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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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Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The Boolean expression P+ P ̅ Q, where P and Q are the inputs of the logic
- The kinetic energy of particle moving along a circle of radius R depends
- A bar magnet with magnetic moment 2.5×10³ JT⁻¹ is rotating in horizontal plane
- Two simple harmonic motions of angular frequency 100 and 1000 rad s⁻¹ have the same
- A parallel plate capacitor with air between the plates has a capacitance of 9 pF.
Topics: Radioactivity (83)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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