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A small particle of mass 036g rests on a...

A small particle of mass `036g` rests on a horizontal turntable at a distance `25cm` from the axis of spindle. The turntable is acceleration at rate of `alpha=(1)/(3)rads^(-2)` . The frictional force that the table exerts on the particle `2s` after the startup is

A

`40muN`

B

`30muN`

C

`50muN`

D

`60muN`

Text Solution

Verified by Experts

The correct Answer is:
C
`f=ma=msqrt(a_(t)^(2)+a_(r)^(2))`
`=msqrt((Ralpha)^(2)+(Romega)^(2))`
`=msqrt((Ralpha)^(2)+[R(alphat)^(2)]^(2))`
`=0.36xx10^(-3)sqrt((0.25xx(1)/(3))^(2)+[0.25((1)/(3)xx2)^(2)]^(2)`
`=50xx10^(-6)N`
`=50muN`
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