A perfectly straight portion of a uniform rope has mass M and length L. At end A of the segment, the tension in the rope is `T_(A)` and at end B it is `T_(B) (T_(B) gt T_(A) )`. Neglect effect of gravity and no contact force acts on the rope in between points A and B. The tension in the rope at a distance L/5 from end A is:-

Options

A. `T_(B)-T_(A)`
B. `(T_(A)+T_(B))^(//5)`
C. `(4T_(A)+T_(B))^(//5)`
D. `(T_(1)-T_(B))^(//5)`

Explanation

To find the tension in the rope at a distance \( \frac{L}{5} \) from end A, we can follow these steps:

### Step 1: Understand the Problem
We have a uniform rope with mass \( M \) and length \( L \). The tension at end A is \( T_A \) and at end B is \( T_B \), where \( T_B > T_A \). We need to find the tension \( T_C \) at a point \( C \) which is located at a distance \( \frac{L}{5} \) from end A.

### Step 2: Set Up the Force Equation
Since the tension increases from \( T_A \) to \( T_B \), there must be a net force acting on the rope, which implies that the rope is accelerating. We can write the equation for the net force acting on the entire rope:

\[ T_B - T_A = M \cdot A \]

where \( A \) is the acceleration of the rope.

### Step 3: Analyze the Segment of the Rope
Now, let's focus on the segment of the rope from point A to point C (length \( \frac{L}{5} \)). The mass of this segment can be calculated as:

\[ \text{Mass of segment} = \frac{M}{L} \cdot \frac{L}{5} = \frac{M}{5} \]

### Step 4: Write the Force Equation for Segment AC
For the segment from A to C, we can write the force equation as follows:

\[ T_C - T_A = \left(\frac{M}{5}\right) \cdot A \]

### Step 5: Substitute for Acceleration
From Step 2, we have \( A = \frac{T_B - T_A}{M} \). Substitute this expression for \( A \) into the equation from Step 4:

\[ T_C - T_A = \left(\frac{M}{5}\right) \cdot \left(\frac{T_B - T_A}{M}\right) \]

### Step 6: Simplify the Equation
This simplifies to:

\[ T_C - T_A = \frac{1}{5}(T_B - T_A) \]

### Step 7: Solve for \( T_C \)
Now, rearranging the equation gives us:

\[ T_C = T_A + \frac{1}{5}(T_B - T_A) \]

Distributing the \( \frac{1}{5} \):

\[ T_C = T_A + \frac{1}{5}T_B - \frac{1}{5}T_A \]

Combining like terms:

\[ T_C = \frac{5}{5}T_A - \frac{1}{5}T_A + \frac{1}{5}T_B \]

\[ T_C = \frac{4}{5}T_A + \frac{1}{5}T_B \]

### Step 8: Final Expression for \( T_C \)
Thus, we can express \( T_C \) as:

\[ T_C = \frac{4T_A + T_B}{5} \]

### Conclusion
The tension in the rope at a distance \( \frac{L}{5} \) from end A is:

\[ T_C = \frac{4T_A + T_B}{5} \]