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JEE AdvancedMedium

Q.A particle moves in a circle under a central force F=krF=kr. The work done over one revolution is

a
2πkR22\pi kR^2
b
πkR2\pi kR^2
c
Zero
d
kR2kR^2

Correct Answer: Option C

The correct solution involves applying the fundamental concept to derive the final value step by step...

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