A block of mass 5 kg and surface area `2m^(2)` just begins to slide down an inclined plane when the angle on inclination is `30^(@)`. Keeping mass same, the surface area of the block is doubled. The angle is which this starts sliding down is

Options

A. `30^(@)`
B. `60^(@)`
C. `15^(@)`
D. none

Explanation

To solve the problem, we need to analyze the forces acting on the block when it is on an inclined plane. The key point here is to understand how the angle of inclination affects the sliding of the block.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block:**
- The weight of the block (W) acts vertically downward and is given by \( W = mg \), where \( m = 5 \, \text{kg} \) and \( g \approx 9.81 \, \text{m/s}^2 \).
- The normal force (N) acts perpendicular to the surface of the inclined plane.
- The frictional force (f) acts parallel to the surface of the inclined plane, opposing the motion of the block.

2. **Calculate the Weight of the Block:**
\[ W = mg = 5 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 49.05 \, \text{N} \]

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

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

5. **Frictional Force:**
- The frictional force can be expressed as \( f = \mu N = \mu mg \cos \theta \), where \( \mu \) is the coefficient of friction.

6. **Condition for Sliding:**
- The block begins to slide when the component of weight parallel to the incline equals the frictional force:
\[ mg \sin \theta = \mu mg \cos \theta \]

7. **Canceling the Mass (m):**
- Since the mass (m) is the same on both sides, we can cancel it out:
\[ g \sin \theta = \mu g \cos \theta \]

8. **Simplifying the Equation:**
- Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ \sin \theta = \mu \cos \theta \]

9. **Dividing by \( \cos \theta \):**
- This gives us:
\[ \tan \theta = \mu \]

10. **Conclusion:**
- The angle at which the block begins to slide down the inclined plane depends only on the coefficient of friction \( \mu \) and not on the surface area of the block. Since the coefficient of friction remains unchanged, the angle of inclination at which the block starts to slide will remain the same, which is \( 30^\circ \).

### Final Answer:
The angle at which the block starts sliding down the inclined plane remains \( 30^\circ \).