A uniform chain has a mass M and length L. It is placed on a frictionless table with length `l_(0)` hanging over the edge. The chain begins to slide down. Ten, the speed v with which the end slides down from the edge is given by

Options

A. `v=sqrt((g)/(L)(L+l_(0)))`
B. `v=sqrt((g)/(L)(L-l_(0)))`
C. `v=sqrt((g)/(L)(L^(2)-l_(0)^(2)))`
D. `v=sqrt(2g`(L-l_(0)))`

Explanation

To solve the problem of the uniform chain sliding off a frictionless table, we can follow these steps:

### Step 1: Understand the System
We have a uniform chain of mass \( M \) and length \( L \) with a portion \( l_0 \) hanging off the edge of a frictionless table. As the chain slides down, we need to find the speed \( v \) of the end of the chain that is sliding down.

### Step 2: Set Up the Energy Conservation Equation
The principle of conservation of energy states that the loss in potential energy (PE) of the chain will equal the gain in kinetic energy (KE).

\[ \Delta U = \Delta KE \]

### Step 3: Calculate the Initial and Final Potential Energy
- **Initial Potential Energy (PE_initial)**: When \( l_0 \) length of the chain is hanging, the mass of the hanging part is \( \frac{M}{L} l_0 \). The height of the center of mass of this hanging part is \( \frac{l_0}{2} \). Thus, the initial potential energy is:

\[ U_i = \left(\frac{M}{L} l_0\right) g \left(\frac{l_0}{2}\right) = \frac{M g l_0^2}{2L} \]

- **Final Potential Energy (PE_final)**: When the chain has completely fallen down, the entire chain is at the same height, and thus the potential energy is zero:

\[ U_f = 0 \]

### Step 4: Calculate the Change in Potential Energy
The change in potential energy is:

\[ \Delta U = U_f - U_i = 0 - \frac{M g l_0^2}{2L} = -\frac{M g l_0^2}{2L} \]

### Step 5: Calculate the Kinetic Energy
The kinetic energy of the chain when it has fallen a distance \( l_0 \) is given by:

\[ KE = \frac{1}{2} M v^2 \]

### Step 6: Set Up the Equation
From the conservation of energy, we have:

\[ -\Delta U = \Delta KE \]

Substituting the expressions we derived:

\[ \frac{M g l_0^2}{2L} = \frac{1}{2} M v^2 \]

### Step 7: Simplify the Equation
We can cancel \( M \) from both sides (assuming \( M \neq 0 \)) and multiply through by 2:

\[ g l_0^2 = v^2 L \]

### Step 8: Solve for \( v \)
Rearranging gives us the expression for \( v \):

\[ v^2 = \frac{g l_0^2}{L} \]

Taking the square root:

\[ v = \sqrt{\frac{g l_0^2}{L}} = \sqrt{\frac{g}{L}} l_0 \]

### Final Answer
Thus, the speed \( v \) with which the end of the chain slides down is:

\[ v = \sqrt{\frac{g}{L}} l_0 \]

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