| ⇦ | 
| ⇨ |
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: - If wavelength of a wave is λ=6000 Å, then wave number will be
- A circular disc rolls down an inclined plane.The ratio of rotational kinetic energy
- A particle moves along the x-axis from x=0 to x=5 m under the influence of a force
- If h is the height of the capillary rise and r be the radius of capillary tube,
- The wavelength of light emitted from second orbit to first orbit in a hydrogen atom is
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If wavelength of a wave is λ=6000 Å, then wave number will be
- A circular disc rolls down an inclined plane.The ratio of rotational kinetic energy
- A particle moves along the x-axis from x=0 to x=5 m under the influence of a force
- If h is the height of the capillary rise and r be the radius of capillary tube,
- The wavelength of light emitted from second orbit to first orbit in a hydrogen atom is
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