A bucket full of water is rotated in a vertical circle of radius R. If the water does not split out, the speed of the bucket at topmost point will be

Options

A. `sqrt(Rg)`
B. `sqrt(5gR)`
C. `sqrt(2Rg)`
D. `sqrt((R)/(g))`

Explanation

To solve the problem of determining the speed of the bucket at the topmost point when the water does not spill out, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Forces at the Topmost Point**:
At the topmost point of the circular motion, the forces acting on the bucket are:
- The gravitational force (weight of the bucket and water) acting downwards, which is \( mg \).
- The centripetal force required to keep the bucket moving in a circle, which is provided by the gravitational force and the normal force from the water.

2. **Setting Up the Equation**:
At the topmost point, the normal force \( N \) can be considered as zero for the water to not spill out. Therefore, the equation for forces can be written as:
\[ mg = \frac{mv^2}{R} + N \]
Since \( N = 0 \) (the water does not exert any upward force), the equation simplifies to:
\[ mg = \frac{mv^2}{R} \]

3. **Cancelling Mass**:
Since mass \( m \) appears on both sides of the equation, we can cancel it out (assuming \( m \neq 0 \)):
\[ g = \frac{v^2}{R} \]

4. **Solving for Velocity**:
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = gR \]
Taking the square root of both sides gives us the speed \( v \):
\[ v = \sqrt{gR} \]

5. **Conclusion**:
Thus, the speed of the bucket at the topmost point, when the water does not spill out, is:
\[ v = \sqrt{gR} \]

### Final Answer:
The speed of the bucket at the topmost point is \( \sqrt{gR} \).

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