A stone of mass of 1 kg is tied to the end of a string 1 m long. It is whirled in a vertical circle. The velocity of the stone at the bottom of the circle is just sufficient to take it to the top of circle without slackening of the string. What is the tension in the string at the top of the circle? (Take, g =10 `ms^(-2)`)

Options

A. zero
B. 1 N
C. `sqrt(10)N`
D. 10 N

Explanation

To solve the problem, we will analyze the motion of the stone at the top of the vertical circle and apply the principles of circular motion and gravitational force.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a stone of mass \( m = 1 \, \text{kg} \) tied to a string of length \( r = 1 \, \text{m} \).
- The stone is whirled in a vertical circle, and we need to find the tension in the string when the stone is at the top of the circle.

2. **Condition for Motion at the Top**:
- At the top of the circle, the forces acting on the stone are the gravitational force \( mg \) acting downward and the tension \( T \) in the string also acting downward.
- For the stone to maintain circular motion at the top, the centripetal force required must be provided by the sum of these forces.

3. **Centripetal Force Equation**:
- The centripetal force \( F_c \) required to keep the stone moving in a circle is given by:
\[ F_c = \frac{mv^2}{r} \]
- At the top of the circle, the net force acting towards the center (which provides the centripetal force) is:
\[ F_c = mg + T \]

4. **Setting Up the Equation**:
- From the above, we can write:
\[ \frac{mv^2}{r} = mg + T \]
- Rearranging gives:
\[ T = \frac{mv^2}{r} - mg \]

5. **Finding the Velocity at the Top**:
- For the stone to just make it to the top without slackening the string, the minimum velocity \( v \) at the top can be derived from the condition that the tension \( T \) can be zero (the stone is just on the verge of falling).
- Setting \( T = 0 \):
\[ 0 = \frac{mv^2}{r} - mg \]
- This leads to:
\[ \frac{mv^2}{r} = mg \]
- Simplifying gives:
\[ v^2 = rg \]

6. **Substituting Values**:
- Substituting \( r = 1 \, \text{m} \) and \( g = 10 \, \text{ms}^{-2} \):
\[ v^2 = 1 \times 10 = 10 \]
- Therefore, \( v = \sqrt{10} \, \text{m/s} \).

7. **Calculating Tension at the Top**:
- Now substituting \( v^2 \) back into the tension equation:
\[ T = \frac{m \cdot 10}{1} - mg \]
- This simplifies to:
\[ T = 10 - 10 = 0 \, \text{N} \]

### Final Answer:
The tension in the string at the top of the circle is \( T = 0 \, \text{N} \).