A block is lying on the horizontal frictionless surface. One end of a uniform rope is fixed to the block which is pulled in the horizontal direction by applying a force F at the other end. If the mass of the rope is half the mass of the block. The tension in the middle of the rope will be

Options

A. F
B. `2F//3`
C. `3 F//5`
D. `5 F//6`

Explanation

To solve the problem, we need to find the tension in the middle of the rope when a force \( F \) is applied to one end of the rope, which is fixed to a block on a frictionless surface. The mass of the rope is half the mass of the block. Let's denote the mass of the block as \( M \) and the mass of the rope as \( \frac{M}{2} \).

### Step-by-Step Solution:

1. **Identify the Masses:**
- Let the mass of the block be \( M \).
- The mass of the rope is given as \( \frac{M}{2} \).

2. **Determine the Total Mass:**
- The total mass of the system (block + rope) is:
\[ M + \frac{M}{2} = \frac{3M}{2} \]

3. **Acceleration of the System:**
- When a force \( F \) is applied to the rope, the entire system will accelerate. Using Newton's second law:
\[ F = \text{Total Mass} \times \text{Acceleration} \]
\[ F = \left(\frac{3M}{2}\right) a \]
- From this, we can express the acceleration \( a \):
\[ a = \frac{2F}{3M} \]

4. **Analyze the Rope:**
- The rope is uniform, and we want to find the tension \( T \) at the midpoint of the rope. The rope can be considered in two segments: the left segment (attached to the block) and the right segment (being pulled).
- The left segment has a mass of \( \frac{M}{4} \) (half of the rope's mass).

5. **Apply Newton's Second Law to the Left Segment:**
- For the left segment of the rope (mass \( \frac{M}{4} \)):
\[ T = \left(\frac{M}{4}\right) a \]
- Substitute \( a \) from step 3:
\[ T = \left(\frac{M}{4}\right) \left(\frac{2F}{3M}\right) \]
- Simplifying this gives:
\[ T = \frac{2F}{12} = \frac{F}{6} \]

6. **Apply Newton's Second Law to the Right Segment:**
- For the right segment of the rope (mass \( \frac{M}{4} \)):
\[ F - T = \left(\frac{M}{4}\right) a \]
- Substitute \( a \):
\[ F - T = \left(\frac{M}{4}\right) \left(\frac{2F}{3M}\right) \]
- This simplifies to:
\[ F - T = \frac{2F}{12} = \frac{F}{6} \]
- Rearranging gives:
\[ T = F - \frac{F}{6} = \frac{6F}{6} - \frac{F}{6} = \frac{5F}{6} \]

7. **Final Result:**
- The tension in the middle of the rope is:
\[ T = \frac{5F}{6} \]

### Conclusion:
The tension in the middle of the rope is \( \frac{5F}{6} \).