A block of base `10cmxx10cm` and height 15cm is kept on an inclined plane. The coefficient of friction between them is `sqrt3`. The inclination `theta` of this inclined plane from the horizontal plane is gradually increased from `0^@` Then

Options

A. at `theta=30^@` the block will start sliding down the plane
B. the block will remain at rest on the plane up to certain `theta` and ten it will topple
C. at `theta=60^@` the block will start slidng down the plane and continue to do so at higher angles
D. at `theta=60^@` the block will start sliding down the plane and on further increasing `theta`, it will topple at certain `theta`

Explanation

To solve the problem, we need to analyze the forces acting on the block as the angle of inclination (theta) of the inclined plane is increased. We will determine the conditions under which the block will start sliding down the plane or topple over.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block**:
- The weight of the block (W = mg) acts vertically downward.
- The normal force (N) acts perpendicular to the inclined surface.
- The frictional force (F_friction) acts parallel to the surface and opposes the motion.

2. **Resolve the Weight into Components**:
- The component of the weight acting down the incline is \( W_{\parallel} = mg \sin(\theta) \).
- The component of the weight acting perpendicular to the incline is \( W_{\perpendicular} = mg \cos(\theta) \).

3. **Determine the Normal Force**:
- The normal force (N) is equal to the perpendicular component of the weight:
\[ N = mg \cos(\theta) \]

4. **Calculate the Frictional Force**:
- The maximum static frictional force can be given by:
\[ F_{\text{friction}} = \mu N = \mu (mg \cos(\theta)) \]
- Given that the coefficient of friction \( \mu = \sqrt{3} \), we have:
\[ F_{\text{friction}} = \sqrt{3} (mg \cos(\theta)) \]

5. **Condition for Sliding**:
- The block will start sliding when the downhill component of the weight exceeds the maximum static friction:
\[ mg \sin(\theta) = \sqrt{3} (mg \cos(\theta)) \]
- Dividing through by \( mg \) (assuming \( mg \neq 0 \)):
\[ \sin(\theta) = \sqrt{3} \cos(\theta) \]
- Dividing both sides by \( \cos(\theta) \):
\[ \tan(\theta) = \sqrt{3} \]
- This implies:
\[ \theta = 60^\circ \]

6. **Condition for Toppling**:
- To determine when the block will topple, we need to analyze the geometry of the block. The block will topple when the center of mass (CM) moves outside the base of the block.
- The height of the block is 15 cm, and the width is 10 cm. The center of mass is at a height of \( \frac{15}{2} = 7.5 \) cm.
- The distance from the edge of the base to the center of mass when tilted can be calculated using the tangent of the angle of inclination:
\[ \tan(\alpha) = \frac{\text{opposite}}{\text{adjacent}} = \frac{7.5}{5} = \frac{3}{2} \]
- Thus, \( \tan(\alpha) = \frac{2}{3} \).

7. **Comparison of Angles**:
- Since \( \tan(60^\circ) = \sqrt{3} \) and \( \tan(\alpha) = \frac{2}{3} \), we see that:
\[ \tan(\alpha) < \tan(\theta) \]
- This implies that the block will topple before it starts sliding.

### Conclusion:
- The block will remain at rest until the angle \( \theta \) reaches \( 60^\circ \), at which point it will start to slide. However, it will topple before it can slide down the incline.