| ⇦ | 
| ⇨ |
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: - An electron in potentiometer experiences a force 2.4×10⁻¹⁹N. The length of potentiometer
- In Young’s double silt experiment, 12 fringes are observed to be formed
- How much work is required to carry a 6 μC charge from the negative terminal
- A capacitor of capacitance 100μF is charged by connecting it to a battery of voltage
- In an inductor when current changes from 2 A to 18 A in 0.05 sec, the e.m.f. induced
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An electron in potentiometer experiences a force 2.4×10⁻¹⁹N. The length of potentiometer
- In Young’s double silt experiment, 12 fringes are observed to be formed
- How much work is required to carry a 6 μC charge from the negative terminal
- A capacitor of capacitance 100μF is charged by connecting it to a battery of voltage
- In an inductor when current changes from 2 A to 18 A in 0.05 sec, the e.m.f. induced
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