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 a block placed on a smooth inclined plane that is accelerating. We will find the contact forces \( F_1 \) and \( F_2 \) in two different scenarios and then calculate the ratio \( \frac{F_1}{F_2} \).

### Step-by-Step Solution:

**Step 1: Analyze the first scenario (Inclined plane moving with acceleration)**

1. **Identify forces acting on the block**:
- Weight of the block: \( mg \) (acting vertically downward).
- Pseudo force due to the acceleration of the inclined plane: \( ma \) (acting opposite to the direction of acceleration).

2. **Resolve the weight and pseudo force into components**:
- The component of weight parallel to the incline: \( mg \sin \theta \).
- The component of weight perpendicular to the incline: \( mg \cos \theta \).
- The component of pseudo force parallel to the incline: \( ma \cos \theta \).
- The component of pseudo force perpendicular to the incline: \( ma \sin \theta \).

3. **Set up the equations for equilibrium**:
- Along the incline (net force = 0):
\[ ma \cos \theta = mg \sin \theta \]
- Perpendicular to the incline (net force = 0):
\[ F_1 = mg \cos \theta + ma \sin \theta \]

4. **Substituting \( a \)**:
From the first equation, we can express \( a \):
\[ a = g \tan \theta \]
Substitute this into the equation for \( F_1 \):
\[ F_1 = mg \cos \theta + m(g \tan \theta) \sin \theta \]
Simplifying:
\[ F_1 = mg \cos \theta + mg \frac{\sin^2 \theta}{\cos \theta} \]
\[ F_1 = mg \left( \cos \theta + \frac{\sin^2 \theta}{\cos \theta} \right) \]
\[ F_1 = mg \left( \frac{\cos^2 \theta + \sin^2 \theta}{\cos \theta} \right) \]
Since \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ F_1 = \frac{mg}{\cos \theta} \]

**Step 2: Analyze the second scenario (Inclined plane at rest)**

1. **Identify forces acting on the block**:
- Weight of the block: \( mg \) (acting vertically downward).
- There is no pseudo force since the inclined plane is at rest.

2. **Set up the equilibrium equation**:
- Perpendicular to the incline:
\[ F_2 = mg \cos \theta \]

**Step 3: Calculate the ratio \( \frac{F_1}{F_2} \)**

1. Substitute the expressions for \( F_1 \) and \( F_2 \):
\[ \frac{F_1}{F_2} = \frac{\frac{mg}{\cos \theta}}{mg \cos \theta} \]
Simplifying:
\[ \frac{F_1}{F_2} = \frac{1}{\cos^2 \theta} \]

2. Substitute \( \theta = 30^\circ \):
\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \]
Therefore:
\[ \frac{F_1}{F_2} = \frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

### Final Answer:
\[ \frac{F_1}{F_2} = \frac{4}{3} \]