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. `((4T_A+T_B))/(5)`
D. `((T_A-T_B))/(5)`

Explanation

To solve the problem step by step, we will analyze the tension in the rope at a distance \( \frac{L}{5} \) from point \( A \).

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

**Hint:** Visualize the rope and label the points clearly to understand the tension distribution.

### Step 2: Define the relationship between tensions
The difference in tension between points \( B \) and \( A \) can be related to the mass and acceleration of the rope segment. We can express this as:
\[ T_B - T_A = M \cdot a \]
where \( a \) is the acceleration of the rope.

**Hint:** Remember that the net force acting on the rope segment is equal to the mass of the segment multiplied by its acceleration.

### Step 3: Analyze the segment from \( A \) to \( C \)
For the segment from \( A \) to \( C \) (length \( \frac{L}{5} \)), the mass of this segment is \( \frac{M}{5} \). The tension difference can be expressed as:
\[ T_C - T_A = \frac{M}{5} \cdot a \]

**Hint:** Use the mass of the segment to relate the tension at point \( C \) to the tension at point \( A \).

### Step 4: Substitute the acceleration
From Step 2, we know that:
\[ a = \frac{T_B - T_A}{M} \]
Substituting this expression for \( a \) into the equation from Step 3 gives:
\[ T_C - T_A = \frac{M}{5} \cdot \left(\frac{T_B - T_A}{M}\right) \]

**Hint:** This substitution allows you to eliminate \( a \) and express everything in terms of \( T_A \) and \( T_B \).

### Step 5: Simplify the equation
Now, simplifying the equation:
\[ T_C - T_A = \frac{1}{5} (T_B - T_A) \]
Rearranging gives:
\[ T_C = T_A + \frac{1}{5} (T_B - T_A) \]

**Hint:** This step combines the tensions effectively to find \( T_C \).

### Step 6: Combine and finalize the expression
Distributing the term on the right-hand side:
\[ T_C = T_A + \frac{1}{5} T_B - \frac{1}{5} T_A \]
Combining like terms results in:
\[ T_C = \frac{4}{5} T_A + \frac{1}{5} T_B \]
This can be rewritten as:
\[ T_C = \frac{4 T_A + T_B}{5} \]

**Hint:** Ensure that you combine the terms correctly to arrive at the final expression for \( T_C \).

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