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A particle of mass 'm' oscillates along the horizontal diameter AB inside a smooth spherical shell of radius R. At any instant K.E. of the particle is K. Then force applied by particle on the shell at this instant is:

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$\frac{3 K}{R}$

$\frac{2 K}{R}$

$\frac{K}{R}$

$\frac{K}{2 R}$

answer is C.

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Detailed Solution

mgR sin $θ$ = $\frac{1}{2} {mv}^{2} = K$

$So, mg \sin θ = \frac{K}{R}$

$N = \frac{m v^{2}}{R} + m g \sin θ = 3 mg \sin θ = \frac{3 K}{R}$

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