A `1kg` stone at the end of `1m` long string is whirled in a vertical circle at a constant speed of `4m//s`. The tension in the string is `6N`, when the stone is at `(g=10m//s^(2))`

Options

A. top of the circle
B. bottom of the circle
C. half way down
D. none of the above

Explanation

To solve the problem, we need to analyze the forces acting on the stone when it is at the highest point of its vertical circular motion.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the stone, \( m = 1 \, \text{kg} \)
- Length of the string (radius of the circle), \( r = 1 \, \text{m} \)
- Speed of the stone, \( v = 4 \, \text{m/s} \)
- Tension in the string at the highest point, \( T = 6 \, \text{N} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

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

3. **Analyze Forces at the Highest Point:**
At the highest point of the circle, the forces acting on the stone are:
- The gravitational force acting downward, \( F_g = mg = 1 \, \text{kg} \times 10 \, \text{m/s}^2 = 10 \, \text{N} \)
- The tension in the string, \( T \), also acting downward.

The net force acting on the stone at the highest point must provide the required centripetal force:
\[ T + mg = F_c \]
Substituting the known values:
\[ T + 10 \, \text{N} = 16 \, \text{N} \]

4. **Solve for Tension \( T \):**
Rearranging the equation to find \( T \):
\[ T = 16 \, \text{N} - 10 \, \text{N} = 6 \, \text{N} \]

5. **Conclusion:**
The tension in the string when the stone is at the highest point is \( 6 \, \text{N} \), which matches the given condition in the problem.