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The potential energy of particle in a force field is U = A/r² – b/r, where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is:
Options
(a) B / 2A
(b) 2A / B
(c) A / B
(d) B / A
Correct Answer:
2A / B
Explanation:
for equilibrium dU/ dr = 0
-2A/r³ + B/r² = 0 r = 2A / B
for stable equilibrium d²U /dr² should be positive for the value of r.
here d²U / dr² = 6A/r⁴ – 2B/r³ is +ve value for r = 2A / B So.
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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A stone of mass 1 kg tied to a light inexensible string of length L=10/3 is whirling
- In a radioactive decay process, the negatively charged emitted β- particles are
- The amplitude of S.H.M. y=2 (sin 5πt+√2 cos 5πt) is
- Certain quantity of water cools from 70⁰ C to 60⁰C in the first 5 minutes and to 54⁰C
- In Wheatstone bridge, three resistors P,Q and R are conncected in three arms
Topics: Electrostatics (146)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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