A small particle of mass `0.36g` rests on a horizontal turntable at a distance `25cm` from the axis of spindle. The turntable is accelerated at rate of `alpha=(1)/(3)rads^(-2)` . The frictional force that the table exerts on the particle `2s` after the startup is

Options

A. (a)`40 mu N`
B. (b)`30 mu N`
C. (c)`50 mu N`
D. (d)`60 mu N`

Answer

C

Explanation

`f = m a = m sqrt(a_(t)^(2) + a_(r )^(2)) = m sqrt((R alpha)^(2) + (R omega^(2))^(2)) = sqrt((R alpha)^(2) + [R (alpha t)^(2)]^(2))`
`= 0.36 xx 10^(-3) sqrt((0.25 xx (1)/(3))^(2) + [0.25 ((1)/(3) xx 2)^(2)]^(-2)) = 5 xx 10^(-5) N = 50 mu N`