A uniform chain of length L lies on a smooth horizontal table with its length perpendicular to the edge of the table and a small portion of the chain is hanging over the edge. The chain starts sliding due to the weight of the hanging part

Options

A. The acceleration of the chain is `(gx//L)`, where x is the length of the hanging part of chain.
B. The acceleration of the chain is `(g//L)(L-x)`, where x is the length of hanging part of chain.
C. The velocity of the chain is `x sqrt((g)/(L))`, where x is the length of hanging part of chain.
D. The velocity of the chain is `(L-x) sqrt((g)/(L))`, where x is the length of hanging part of chain.

Answer

To solve the problem of a uniform chain of length \( L \) lying on a smooth horizontal table with a portion hanging over the edge, we will analyze the forces acting on the chain and derive expressions for both the acceleration and velocity of the hanging part.

### Step-by-Step Solution:

1. **Define Variables**:
- Let the total length of the chain be \( L \).
- Let the length of the chain hanging over the edge at any time \( t \) be \( x \).
- The length of the chain on the table will then be \( L - x \).
- Let the total mass of the chain be \( m \).

2. **Determine the Mass of the Hanging Portion**:
- The mass of the entire chain is \( m \) for length \( L \).
- The mass per unit length of the chain is \( \frac{m}{L} \).
- Therefore, the mass of the hanging portion \( x \) is given by:
\[ m_x = \frac{m}{L} \cdot x \]

3. **Calculate the Weight of the Hanging Portion**:
- The weight \( W \) of the hanging portion is:
\[ W = m_x \cdot g = \left(\frac{m}{L} \cdot x\right) \cdot g = \frac{mgx}{L} \]

4. **Apply Newton's Second Law**:
- The net force acting on the hanging portion is equal to the weight of the hanging part, which causes the acceleration \( a \) of the chain.
- According to Newton's second law:
\[ F = m \cdot a \]
- Here, the force \( F \) is the weight of the hanging part:
\[ \frac{mgx}{L} = m \cdot a \]
- Dividing both sides by \( m \):
\[ \frac{gx}{L} = a \]
- Thus, the acceleration \( a \) of the hanging part is:
\[ a = \frac{g}{L} \cdot x \]

5. **Determine the Velocity of the Hanging Part**:
- To find the velocity \( v \) of the hanging part, we can use the relationship between acceleration and velocity:
\[ a = v \frac{dv}{dx} \]
- Substituting \( a = \frac{g}{L} \cdot x \):
\[ \frac{g}{L} \cdot x = v \frac{dv}{dx} \]
- Rearranging gives:
\[ v \, dv = \frac{g}{L} \cdot x \, dx \]
- Integrating both sides:
\[ \int v \, dv = \int \frac{g}{L} \cdot x \, dx \]
- This results in:
\[ \frac{v^2}{2} = \frac{g}{L} \cdot \frac{x^2}{2} + C \]
- Assuming the initial condition \( v = 0 \) when \( x = 0 \), we find \( C = 0 \):
\[ v^2 = \frac{g}{L} \cdot x^2 \]
- Therefore, the velocity \( v \) of the hanging part is:
\[ v = \sqrt{\frac{g}{L}} \cdot x \]

### Summary of Results:
- The acceleration of the hanging part is:
\[ a = \frac{g}{L} \cdot x \]
- The velocity of the hanging part is:
\[ v = \sqrt{\frac{g}{L}} \cdot x \]