The minimum speed of a bucket full of water whirled in a vertical circle of radius 10 m at the highest point so that the water may not fall is `(g=10ms^(-2))`

Options

A. `1ms^(-1)`
B. `2ms^(-1)`
C. `3ms^(-1)`
D. `10ms^(-1)`

Answer

To find the minimum speed of a bucket full of water whirled in a vertical circle at the highest point so that the water does not fall, we can follow these steps:

### Step 1: Understand the Forces at the Highest Point
At the highest point of the vertical circle, the forces acting on the bucket are:
1. The gravitational force acting downwards (weight of the water).
2. The centripetal force required to keep the bucket moving in a circle.

For the water to stay in the bucket, the centripetal force must be at least equal to the weight of the water.

### Step 2: Write the Equation for Centripetal Force
The centripetal force (\(F_c\)) required to keep the bucket moving in a circle is given by:
\[ F_c = \frac{mv^2}{r} \]
where:
- \(m\) is the mass of the bucket (and water),
- \(v\) is the speed of the bucket at the highest point,
- \(r\) is the radius of the circle.

### Step 3: Write the Equation for Weight
The weight (\(W\)) of the water in the bucket is given by:
\[ W = mg \]
where:
- \(g\) is the acceleration due to gravity.

### Step 4: Set Up the Equation
At the highest point, for the water to not fall, the centripetal force must be equal to the weight of the water:
\[ \frac{mv^2}{r} = mg \]

### Step 5: Simplify the Equation
We can cancel \(m\) from both sides (assuming \(m \neq 0\)):
\[ \frac{v^2}{r} = g \]

### Step 6: Solve for Speed
Rearranging the equation gives:
\[ v^2 = rg \]
Taking the square root of both sides:
\[ v = \sqrt{rg} \]

### Step 7: Substitute Values
Now, substitute the given values:
- \(r = 10 \, m\)
- \(g = 10 \, m/s^2\)

So,
\[ v = \sqrt{10 \times 10} = \sqrt{100} = 10 \, m/s \]

### Conclusion
The minimum speed of the bucket at the highest point so that the water does not fall is:
\[ \boxed{10 \, m/s} \]

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