A particle is moving in a vertical circle with constant speed. The tansions in the string when passing through two positions at angles `30^@` and `60^@` from vertical (lowest position) are `T_1` and `T_2` respectively. Then

Options

A. `T_(1)=T_(2)`
B. `T_(2)gtT_(1)`
C. `T_(1)gtT_(2)`
D. Data insufficient

Explanation

To solve the problem, we need to analyze the forces acting on the particle at the two given positions in the vertical circle.

### Step-by-Step Solution:

1. **Understanding the Forces**:
When the particle is at an angle θ from the vertical, two forces act on it: the gravitational force (mg) acting downward and the tension (T) in the string acting upward along the direction of the string.

2. **Centripetal Force Requirement**:
For the particle to move in a circular path, the net force towards the center of the circle must provide the necessary centripetal force. This net force is the difference between the tension in the string and the component of gravitational force acting along the direction of the string.

3. **For θ = 30°**:
- The component of the gravitational force acting along the direction of the string is \( mg \cos(30°) \).
- The centripetal force required is given by \( \frac{mv^2}{r} \).
- Therefore, the equation for tension \( T_1 \) at θ = 30° is:
\[ T_1 = mg \cos(30°) + \frac{mv^2}{r} \]

4. **For θ = 60°**:
- Similarly, for θ = 60°, the component of the gravitational force is \( mg \cos(60°) \).
- The equation for tension \( T_2 \) at θ = 60° is:
\[ T_2 = mg \cos(60°) + \frac{mv^2}{r} \]

5. **Relating T1 and T2**:
- We know that \( \cos(30°) = \frac{\sqrt{3}}{2} \) and \( \cos(60°) = \frac{1}{2} \).
- Therefore:
\[ T_1 = mg \cdot \frac{\sqrt{3}}{2} + \frac{mv^2}{r} \]
\[ T_2 = mg \cdot \frac{1}{2} + \frac{mv^2}{r} \]

6. **Comparing T1 and T2**:
- Since \( \frac{\sqrt{3}}{2} > \frac{1}{2} \), it follows that:
\[ T_1 > T_2 \]

### Conclusion:
The relationship between the tensions in the string at the two positions is:
\[ T_1 > T_2 \]