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. `5:3`
B. `1:4`
C. `1:2`
D. `1:3`

Explanation

To solve the problem, we need to analyze the system of two particles connected by a rope and rotating about a fixed point. We will denote the mass of each particle as \( m \) and the total length of the rope as \( L \). The two particles are positioned such that one is at point A and the other divides the length of the rope in the ratio 1:2 from A.

### Step-by-Step Solution:

1. **Identify the Lengths**:
- Let the total length of the rope \( AB = L \).
- The first particle at point A is at distance \( 0 \) from A.
- The second particle divides the rope in the ratio 1:2, so it is located at \( \frac{L}{3} \) from A and \( \frac{2L}{3} \) from B.

2. **Determine the Radius of Rotation**:
- The radius of rotation for the particle at A is \( r_A = L \).
- The radius of rotation for the particle at the second position (let's call it P) is \( r_P = \frac{2L}{3} \).

3. **Centrifugal Force Calculation**:
- The centrifugal force acting on the particle at A is given by:
\[ F_A = m \omega^2 r_A = m \omega^2 L \]
- The centrifugal force acting on the particle at P is given by:
\[ F_P = m \omega^2 r_P = m \omega^2 \left(\frac{2L}{3}\right) = \frac{2}{3} m \omega^2 L \]

4. **Tension Calculation**:
- For the particle at A, the tension \( T_2 \) is equal to the centrifugal force:
\[ T_2 = m \omega^2 L \]
- For the particle at P, the tension \( T_1 \) must balance the centrifugal force acting on it and the tension in the rope:
\[ T_1 = T_2 + F_P \]
Substituting the values:
\[ T_1 = m \omega^2 L + \frac{2}{3} m \omega^2 L = m \omega^2 L \left(1 + \frac{2}{3}\right) = m \omega^2 L \cdot \frac{5}{3} \]

5. **Ratio of Tensions**:
- Now, we can find the ratio of the tensions \( \frac{T_1}{T_2} \):
\[ \frac{T_1}{T_2} = \frac{\frac{5}{3} m \omega^2 L}{m \omega^2 L} = \frac{5}{3} \]

### Final Answer:
The ratio of the tensions in the smaller part to the other is \( \frac{5}{3} \).