A block is kept on a smooth inclined plane of angle of inclination `30^(@)` that moves with a constant acceleration so that the block does not slide relative to the inclined plane. Let `F_(1)` be the the contact force between the block and the plane . Now the inclined plane stops and let `F_(2)` be the contact force between the two in this case. Then, `F_(1)//F_(2)` is

Options

A. `1`
B. `(4)/(3)`
C. `2`
D. `(3)/(2)`

Answer

To solve the problem, we need to analyze the forces acting on the block in both scenarios: when the inclined plane is accelerating and when it stops.

### Step 1: Analyze the forces when the inclined plane is accelerating (F1)

1. **Identify the forces acting on the block**:
- Gravitational force (weight) acting downwards: \( W = mg \)
- Normal force \( F_1 \) acting perpendicular to the inclined plane.
- The inclined plane is accelerating with some acceleration \( a \).

2. **Resolve the weight into components**:
- The component of weight acting parallel to the inclined plane: \( W_{\parallel} = mg \sin(30^\circ) = \frac{mg}{2} \)
- The component of weight acting perpendicular to the inclined plane: \( W_{\perpendicular} = mg \cos(30^\circ) = mg \cdot \frac{\sqrt{3}}{2} \)

3. **Apply Newton's second law**:
- In the direction parallel to the incline, since the block does not slide, the net force is zero:
\[ F_{\text{net}} = ma = F_1 - W_{\parallel} \]
\[ ma = F_1 - \frac{mg}{2} \]
- Rearranging gives:
\[ F_1 = ma + \frac{mg}{2} \]

### Step 2: Analyze the forces when the inclined plane stops (F2)

1. **Identify the forces acting on the block when the inclined plane stops**:
- The gravitational force \( W = mg \) acting downwards.
- Normal force \( F_2 \) acting perpendicular to the inclined plane.

2. **Resolve the weight into components**:
- The component of weight acting parallel to the inclined plane remains the same: \( W_{\parallel} = \frac{mg}{2} \)
- The component of weight acting perpendicular to the inclined plane remains the same: \( W_{\perpendicular} = mg \cdot \frac{\sqrt{3}}{2} \)

3. **Apply Newton's second law**:
- In the direction perpendicular to the incline, the net force is zero when the inclined plane stops:
\[ F_2 = W_{\perpendicular} = mg \cdot \frac{\sqrt{3}}{2} \]

### Step 3: Compare F1 and F2

1. **Find the ratio \( \frac{F_1}{F_2} \)**:
- From the expressions derived:
\[ F_1 = ma + \frac{mg}{2} \]
\[ F_2 = mg \cdot \frac{\sqrt{3}}{2} \]

2. **Express \( F_1 \) in terms of \( F_2 \)**:
- Since \( F_2 \) is dependent on \( mg \), we can express \( F_1 \) in terms of \( F_2 \):
\[ \frac{F_1}{F_2} = \frac{ma + \frac{mg}{2}}{mg \cdot \frac{\sqrt{3}}{2}} \]

### Conclusion

The ratio \( \frac{F_1}{F_2} \) will depend on the acceleration \( a \) of the inclined plane. However, since \( F_1 \) includes the term \( ma \), it suggests that \( F_1 \) will generally be greater than \( F_2 \) unless \( a \) is zero.