A uniform chain of length l is placed on a smooth horizontal table, such that half of its length hangs over one edge. It is releasedfrom rest, the velocity with which it leaves the table is

Options

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

Explanation

To solve the problem of finding the velocity with which the chain leaves the table, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a uniform chain of length \( L \).
- Half of the chain, which is \( \frac{L}{2} \), is hanging off the edge of the table.
- The chain is released from rest.

2. **Identifying the Mass of the Chain**:
- Let the total mass of the chain be \( M \).
- The mass of the hanging part of the chain is \( \frac{M}{2} \) since half the length is hanging.

3. **Calculating the Change in Gravitational Potential Energy**:
- The center of mass of the hanging part of the chain is at a distance of \( \frac{L}{4} \) below the table (since it is uniformly distributed).
- The change in gravitational potential energy (\( \Delta PE \)) when the chain is released can be calculated as:
\[ \Delta PE = -\left(\text{mass of hanging part}\right) \times g \times \left(\text{height of center of mass}\right) \]
\[ \Delta PE = -\left(\frac{M}{2}\right) \times g \times \left(\frac{L}{4}\right) = -\frac{MgL}{8} \]

4. **Relating Potential Energy to Kinetic Energy**:
- The loss in gravitational potential energy is converted into kinetic energy (\( KE \)) of the chain as it falls.
- The kinetic energy of the chain when it leaves the table is given by:
\[ KE = \frac{1}{2} M v^2 \]
- Setting the loss in potential energy equal to the kinetic energy:
\[ -\frac{MgL}{8} = \frac{1}{2} M v^2 \]

5. **Solving for Velocity**:
- We can simplify the equation by canceling \( M \) (assuming \( M \neq 0 \)):
\[ -\frac{gL}{8} = \frac{1}{2} v^2 \]
- Rearranging gives:
\[ v^2 = \frac{gL}{4} \]
- Taking the square root of both sides, we find:
\[ v = \sqrt{\frac{gL}{4}} = \frac{1}{2} \sqrt{gL} \]

### Final Answer:
The velocity with which the chain leaves the table is:
\[ v = \frac{1}{2} \sqrt{gL} \]