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Question: Answered & Verified by Expert
A particle of mass m oscillates along the horizontal diameter AB inside a smooth spherical shell of the radius R . At any instant the kinetic energy of the particle is K. Then the force applied by particle on the shell at this instant is 

PhysicsLaws of MotionNEET
Options:
  • A KR
  • B 2KR
  • C 3KR
  • D K2R
Solution:
2253 Upvotes Verified Answer
The correct answer is: 3KR
Let the velocity of the particle at the point P is v.

From the conservation of mechanical energy




12mv2=K=mgh

Let N be the normal reaction between the particle and the shell at this instant. Then

N- mg sinθ=mv2Rmv2R=2KR

or N=mghR+2KR=KR+2KR   mgh= K

 N=3KR= force on shell

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