MHT CET · Physics · Mechanical Properties of Fluids
A bucket containing water is revolved in a vertical circle of radius \(r\) .To prevent the water from falling down, the minimum frequency of revolution required is ( \(g=\) acceleration due to gravity)
- A \(2 \pi \sqrt{\frac{r}{g}}\)
- B \(\frac{1}{2 \pi} \sqrt{\frac{r}{g}}\)
- C \(\frac{1}{2 \pi} \sqrt{\frac{g}{r}}\)
- D \(2 \pi \sqrt{\frac{g}{r}}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2 \pi} \sqrt{\frac{g}{r}}\)
Step-by-step Solution
Detailed explanation
Let its angular velocity be \(\omega\) at all points (uniform motion). At the highest point, weight of the body is balanced centrifugal force, so
\(
\begin{aligned}
& m \omega^2 r=m g \\
& \Rightarrow \omega=\sqrt{\frac{g}{r}}
\end{aligned}
\)
Angular frequency is related to frequency as follows:
\(\omega=2 \pi f\)
\(\therefore f=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{g}{r}}\)
\(
\begin{aligned}
& m \omega^2 r=m g \\
& \Rightarrow \omega=\sqrt{\frac{g}{r}}
\end{aligned}
\)
Angular frequency is related to frequency as follows:
\(\omega=2 \pi f\)
\(\therefore f=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{g}{r}}\)
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