A coin, placed on a rotating turn-table slips, when it is  placed at a distance of 9 cm from the centre. If the angular velocity of the turn-table is tripled, it will just slip, if its distance from the centre is

The centrifugal force on the coin in first case is given by 
${\mathrm{F}}_1={\mathrm{mr}}_1{\mathrm{\omega }}_1^2=\mathrm{m}\times 9\times {\mathrm{\omega }}_1^2=9{\mathrm{m\omega }}_1^2$
Similarly, the centrifugal force on the coin in the second case is given by
${\mathrm{F}}_2={\mathrm{mr}}_2{\mathrm{\omega }}_2^2={\mathrm{mr}}_2{\left(3{\mathrm{\omega }}_1\right)}^2=9{\mathrm{mr}}_2{\mathrm{\omega }}_1^2$
Equating the two forces, we get,
$\begin{array}{r}9{\mathrm{m}}_1^2=9{\mathrm{mr}}_2{\mathrm{\omega }}_1^2\\ {\mathrm{r}}_2=1\mathrm{cm}\end{array}$
where r₂, is the final distance of the coin from the centre