| ⇦ |
| ⇨ |
The energy released in the fission of 1 kg of ₉₂U²³⁵ is (energy per fission = 200 MeV)
Options
(a) 5.1 x 10²⁶ eV
(b) 5.1 x 10²⁶ J
(c) 8.2 x 10¹³ J
(d) 8.2 x 10¹³ MeV
Correct Answer:
8.2 x 10¹³ J
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - A potentiometer wire of length 10 m and resistance 10 Ω per metre is connected in series
- A particle executing simple harmonic motion of amplitude 5 cm has maximum
- The magnetic field due to a current carrying circular loop of radius 3 cm at a point
- The component of vector A=2i+3j along the vector i+j is
- The displacement ‘x’ (in meter) of a particle of mass ‘m’ (in kg) moving in one
Topics: Radioactivity
(83)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A potentiometer wire of length 10 m and resistance 10 Ω per metre is connected in series
- A particle executing simple harmonic motion of amplitude 5 cm has maximum
- The magnetic field due to a current carrying circular loop of radius 3 cm at a point
- The component of vector A=2i+3j along the vector i+j is
- The displacement ‘x’ (in meter) of a particle of mass ‘m’ (in kg) moving in one
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