A coin placed on a rotating turn table just slips if it is placed at a distance of $4 \mathrm{~cm}$ from the centre. If the angular velocity of the turn table is doubled it will just slip at a distance of

When the coin placed on a rotating table is first about to slips, then till the time the coin rotates along with the table, necessary centripetal force for rotation will be provided by the frictional force, i.e., $\quad m r \omega^{2}=\mu m g$ where, $m=$ mass of the coin, $r=$ distance of the coin from the centre of the table, $\omega=$ angular velocity, $\mu=$ coefficient of friction and $g=$ acceleration due to gravity. Since, $m, \mu$ and $g$ remains constant, so $r \omega^{2}=$ constant $\Rightarrow \quad r \propto \frac{1}{\omega^{2}}$ or $\quad \frac{r_{2}}{r_{1}}=\left(\frac{\omega_{1}}{\omega_{2}}\right)^{2}$ Given, $\omega_{2}=2 \omega_{1}, r_{1}=4 \mathrm{~cm}$ $\Rightarrow \quad \frac{r_{2}}{r_{1}}=\left(\frac{\omega_{1}}{2 \omega_{1}}\right)^{2}$ or $r_{2}=\frac{r_{1}}{4}=\frac{4}{4}=1 \mathrm{~cm}$