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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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Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A particle under the action of a constant force moves from rest upto 20 seconds
- In atom bomb, the reaction which occurs is
- The induced emf in a coil of 10 H inductance in which current varies
- A tyre filled with air (27° C and 2 atm) bursts. The temperature of air is (ˠ=1.5)
- When a mass m is attached to a spring, it normally extends by 0.2 m.
Topics: Electrostatics (146)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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