A block of mass M is resting on a smooth horizontal plane. One end of a uniform rope of mass `M/4` is fixed to the block , which pulled it in the horizontal direction by applying a force F at the other ends . The tension in the middle of the rope is

Options

A. `F/2`
B. `F/5`
C. `9/10F`
D. F

Answer

To solve the problem of finding the tension at the midpoint of a rope attached to a block being pulled, we can follow these steps:

### Step-by-Step Solution

1. **Identify the System**:
- We have a block of mass \( M \) resting on a smooth horizontal plane.
- A uniform rope of mass \( \frac{M}{4} \) is attached to the block, and a force \( F \) is applied to the other end of the rope.

2. **Draw Free Body Diagrams (FBD)**:
- For the block of mass \( M \): The only horizontal forces acting on it are the tension \( T \) in the rope and the applied force \( F \).
- For the rope: The rope can be divided into two halves. The tension at the end attached to the block is \( T \), and the tension at the midpoint is \( T' \).

3. **Apply Newton's Second Law**:
- For the block:
\[ F - T = M \cdot a \quad \text{(1)} \]
- For the first half of the rope (mass = \( \frac{M}{8} \)):
\[ T' - T = \frac{M}{8} \cdot a \quad \text{(2)} \]
- For the second half of the rope (mass = \( \frac{M}{8} \)):
\[ F - T' = \frac{M}{8} \cdot a \quad \text{(3)} \]

4. **Combine the Equations**:
- From equations (1), (2), and (3), we can express \( F \) in terms of \( a \):
- From (1): \( F = T + M \cdot a \)
- From (2): \( T' = T + \frac{M}{8} \cdot a \)
- From (3): \( F = T' + \frac{M}{8} \cdot a \)

5. **Substituting and Solving for Acceleration**:
- Substitute \( T' \) from equation (2) into equation (3):
\[ F = (T + \frac{M}{8} \cdot a) + \frac{M}{8} \cdot a \]
\[ F = T + \frac{M}{4} \cdot a \]
- Set the two expressions for \( F \) equal:
\[ T + M \cdot a = T + \frac{M}{4} \cdot a \]
- Rearranging gives:
\[ M \cdot a = \frac{M}{4} \cdot a \]
- This implies:
\[ a = \frac{4F}{5M} \]

6. **Finding the Tension at Midpoint**:
- Substitute \( a \) back into equation (2) to find \( T' \):
\[ T' = T + \frac{M}{8} \cdot \frac{4F}{5M} \]
- From equation (1), we know \( T = M \cdot a = M \cdot \frac{4F}{5M} = \frac{4F}{5} \).
- Substitute \( T \) into the equation for \( T' \):
\[ T' = \frac{4F}{5} + \frac{M}{8} \cdot \frac{4F}{5M} = \frac{4F}{5} + \frac{4F}{40} = \frac{4F}{5} + \frac{F}{10} = \frac{8F}{10} + \frac{F}{10} = \frac{9F}{10} \]

### Final Answer:
The tension at the midpoint of the rope is:
\[ T' = \frac{9F}{10} \]