A small object placed on a rotating horizontal turn table just slips when it is placed at a distance $4\, cm$ from the axis of rotation. If the angular velocity of the turn-table is doubled, the object slips when its distance from the axis of rotation is
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The object will slip if centripetal force $\geq$ force of
friction
${\operatorname{mr} \omega^{2} \geq \mu m g} $
${r \omega^{2} \geq \mu g}$
$\mathrm{r} \omega^{2} \geq \mathrm{constant}, \mathrm{or}\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)=\left(\frac{\omega_{2}}{\omega_{1}}\right)^{2}$
$\frac{4 \mathrm{cm}}{\mathrm{r}_{2}}=\left(\frac{2 \omega}{\omega}\right)^{2} \quad \therefore \mathrm{r}_{2}=1 \mathrm{cm}$
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