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 1: Understand the System
We have a uniform chain of length \( L \) placed on a smooth horizontal table, with half of its length (\( \frac{L}{2} \)) hanging off the edge. When the chain is released from rest, the hanging part will fall due to gravity.

### Step 2: Identify the Mass and Center of Mass
The total mass of the chain can be denoted as \( M \). Since half of the chain is hanging, the mass of the hanging part is \( \frac{M}{2} \). The center of mass of the hanging part, which is \( \frac{L}{4} \) from the edge of the table, will fall a distance of \( \frac{L}{4} \) when the chain is released.

### Step 3: Calculate the Change in Potential Energy
The change in potential energy (\( \Delta PE \)) when the chain falls can be calculated as:
\[ \Delta PE = \text{mass} \times g \times \text{height} \]
For the hanging part:
\[ \Delta PE = \left(\frac{M}{2}\right) \cdot g \cdot \left(\frac{L}{4}\right) = \frac{MgL}{8} \]

### Step 4: Relate Potential Energy to Kinetic Energy
When the chain falls, this potential energy converts into kinetic energy (\( KE \)). The kinetic energy of the chain when it leaves the table is given by:
\[ KE = \frac{1}{2} M v^2 \]
Since only half of the mass is falling, the kinetic energy at the point of leaving will be:
\[ KE = \frac{1}{2} \left(\frac{M}{2}\right) v^2 = \frac{Mv^2}{4} \]

### Step 5: Set Up the Energy Conservation Equation
Setting the change in potential energy equal to the kinetic energy gives us:
\[ \frac{MgL}{8} = \frac{Mv^2}{4} \]

### Step 6: Solve for Velocity
We can cancel \( M \) from both sides (assuming \( M \neq 0 \)):
\[ \frac{gL}{8} = \frac{v^2}{4} \]
Multiplying both sides by 4:
\[ \frac{gL}{2} = v^2 \]
Taking the square root gives us:
\[ v = \sqrt{\frac{gL}{2}} \]

### Final Step: Conclusion
Thus, the velocity with which the chain leaves the table is:
\[ v = \sqrt{\frac{gL}{2}} \]