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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
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A source of sound S emitting waves of frequency 100 Hz and an observer
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- The specific charge of a proton is 9.6×10⁷ C/kg. The specific charge of an alpha particle
- A person has a minimum distance of distinct vision as 50 cm. The power of lenses
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Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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