A bucket containing water is revolved in a vertical circle of radius ' $\mathrm{r}^{\prime}$. To prevent the water from falling down, the minimum frequency of revolution required is $[\mathrm{g}=$ acceleration due to gravity $]$

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Aicharya · Accepted Answer

For minimum velocity at the highest point we should have $\mathrm{mr} \omega^{2}=\mathrm{mg}$ $	herefore \omega^{2}=\frac{g}{r} \quad$ or $\omega=\sqrt{\frac{g}{r}}$ $\begin{aligned} 2 \pi f &=\sqrt{\frac{g}{r}} \ f &=\frac{1}{2 \pi} \sqrt{\frac{g}{r}} \end{aligned}$

A bucket containing water is revolved in a vertical circle of radius ' $\mathrm{r}^{\prime}$. To prevent the…

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