MHT CET · Physics · Rotational Motion
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 \(]\)
- A \(\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\mathrm{r}}}\)
- B \(2 \pi \sqrt{\frac{\mathrm{g}}{\mathrm{r}}}\)
- C \(\frac{2 \pi g}{r}\)
- D \(\frac{1}{2 \pi} \sqrt{\frac{r}{g}}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\mathrm{r}}}\)
Step-by-step Solution
Detailed explanation
For minimum velocity at the highest point we should have
\(\mathrm{mr} \omega^{2}=\mathrm{mg}\)
\(\therefore \omega^{2}=\frac{g}{r} \quad\) or \(\omega=\sqrt{\frac{g}{r}}\)
\(\begin{aligned} 2 \pi f &=\sqrt{\frac{g}{r}} \\ f &=\frac{1}{2 \pi} \sqrt{\frac{g}{r}} \end{aligned}\)
\(\mathrm{mr} \omega^{2}=\mathrm{mg}\)
\(\therefore \omega^{2}=\frac{g}{r} \quad\) or \(\omega=\sqrt{\frac{g}{r}}\)
\(\begin{aligned} 2 \pi f &=\sqrt{\frac{g}{r}} \\ f &=\frac{1}{2 \pi} \sqrt{\frac{g}{r}} \end{aligned}\)
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