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Question

A bead of mass 10 g oscillates along the horizontal diameter AB of the ring of radius R=5 cm. At an instant shown in the figure, kinetic energy of the particle is 0.015 J. Find the force applied (in Newtons) by the ring on the bead.


Your Answer
A
0.33
Your Answer
B
0.75
Correct Answer
C
0.66
Your Answer
D
0.60

Solution

The correct option is C 0.66
Let N be the normal reaction between the bead and the ring at given instant, then from FBD of the bead 


Nmgsin37o=mv2R
N=mgsin37o+2KR
where K is kinetic energy of the bead.
N=0.01×10×35+2×0.0155×102
N=0.35+35
N=3.35=0.66 N
Hence option C is the correct answer

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