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A cone filled with water is revolved in a vertical circle of radius 4 m and the water does not fall down. What must be the maximum period of revolution?
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1109 Upvotes
Verified Answer
The correct answer is:
4 s
When a cone filled with water is revolved in a vertical circle, then the velocity at the highest point is given by
Also
$\begin{aligned}
& v \geq \sqrt{g r} \\
& v=r w=r \frac{2 \pi}{T}
\end{aligned}$
Hence, $\quad \frac{2 r \pi}{T} \geq \sqrt{r g}$
$\begin{aligned}
T & \leq \frac{2 \pi r}{\sqrt{r g}}=2 \pi \sqrt{\frac{r}{g}}=2 \pi \sqrt{\frac{4}{9.8}} \\
& =2 \times 3.14 \times 0.6389=4.0 \mathrm{sec}
\end{aligned}$
So, $T_{\max }=4 \mathrm{sec}$
Also
$\begin{aligned}
& v \geq \sqrt{g r} \\
& v=r w=r \frac{2 \pi}{T}
\end{aligned}$
Hence, $\quad \frac{2 r \pi}{T} \geq \sqrt{r g}$
$\begin{aligned}
T & \leq \frac{2 \pi r}{\sqrt{r g}}=2 \pi \sqrt{\frac{r}{g}}=2 \pi \sqrt{\frac{4}{9.8}} \\
& =2 \times 3.14 \times 0.6389=4.0 \mathrm{sec}
\end{aligned}$
So, $T_{\max }=4 \mathrm{sec}$
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