Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        The potential energy of a particle in a certain field has the formU =a/r²-b/r,where  A and B are positive constants, r is the distance from the centre of the field. the value of r0 corresponding to the equilibrium position of the particle (state if it is stable or unstable )is`
one year ago

Arun
23382 Points
```							for equilibrium dU/ dr = 0-2A/r³ + B/r² = 0 r = 2A / Bfor 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.
```
one year ago
Think You Can Provide A Better Answer ?

## Other Related Questions on IIT JEE Entrance Exam

View all Questions »

### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions