A uniform flexible chain of mass m and length 2l hangs in equilibrium over a smooth horizontal pin of negligible diameter. One end of the chain is given a small vertical displacement so that the chain slips over the pin. The speed of chain when it leaves the pin is

Options

A. `sqrt(2gl)`
B. `sqrt(gl)`
C. `sqrt(4gl)`
D. `sqrt(3gl)`

Answer

To solve the problem of finding the speed of the chain when it leaves the pin, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Setup**:
- We have a uniform flexible chain of mass \( m \) and length \( 2l \) hanging in equilibrium over a smooth horizontal pin.
- The chain is initially at rest, and one end is given a small vertical displacement.

2. **Identify the Displacement of the Chain**:
- When the chain is displaced, the center of mass of the chain will also move.
- The center of mass of the hanging chain initially is at a height of \( l \) (half of its length). After the small displacement, the center of mass will drop by \( \frac{l}{2} \).

3. **Calculate the Change in Potential Energy**:
- The initial potential energy (PE) of the chain when it is at rest is given by:
\[ PE_{\text{initial}} = mgh = mg \cdot l \]
- After the displacement, the height of the center of mass is reduced by \( \frac{l}{2} \), so the new potential energy is:
\[ PE_{\text{final}} = mg \left(l - \frac{l}{2}\right) = mg \cdot \frac{l}{2} \]
- The loss in potential energy (\( \Delta PE \)) as the chain moves down is:
\[ \Delta PE = PE_{\text{initial}} - PE_{\text{final}} = mg \cdot l - mg \cdot \frac{l}{2} = mg \cdot \frac{l}{2} \]

4. **Relate Potential Energy Loss to Kinetic Energy Gain**:
- According to the principle of conservation of energy, the loss in potential energy is equal to the gain in kinetic energy (KE):
\[ \Delta PE = KE \]
- The kinetic energy of the chain when it leaves the pin can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]

5. **Set Up the Equation**:
- Equating the loss in potential energy to the kinetic energy gives:
\[ mg \cdot \frac{l}{2} = \frac{1}{2} mv^2 \]

6. **Solve for the Speed \( v \)**:
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g \cdot \frac{l}{2} = \frac{1}{2} v^2 \]
- Multiply both sides by 2:
\[ gl = v^2 \]
- Taking the square root of both sides gives:
\[ v = \sqrt{gl} \]

### Final Answer:
The speed of the chain when it leaves the pin is:
\[ v = \sqrt{gl} \]