A particle is projected so as to just move along a vertical circle of radius r. The ratio of the tension in the string when the particle is at the lowest and highest point on the circle is infinite

Options

A. Statement -1 True, Statement -2 is True , Statement - 2 is a correct explanation for Statement -6
B. Statement -1 is True , Statement -2 is True , Statement -2 is NOT a correct explanation for statement -6
C. Statement -1 is true, Statement -2 is False
D. Statement - 1 is False , Statement - 2 is True

Answer

To solve the problem, we need to analyze the motion of a particle moving in a vertical circle and determine the tensions at the lowest and highest points of the circle.

### Step-by-Step Solution:

1. **Understand the Problem**:
A particle is projected to move along a vertical circle of radius \( r \). We need to find the tension in the string at the lowest point (point A) and the highest point (point B) of the circle.

2. **Identify Forces at the Lowest Point (Point A)**:
At the lowest point, the forces acting on the particle are:
- The gravitational force \( mg \) acting downwards.
- The tension \( T_A \) in the string acting upwards.

According to Newton's second law, the net force acting on the particle is equal to the centripetal force required to keep it moving in a circle:
\[ T_A - mg = \frac{mv_A^2}{r} \]
Rearranging gives:
\[ T_A = \frac{mv_A^2}{r} + mg \]

3. **Identify Forces at the Highest Point (Point B)**:
At the highest point, the forces acting on the particle are:
- The gravitational force \( mg \) acting downwards.
- The tension \( T_B \) in the string also acting downwards.

The net force must provide the centripetal force:
\[ T_B + mg = \frac{mv_B^2}{r} \]
Rearranging gives:
\[ T_B = \frac{mv_B^2}{r} - mg \]

4. **Condition for Just Completing the Circle**:
For the particle to just complete the vertical circle, the tension at the highest point must be zero (i.e., \( T_B = 0 \)). Thus, we set:
\[ 0 = \frac{mv_B^2}{r} - mg \]
This leads to:
\[ \frac{mv_B^2}{r} = mg \quad \Rightarrow \quad v_B^2 = rg \]

5. **Calculate Tension at the Lowest Point**:
Now, substituting \( v_B^2 \) into the equation for \( T_A \):
\[ T_A = \frac{m(v_A^2)}{r} + mg \]
To find \( v_A \), we can use conservation of energy between the lowest and highest points:
\[ \frac{1}{2}mv_A^2 + mg(2r) = \frac{1}{2}mv_B^2 + mg(0) \]
Substituting \( v_B^2 = rg \):
\[ \frac{1}{2}mv_A^2 + 2mgr = \frac{1}{2}m(rg) \]
Rearranging gives:
\[ \frac{1}{2}mv_A^2 = \frac{1}{2}mgr - 2mgr = -\frac{3}{2}mgr \]
This indicates that \( v_A^2 \) must be sufficient to maintain circular motion, leading to:
\[ T_A = \frac{m(4g)}{r} + mg = \frac{4mg}{r} + mg = \frac{5mg}{r} \]

6. **Calculate the Ratio of Tensions**:
Now, we can find the ratio of tensions:
\[ \text{Ratio} = \frac{T_A}{T_B} = \frac{\frac{5mg}{r}}{0} \]
Since \( T_B = 0 \), the ratio is infinite.

### Conclusion:
The ratio of the tension in the string when the particle is at the lowest point to the highest point is indeed infinite.