Two particles of equal mass are connected to a rope AB of negligible mass, such that one is at end A and the other dividing the length of the rope in the ratio 1:2 from A.The rope is rotated about end B is a horizontal plane. Ratio of the tensions in the smaller part to the other is (ignore effect of gravity)

Options

A. `4:3`
B. `1:4`
C. `1:2`
D. `1:3`

Explanation

To solve the problem, we need to analyze the forces acting on the two masses connected by the rope and derive the ratio of the tensions in the two segments of the rope.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two particles of equal mass \( m \).
- The rope AB is divided into two parts: one part (AC) is \( \frac{L}{3} \) and the other part (CB) is \( \frac{2L}{3} \).
- The rope is rotated about point B in a horizontal plane.

2. **Identifying the Lengths**:
- Let the total length of the rope \( L \).
- The length of segment AC (the shorter part) is \( \frac{L}{3} \).
- The length of segment CB (the longer part) is \( \frac{2L}{3} \).

3. **Centripetal Force for Mass at C**:
- For the mass at point C (which is at distance \( \frac{2L}{3} \) from point B), the centripetal force required is provided by the tension \( T_1 \) in the rope segment CB.
- The equation for the centripetal force is:
\[ T_1 = m \omega^2 \left(\frac{2L}{3}\right) \]

4. **Centripetal Force for Mass at A**:
- For the mass at point A (which is at distance \( \frac{L}{3} \) from point B), the centripetal force required is provided by the tension \( T_2 \) in the rope segment AC.
- The equation for the centripetal force is:
\[ T_2 = m \omega^2 \left(\frac{L}{3}\right) \]

5. **Finding the Ratio of Tensions**:
- Now, we have:
\[ T_1 = m \omega^2 \left(\frac{2L}{3}\right) \]
\[ T_2 = m \omega^2 \left(\frac{L}{3}\right) \]
- To find the ratio \( \frac{T_2}{T_1} \):
\[ \frac{T_2}{T_1} = \frac{m \omega^2 \left(\frac{L}{3}\right)}{m \omega^2 \left(\frac{2L}{3}\right)} = \frac{\frac{L}{3}}{\frac{2L}{3}} = \frac{1}{2} \]

6. **Final Ratio**:
- Therefore, the ratio of the tensions in the smaller part (T2) to the larger part (T1) is:
\[ \frac{T_2}{T_1} = \frac{1}{2} \]

### Conclusion:
The ratio of the tensions in the smaller part to the larger part is \( 1:2 \).