a
The object will slip if centripetal force $\geq$ force of
friction
${\operatorname{mr} \omega^{2} \geq \mu m g} $
${r \omega^{2} \geq \mu g}$
$\mathrm{r} \omega^{2} \geq \mathrm{constant}, \mathrm{or}\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)=\left(\frac{\omega_{2}}{\omega_{1}}\right)^{2}$
$\frac{4 \mathrm{cm}}{\mathrm{r}_{2}}=\left(\frac{2 \omega}{\omega}\right)^{2} \quad \therefore \mathrm{r}_{2}=1 \mathrm{cm}$