Question

A bucket filled with water is revolved in a vertical circle of radius 10 m and water just not fall down. The time period of revolution will be (acceleration due to gravity \(=10 \mathrm{~m} / \mathrm{s}^2\) )

• A. \(\pi \mathrm{s}\)
• B. \(2\pi \text{ }\mathrm{s}\)
• C. \(10\pi \text{ }\mathrm{s}\)
• D. \(20\pi \text{ }\mathrm{s}\)

Hint:

Water just stays in the bucket at the top when gravity alone provides the centripetal force, giving the minimum speed there.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
At the top of the circle, water just does not fall when the weight supplies the centripetal force: \(mg = \dfrac{mv^2}{r}\).

Step 2: Speed at the Top
\(v = \sqrt{gr} = \sqrt{10\times 10} = 10\ \text{m/s}\)

Step 3: Time Period
\(T = \dfrac{2\pi r}{v} = \dfrac{2\pi\times 10}{10} = 2\pi\ \text{s}\)

Step 4: Compare the options
\(2\pi\) s is option B.

Final Answer:
The time period of revolution is 2 pi seconds. \[ \boxed{\text{(B) }2\pi\ \text{s}} \]