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

Options

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

Explanation

To solve the problem of finding the speed of the chain when it leaves the pin, we will use the principle of conservation of energy. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the Initial and Final Positions of the Chain
- Initially, the center of mass of the chain is at a distance of \( \frac{l}{4} \) below the pin.
- After the small vertical displacement, the center of mass moves to a distance of \( \frac{l}{2} \) below the pin.

### Step 2: Calculate the Change in Height of the Center of Mass
- The change in height of the center of mass (\( h \)) can be calculated as:
\[ h = \text{final position} - \text{initial position} = \frac{l}{2} - \frac{l}{4} = \frac{l}{4} \]

### Step 3: Calculate the Change in Gravitational Potential Energy
- The decrease in gravitational potential energy (\( \Delta PE \)) when the chain moves down by \( h \) is given by:
\[ \Delta PE = mgh = mg \left(\frac{l}{4}\right) = \frac{mgl}{4} \]

### Step 4: Apply the Conservation of Energy Principle
- According to the conservation of energy, the decrease in potential energy is equal to the increase in kinetic energy (\( KE \)):
\[ \Delta PE = \Delta KE \]
- The kinetic energy of the chain when it leaves the pin can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]

### Step 5: Set Up the Equation
- Setting the decrease in potential energy equal to the increase in kinetic energy gives us:
\[ \frac{mgl}{4} = \frac{1}{2} mv^2 \]

### Step 6: Simplify the Equation
- We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{gl}{4} = \frac{1}{2} v^2 \]

### Step 7: Solve for \( v \)
- Rearranging the equation to solve for \( v^2 \):
\[ v^2 = \frac{gl}{2} \]
- Taking the square root of both sides gives:
\[ v = \sqrt{\frac{gl}{2}} \]

### Final Answer
- The speed of the chain when it leaves the pin is:
\[ v = \sqrt{\frac{gl}{2}} \]