A narrow but tall cabin is falling freely near the earth's surface. Inside the cabin, two small stones A and B are released from rest (relative to the cabin). Initially A is much above the centre of mass and B much below the centre of mass of the cabin. A close observation of the motion of A and B will reveal that -

Options

A. both A and B continue to be exactly at rest relative to the cabin
B. A moves slowly upward and B moves slowly downward relative to the cabin
C. both A and B fall to the bottom of the cabin with constant acceleration due to gravity
D. A and B move slightly towards each other vertically

Answer

To solve the problem, we need to analyze the situation of the falling cabin and the stones A and B inside it.

### Step-by-Step Solution:

1. **Understanding the System**:
- The cabin is falling freely under the influence of gravity, which means it is in free fall.
- Inside the cabin, two stones A and B are released from rest relative to the cabin.

2. **Position of Stones**:
- Stone A is positioned much above the center of mass of the cabin.
- Stone B is positioned much below the center of mass of the cabin.

3. **Acceleration of the Cabin**:
- Since the cabin is in free fall, it has an acceleration equal to the acceleration due to gravity (g) directed downwards.

4. **Relative Motion of Stones**:
- When stones A and B are released, they are initially at rest relative to the cabin. However, due to their positions relative to the center of mass of the cabin, their accelerations will differ from that of the cabin.
- Stone A, being above the center of mass, will experience a slightly lesser gravitational pull compared to the cabin, while stone B, being below the center of mass, will experience a slightly greater gravitational pull.

5. **Acceleration Comparison**:
- The acceleration of stone A (let's denote it as \( a_A \)) will be less than the acceleration of the cabin (denoted as \( a_{CM} \)).
- The acceleration of stone B (denote it as \( a_B \)) will be greater than \( a_{CM} \).
- Therefore, we have:
\[ a_A < a_{CM} < a_B \]

6. **Relative Motion**:
- Since stone A has a lesser downward acceleration, it will appear to move upwards relative to the cabin.
- Conversely, stone B, having a greater downward acceleration, will appear to move downwards relative to the cabin.

7. **Conclusion**:
- As a result, stone A moves slowly upward and stone B moves slowly downward relative to the cabin.

### Final Answer:
- The correct observation is: **A moves slowly upward and B moves slowly downward relative to the cabin.**