| ⇦ | 
| ⇨ |
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: - The electromagnetic theory of light failed to explain
- Two racing cars of masses m₁ and m₂ are moving in circles of radii
- The interferance pattern is obtained with two coherent light sources of intensity
- An ideal spring with spring constant K is hung from the ceiling and a block of mass
- A ray of light is incident at an angle of incidence i, on one face of a prism
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The electromagnetic theory of light failed to explain
- Two racing cars of masses m₁ and m₂ are moving in circles of radii
- The interferance pattern is obtained with two coherent light sources of intensity
- An ideal spring with spring constant K is hung from the ceiling and a block of mass
- A ray of light is incident at an angle of incidence i, on one face of a prism
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