Two blocks `A` and `B` of same masses are resting in equilibrium on an inclined plane having inclination with horizontal `=alpha(gt0)`. The blocks are touching and exerting no zero normal force on each other with block `B` higher than `A`. Coeffcient of static friction of `A` with incline `=1.2` and of `B=0.8`. If motion is not imminent

Options

A. `alphagt30^(@)`
B. `("Friction")_(A)gt("Friction")_(B)`
C. `alphagt45^(@)`
D. `("Friction")_(A)=("Friction")_(B)`

Explanation

To solve the problem, we need to analyze the forces acting on the two blocks \( A \) and \( B \) resting on the inclined plane. Here are the steps to arrive at the solution:

### Step 1: Identify the Forces Acting on Each Block
For both blocks \( A \) and \( B \), the forces acting on them include:
- The weight of the blocks \( W = mg \) acting downwards.
- The normal force \( N \) acting perpendicular to the inclined plane.
- The frictional force \( F_f \) acting parallel to the inclined plane, opposing the motion.

### Step 2: Break Down the Weight into Components
The weight can be resolved into two components:
- Perpendicular to the incline: \( W \cos \alpha \)
- Parallel to the incline: \( W \sin \alpha \)

### Step 3: Determine the Normal Force
For both blocks, the normal force \( N \) is equal to the perpendicular component of their weight:
\[ N = W \cos \alpha \]

### Step 4: Calculate the Frictional Forces
The frictional force for each block can be calculated using the formula:
\[ F_f = \mu N \]
Where \( \mu \) is the coefficient of static friction.

For block \( A \):
\[ F_{fA} = \mu_A N = 1.2 \times W \cos \alpha \]

For block \( B \):
\[ F_{fB} = \mu_B N = 0.8 \times W \cos \alpha \]

### Step 5: Analyze the Condition for Equilibrium
Since the blocks are in equilibrium and motion is not imminent, the frictional force must be greater than or equal to the component of weight acting down the incline.

For block \( B \):
\[ F_{fB} \geq W \sin \alpha \]
Substituting the expression for \( F_{fB} \):
\[ 0.8 W \cos \alpha \geq W \sin \alpha \]

### Step 6: Simplify the Inequality
Dividing both sides by \( W \) (assuming \( W \neq 0 \)):
\[ 0.8 \cos \alpha \geq \sin \alpha \]

### Step 7: Rearranging the Inequality
Rearranging gives:
\[ \frac{\sin \alpha}{\cos \alpha} \leq 0.8 \]
This can be rewritten as:
\[ \tan \alpha \leq 0.8 \]

### Step 8: Determine the Range of \( \alpha \)
To find the angle \( \alpha \), we can use the arctangent function:
\[ \alpha \leq \tan^{-1}(0.8) \]

### Step 9: Analyze the Options
Now, we check the options given in the question:
- Option A: \( \alpha > 30^\circ \) (This is valid since \( \tan(30^\circ) \approx 0.577 \))
- Option B: Not specified.
- Option C: \( \alpha > 45^\circ \) (This is invalid since \( \tan(45^\circ) = 1 \))
- Option D: Not specified.

### Conclusion
From the analysis, we conclude that:
- The condition \( \tan \alpha \leq 0.8 \) implies that \( \alpha \) must be less than \( \tan^{-1}(0.8) \) which is approximately \( 38.66^\circ \). Therefore, \( \alpha \) cannot be greater than \( 45^\circ \).

Thus, the correct options based on the analysis are:
- Option A is valid.
- Option C is invalid.