| ⇦ |
| ⇨ |
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: - A galvanometer acting as a voltmeter will have
- A susceptibility of a certain magnetic material is 400. What is the class
- In adiabatic expansion, product of PV
- A particle of mass 1.96×10⁻¹⁵ kg is kept in equilibrium between two horizontal metal
- The excess pressure inside one soap bubble is three times that inside a second soap bubble
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A galvanometer acting as a voltmeter will have
- A susceptibility of a certain magnetic material is 400. What is the class
- In adiabatic expansion, product of PV
- A particle of mass 1.96×10⁻¹⁵ kg is kept in equilibrium between two horizontal metal
- The excess pressure inside one soap bubble is three times that inside a second soap bubble
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