| ⇦ | 
| ⇨ |
The energy released by the fission of one uranium atom is 200 MeV. The number of fission per second required to produce 6.4 W power is
Options
(a) 2×10¹¹
(b) 10¹¹
(c) 10¹⁰
(d) 2×10¹⁰
Correct Answer:
2×10¹¹
Explanation:
Let there is n number of fission per second produces a power of 6.4 W, then
n × 200 × 10⁶ × 1.6 × 10⁻¹⁹ Js⁻¹ = 6.4 Js⁻¹
.·. n = 6.4 / 200 × 10⁻¹³ × 1.6 = 4 / 2 × 10¹¹
Related Questions: - A body of mass 1.0 kg is falling with an acceleration of 10 ms⁻².
- A wire is suspended vertically from one of its ends is stretched by attaching a weight
- An electron is moving in a circular path under the influence of a transverse magnetic
- In common base circuit of a transistor, current amplification factor is 0.95.
- An electron moving in a circular orbit of radius r makes n rotations per second.
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A body of mass 1.0 kg is falling with an acceleration of 10 ms⁻².
- A wire is suspended vertically from one of its ends is stretched by attaching a weight
- An electron is moving in a circular path under the influence of a transverse magnetic
- In common base circuit of a transistor, current amplification factor is 0.95.
- An electron moving in a circular orbit of radius r makes n rotations per second.
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