A particle of mass `2m` is connected by an inextensible string of length `1.2 m` to a ring of mass m which is free to slide on a horizontal smooth rod. Initially the ring and the particle are at the same level with the string taut. Both are then released simultaneously. The distance in meter moved by the ring when the string becomes veritcal is :-

Options

A. 0
B. `0.4`
C. `0.8`
D. `1.2`

Explanation

To solve the problem, we need to determine how far the ring moves when the string becomes vertical. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the System
We have a particle of mass \(2m\) hanging from a ring of mass \(m\) via an inextensible string of length \(1.2\) m. The system is initially at rest with the particle and the ring at the same horizontal level.

### Step 2: Identify the Center of Mass (COM)
The center of mass of the system can be calculated using the formula:
\[ x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
where:
- \(m_1 = 2m\) (mass of the particle)
- \(m_2 = m\) (mass of the ring)
- \(x_1\) is the position of the particle (which we will take as \(1.2\) m when the string is vertical)
- \(x_2\) is the position of the ring (which we will denote as \(x\))

### Step 3: Set Up the Equation
When the string becomes vertical, the distance from the ring to the center of mass can be expressed as:
\[ x_{cm} = \frac{(2m)(1.2) + (m)(x)}{2m + m} \]
This simplifies to:
\[ x_{cm} = \frac{2.4m + mx}{3m} \]
\[ x_{cm} = \frac{2.4 + x}{3} \]

### Step 4: Calculate the Position of the Center of Mass
Since the string is vertical, the center of mass will remain at the same horizontal position due to the absence of horizontal forces. The center of mass will be at a distance of \(0.8\) m from the ring when the string is vertical. Thus, we set:
\[ x_{cm} = 0.8 \]

### Step 5: Solve for \(x\)
Now we can set up the equation:
\[ 0.8 = \frac{2.4 + x}{3} \]
Multiplying both sides by \(3\):
\[ 2.4 + x = 2.4 \]
Subtracting \(2.4\) from both sides:
\[ x = 0.8 \]

### Conclusion
The distance moved by the ring when the string becomes vertical is \(0.8\) m.

### Final Answer
The distance moved by the ring is \(0.8\) m.

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