| ⇦ | 
| ⇨ |
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.
Related Questions: - Young’s modulus of the material of a wire is 18×10¹¹ dyne cm⁻².its value in SI is
- In Young’s experiment, the ratio of maximum to minimum intensities of the fringe
- A circular disc rolls down an inclined plane.The ratio of rotational kinetic energy
- Three solids of masses m₁,m₂ and m₃ are connected with weightless string
- The average power dissipated in a pure inductor is
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Young’s modulus of the material of a wire is 18×10¹¹ dyne cm⁻².its value in SI is
- In Young’s experiment, the ratio of maximum to minimum intensities of the fringe
- A circular disc rolls down an inclined plane.The ratio of rotational kinetic energy
- Three solids of masses m₁,m₂ and m₃ are connected with weightless string
- The average power dissipated in a pure inductor is
Topics: Electrostatics (146)
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