Circular disc rotates at 60 rpm. A coin of 18 g is placed at a distance of 8 cm, from the centre. The centrifugal force on  the coin becomes

Here, $n=60rpm=\frac{60}{60}=1rps$

$m=18g=18\times {10}^{-3}kg$

$r=8cm=8\times {10}^{-2}m$

Centrifugal force, $F=mr{\omega }^2=mr{\left(2\pi n\right)}^2$

= $4{\pi }^{2}mr{n}^2$

$=4\times \frac{22}{7}\times \frac{22}{7}\times (18\times {10}^{-3})\times (8\times {10}^{-2})\times 1^2$

$F=5.689\times {10}^{-2}N$