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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
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Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A current of 5 A is passing through a metallic wire of cross-sectional area 4×10⁻⁶ m².
- By sucking through a straw, a student can reduce the pressure of his lungs
- An object of mass 3kg is at rest. Now a force F=6t² i⃗+4t j⃗ is applied on the object
- A metal bar of length L and area of cross-section A is clamped between two rigid
- At 10⁰C the value of the density of a fixed mass of an ideal gas divided by its pressure
Topics: Electrostatics (146)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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