy from one atom is 200 \times 10^{6} \ e.v.

hence total energy from 2 kg uranium=\frac{2 \times 6.023 \times 10^{26}}{235} \times 200 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J}
And 2kg uranium is used in 30 days hence this energy is received in 30 days hence energy received per second or power is
\text { Power }=\frac{2 \times 6.023 \times 10^{26} \times 200 \times 10^{6} \times 1.6 \times 10^{-19}}{235 \times 30 \times 24 \times 3600}
Power =63.2 \times 10^{6} \ Watt\ \ or \ \ 63.2 \ Mega \ \ Watt

So  60 MW is the answer as it is closest to 63.2 MW

Most Viewed Questions

Preparation Products

Knockout JEE Main (Six Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 9999/- ₹ 8499/-
Buy Now
Knockout JEE Main (Nine Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 13999/- ₹ 12499/-
Buy Now
Test Series JEE Main 2024

Chapter/Subject/Full Mock Tests for JEE Main, Personalized Performance Report, Weakness Sheet, Complete Answer Key,.

₹ 7999/- ₹ 4999/-
Buy Now
JEE Main Rank Booster 2023

Booster and Kadha Video Lectures, Unlimited Full Mock Test, Adaptive Time Table, Faculty Support.

₹ 9999/- ₹ 6999/-
Buy Now
Knockout JEE Main (One Month Subscription)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, Faculty Support.

₹ 7999/- ₹ 4999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions