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A small particle of mass $0.36 g$ rests on a horizontal turn-table at a distance $25 cm$ from the axis of spindle. The turn-table is accelerated at a rate of $α = \frac{1}{3}$ rad $s^{- 2}$ . The frictional force that the table exerts on the particle 2 sec after the startup is

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$40 μN$

$30 μN$

$50 μN$

$60 μN$

answer is C.

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Detailed Solution

$f = ma = m \sqrt{a_{t}^{2} + a_{r}^{2}}$

$= m \sqrt{(Rα)^{2} + {({Rω}^{2})}^{2}}$

$= m \sqrt{(Rα)^{2} + {[R (αt)^{2}]}^{2}}$

= $0.36 \times 10^{- 3} \sqrt{{(0.25 \times \frac{1}{3})}^{2} + {[0.25 {(\frac{1}{3} \times 2)}^{2}]}^{2}}$

$= 50 \times 10^{- 6} N$

$= 50 μN$

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