A stone of mass 5 kg is tied to a string of length 10 m is whirled round in a horizontal circle. What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

Options

A. `10 "ms"^(-1)`
B. `15 "ms"^(-1)`
C. `20 "ms"^(-1)`
D. `25 "ms"^(-1)`

Explanation

To solve the problem, we need to determine the maximum speed at which a stone can be whirled around in a horizontal circle, given the mass of the stone, the length of the string, and the maximum tension the string can withstand.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the stone (m) = 5 kg
- Length of the string (R) = 10 m (This is also the radius of the circular path)
- Maximum tension in the string (T) = 200 N

2. **Understand the Relationship Between Tension and Centripetal Force:**
- When an object is moving in a circular path, the tension in the string provides the necessary centripetal force to keep the object moving in that path.
- The formula for centripetal force (F_c) is given by:
\[ F_c = \frac{mv^2}{R} \]
- Here, \(v\) is the speed of the stone, \(m\) is the mass, and \(R\) is the radius of the circular path.

3. **Set Up the Equation:**
- Since the maximum tension in the string is equal to the centripetal force required to keep the stone moving in a circle, we can set the two equal to each other:
\[ T = \frac{mv^2}{R} \]
- Substituting the known values:
\[ 200 = \frac{5v^2}{10} \]

4. **Rearrange the Equation to Solve for \(v^2\):**
- Multiply both sides by 10 to eliminate the denominator:
\[ 2000 = 5v^2 \]
- Now, divide both sides by 5:
\[ v^2 = \frac{2000}{5} = 400 \]

5. **Calculate the Maximum Speed \(v\):**
- Take the square root of both sides to find \(v\):
\[ v = \sqrt{400} = 20 \text{ m/s} \]

### Final Answer:
The maximum speed with which the stone can be whirled around is **20 m/s**.

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