| ⇦ | 
| ⇨ |
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 pair of quantities having same dimensions is
- In potentiometer experiment, a cell of emf 1.25V gives balancing length of 30 cm
- When an electron beam passes through an electric field they gain kinetic energy.
- If 10000 V is applied across an X-ray tube, what will be the ratio of de-Broglie
- The time by a photoelectron to come out after the photon strikes is approximately
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The pair of quantities having same dimensions is
- In potentiometer experiment, a cell of emf 1.25V gives balancing length of 30 cm
- When an electron beam passes through an electric field they gain kinetic energy.
- If 10000 V is applied across an X-ray tube, what will be the ratio of de-Broglie
- The time by a photoelectron to come out after the photon strikes is approximately
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