A perfectly straight protion 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)gtT_(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. `(4 T_(A) + T_(B))//5`
D. `(T_(A) - 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 Given Information
We have a uniform rope of 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 distance \( \frac{L}{5} \) from end \( A \).

### Step 2: Set Up the Problem
Since the rope is uniform, the mass of the segment from \( A \) to \( \frac{L}{5} \) is:
\[ m = \frac{M}{L} \cdot \frac{L}{5} = \frac{M}{5} \]

### Step 3: Apply Newton's Second Law
Consider the segment of the rope from \( A \) to \( \frac{L}{5} \). Let the acceleration of the rope be \( a \) in the direction of \( T_B \).

Using Newton's second law for this segment:
\[ T_C - T_A = \text{mass} \times \text{acceleration} \]
Substituting the mass:
\[ T_C - T_A = \left(\frac{M}{5}\right) a \]

### Step 4: Relate the Acceleration to Tensions
From the whole rope, we can also write:
\[ T_B - T_A = M a \]
From this, we can express \( a \):
\[ a = \frac{T_B - T_A}{M} \]

### Step 5: Substitute \( a \) Back into the Equation
Substituting \( a \) back into the equation for \( T_C \):
\[ T_C - T_A = \left(\frac{M}{5}\right) \left(\frac{T_B - T_A}{M}\right) \]
This simplifies to:
\[ T_C - T_A = \frac{1}{5}(T_B - T_A) \]

### Step 6: Solve for \( T_C \)
Rearranging gives:
\[ T_C = T_A + \frac{1}{5}(T_B - T_A) \]
\[ T_C = T_A + \frac{1}{5}T_B - \frac{1}{5}T_A \]
\[ T_C = \frac{4}{5}T_A + \frac{1}{5}T_B \]

### Final Expression
Thus, the tension in the rope at a distance \( \frac{L}{5} \) from end \( A \) is:
\[ T_C = \frac{4T_A + T_B}{5} \]