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.

A
0.33
B
0.75
C
0.66
D
0.60

Solution

## The correct option is C 0.66Let N be the normal reaction between the bead and the ring at given instant, then from FBD of the bead  N−mgsin37o=mv2R ⟹N=mgsin37o+2KR where K is kinetic energy of the bead. ⟹N=0.01×10×35+2×0.0155×10−2 ⟹N=0.35+35 ⟹N=3.35=0.66 N Hence option C is the correct answer

Suggest corrections