| ⇦ | 
| ⇨ |
In a nuclear reactor, the number of U²³⁵ nuclei undergoing fissions per second is 4×10²⁰. If the energy released per fission is 250 MeV, then the total energy released in 10 h is (1 eV=1.6×10⁻¹⁹ J)
Options
(a) 576×10⁶ J
(b) 576×10¹² J
(c) 576×10¹⁵ J
(d) 576×10¹⁸ J
Correct Answer:
576×10¹² J
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - A planet moving along an elliptical orbit is closest to the sun at a distance of r₁
- The particle executing simple harmonic motion has a kinetic energy K₀cos²ωt.
- The density of ice is 0.9 g/cc and that of sea water is 1.1 g/cc
- Young’s modulus of the material of a wire is 18×10¹¹ dyne cm⁻².its value in SI is
- The wing span of an aeroplane is 20 m. It is flying in a field where the vertical
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A planet moving along an elliptical orbit is closest to the sun at a distance of r₁
- The particle executing simple harmonic motion has a kinetic energy K₀cos²ωt.
- The density of ice is 0.9 g/cc and that of sea water is 1.1 g/cc
- Young’s modulus of the material of a wire is 18×10¹¹ dyne cm⁻².its value in SI is
- The wing span of an aeroplane is 20 m. It is flying in a field where the vertical
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply