A stone of mass 1 kg tied at the end of a string of length 1 m and is whirled in a verticle circle at a constant speed of `3 ms^(-1)`. The tension in the string will be 19 N when the stone is `(g=10 ms^(-1))`

Options

A. Top of the circle
B. Bottom of the circle
C. Half way down
D. None of these

Answer

To solve the problem, we need to analyze the forces acting on the stone when it is at the bottom of the vertical circle. We'll use the concepts of centripetal force and gravitational force to find the tension in the string.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Stone at the Bottom of the Circle:**
At the bottom of the vertical circle, the forces acting on the stone are:
- The tension (T) in the string acting upwards.
- The weight (mg) of the stone acting downwards.

2. **Calculate the Weight of the Stone:**
The weight of the stone can be calculated using the formula:
\[ mg = m \cdot g \]
Given that the mass (m) is 1 kg and the acceleration due to gravity (g) is \(10 \, \text{ms}^{-2}\):
\[ mg = 1 \, \text{kg} \cdot 10 \, \text{ms}^{-2} = 10 \, \text{N} \]

3. **Calculate the Centripetal Force Required:**
The centripetal force (Fc) required to keep the stone moving in a circle is given by:
\[ F_c = \frac{mv^2}{r} \]
Where:
- \(m = 1 \, \text{kg}\)
- \(v = 3 \, \text{ms}^{-1}\)
- \(r = 1 \, \text{m}\)

Substituting the values:
\[ F_c = \frac{1 \, \text{kg} \cdot (3 \, \text{ms}^{-1})^2}{1 \, \text{m}} = \frac{1 \cdot 9}{1} = 9 \, \text{N} \]

4. **Apply Newton's Second Law:**
At the bottom of the circle, the net force acting on the stone is the difference between the tension and the weight:
\[ T - mg = F_c \]
Rearranging gives:
\[ T = F_c + mg \]

5. **Substitute the Values to Find Tension:**
Now substitute the values of \(F_c\) and \(mg\):
\[ T = 9 \, \text{N} + 10 \, \text{N} = 19 \, \text{N} \]

Thus, the tension in the string when the stone is at the bottom of the circle is **19 N**.

### Summary:
- The tension in the string is found to be 19 N when the stone is at the bottom of the vertical circle.