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`        Three identical particles of mass m are present at the corner of equilateral triangle of side a ,than find out work done(gravitational potential energy) in increasing side of the triangle from a to 2a.`
one year ago

Kapil Khare
80 Points
```							Initially each particle is at distance a from any other particle. So, initial potential energy                   U1 = [(-Gmm)/a] + [(-Gmm)/a] + [(-Gmm)/a]                   U1 = (-3Gmm)/aThen distance between any two particles is 2a. So, final potential energy                   U2 =  [(-Gmm)/2a] + [(-Gmm)/2a] + [(-Gmm)/2a]                   U2 = (-3Gmm)/2aWork done in increasing the side of the triangle = (Initial gravitational potential energy) – (Final gravitational potential energy)
```
one year ago
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### Course Features

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