A 2 kg stone tied at the end of a string of l m len"gt"h, is whirled along a vertical circle at a constant speed of `4 ms^-1`. The tension in the string has a value of 52 N when the stone is

Options

A. At the top of the circle
B. Half way down
C. At the bottom of the circle
D. Half way up

Answer

To solve the problem, we need to analyze the forces acting on the stone when it is whirled in a vertical circle. We will consider the position of the stone at the bottom of the circle since the tension is highest at that point.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the stone, \( m = 2 \, \text{kg} \)
- Length of the string (radius of the circle), \( r = 1 \, \text{m} \)
- Speed of the stone, \( v = 4 \, \text{m/s} \)

2. **Calculate the Centripetal Force:**
The centripetal force required to keep the stone moving in a circle is given by the formula:
\[ F_c = \frac{mv^2}{r} \]
Substituting the values:
\[ F_c = \frac{2 \, \text{kg} \cdot (4 \, \text{m/s})^2}{1 \, \text{m}} = \frac{2 \cdot 16}{1} = 32 \, \text{N} \]

3. **Calculate the Weight of the Stone:**
The weight of the stone (force due to gravity) is given by:
\[ F_g = mg \]
where \( g \approx 10 \, \text{m/s}^2 \). Therefore,
\[ F_g = 2 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20 \, \text{N} \]

4. **Determine the Tension in the String:**
At the bottom of the circle, the tension in the string \( T \) must provide the centripetal force and also balance the weight of the stone. The equation at the bottom of the circle is:
\[ T = F_c + F_g \]
Substituting the values we calculated:
\[ T = 32 \, \text{N} + 20 \, \text{N} = 52 \, \text{N} \]

5. **Conclusion:**
The tension in the string when the stone is at the bottom of the circle is \( 52 \, \text{N} \).

### Answer:
The tension in the string has a value of \( 52 \, \text{N} \) when the stone is at the **bottom of the circle**.