Question

A bucket containing water is revolved in vertical circle of radius r. To prevent the water from falling down, the period of revolution required is (\(g\) = gravitational acceleration)

• A. \(2π\sqrt{\frac{g}{r}}\)
• B. \(2π\sqrt{\frac{r}{g}}\)
• C. \(2π\sqrt{rg}\)
• D. \(\frac{\sqrt{rg}}{2π}\)

Hint:

At the top, minimum speed satisfies mg = mv^2/r.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Water stays in the bucket at the top only if gravity alone is enough to supply the centripetal force, so the minimum speed there is \(v = \sqrt{rg}\).

Step 2: Compute period
\[ T = \frac{2\pi r}{v} = \frac{2\pi r}{\sqrt{rg}} = 2\pi\sqrt{\frac{r}{g}} \]
This is the maximum period (a slower revolution would let the water fall). Option (A) has \(g/r\) inverted, and (D) is not a time.

Final Answer:
The period must be at most \(2\pi\sqrt{r/g}\), option (B). \[ \boxed{2\pi\sqrt{\frac{r}{g}}} \]