Half of the chain is lying on a table. The rest of the chain hangs in equilibrium from both of the ends of the table. Extra length of the hanging chain at one end is 1/8 th of the total length of the chain than that hanging at the other end. The coefficient of friction between table and chain is `25 x *10^(-4)` the value of x is

Answer

To solve the problem step by step, we will analyze the situation described and apply the principles of equilibrium and friction.

### Step 1: Understand the problem setup
We have a chain of total length \( L \). Half of the chain, which is \( \frac{L}{2} \), is lying on the table. The remaining half is hanging off the table, with one end hanging \( \frac{1}{8} \) of the total length of the chain less than the other end.

### Step 2: Define the lengths of the hanging parts
Let the length of the chain hanging from one end be \( x \). According to the problem, the length hanging from the other end will be \( x + \frac{L}{8} \). Since the total length of the chain hanging must equal \( \frac{L}{2} \), we can set up the equation:

\[ x + \left(x + \frac{L}{8}\right) = \frac{L}{2} \]

### Step 3: Simplify the equation
Combining the terms gives:

\[ 2x + \frac{L}{8} = \frac{L}{2} \]

### Step 4: Solve for \( x \)
To isolate \( x \), first subtract \( \frac{L}{8} \) from both sides:

\[ 2x = \frac{L}{2} - \frac{L}{8} \]

Finding a common denominator (which is 8) allows us to rewrite \( \frac{L}{2} \) as \( \frac{4L}{8} \):

\[ 2x = \frac{4L}{8} - \frac{L}{8} = \frac{3L}{8} \]

Now, divide both sides by 2:

\[ x = \frac{3L}{16} \]

### Step 5: Find the lengths of the hanging parts
The lengths of the hanging parts are now:

- One side: \( x = \frac{3L}{16} \)
- The other side: \( x + \frac{L}{8} = \frac{3L}{16} + \frac{2L}{16} = \frac{5L}{16} \)

### Step 6: Calculate the masses of the hanging parts
Assuming the mass per unit length of the chain is \( \lambda \), the masses of the hanging parts are:

- Mass on one side: \( m_1 = \lambda \cdot \frac{3L}{16} \)
- Mass on the other side: \( m_2 = \lambda \cdot \frac{5L}{16} \)

### Step 7: Apply the equilibrium condition
At equilibrium, the frictional force must balance the difference in the weights of the hanging parts. The frictional force \( F_f \) acting on the chain on the table is given by:

\[ F_f = \mu \cdot m_t \cdot g \]

where \( m_t \) is the mass of the chain on the table, which is \( \lambda \cdot \frac{L}{2} \).

### Step 8: Set up the equation
The weight of the hanging parts gives us:

\[ m_2 \cdot g - m_1 \cdot g = F_f \]

Substituting the values gives:

\[ \left(\lambda \cdot \frac{5L}{16} - \lambda \cdot \frac{3L}{16}\right) g = \mu \cdot \left(\lambda \cdot \frac{L}{2}\right) g \]

### Step 9: Simplify the equation
Canceling \( \lambda g \) from both sides results in:

\[ \frac{2L}{16} = \mu \cdot \frac{L}{2} \]

### Step 10: Solve for the coefficient of friction \( \mu \)
This simplifies to:

\[ \frac{1}{8} = \mu \cdot \frac{1}{2} \]

Thus, solving for \( \mu \):

\[ \mu = \frac{1}{8} \cdot 2 = \frac{1}{4} \]

### Step 11: Relate \( \mu \) to the given coefficient
The problem states that:

\[ \mu = 25x \cdot 10^{-4} \]

Setting the two expressions for \( \mu \) equal gives:

\[ \frac{1}{4} = 25x \cdot 10^{-4} \]

### Step 12: Solve for \( x \)
Rearranging gives:

\[ x = \frac{1}{4} \cdot \frac{10^4}{25} = \frac{10^4}{100} = 100 \]

Thus, the value of \( x \) is:

\[ \boxed{100} \]