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A can filled with water is revolved in a vertical circle of radius 4 m, and the water just does not fall down. The time period of revolution will be:

1. 1 s
2. 10 s
3. 8 s
4. 4 s

Correct answer: 4 s

Solution

For water to just not fall at the top, the normal reaction there must be zero, so \(v^2/r = g\). With \(r=4\) m and \(g=10\,\text{m/s}^2\), the speed is \(v=\sqrt{40}\). Then \(T = 2\pi r/v\), which gives approximately 4 s.

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