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 and determine the angle at which it begins to slide down.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the block, \( m = 5 \, \text{kg} \)
- Initial surface area of the block, \( A = 2 \, \text{m}^2 \)
- Angle of inclination at which the block just begins to slide, \( \theta = 30^\circ \)

2. **Understand the Forces Acting on the Block:**
- The gravitational force acting on the block is \( F_g = mg \), where \( g \approx 9.81 \, \text{m/s}^2 \).
- The component of gravitational force acting down the incline is \( F_{\text{down}} = mg \sin \theta \).
- The normal force acting on the block is \( F_N = mg \cos \theta \).
- The frictional force opposing the motion is given by \( F_f = \mu F_N = \mu mg \cos \theta \), where \( \mu \) is the coefficient of friction.

3. **Set Up the Condition for Sliding:**
- The block just begins to slide when the downhill force equals the frictional force:
\[ mg \sin \theta = \mu mg \cos \theta \]
- We can simplify this equation by canceling \( mg \) from both sides (since \( m \) and \( g \) are non-zero):
\[ \sin \theta = \mu \cos \theta \]

4. **Determine the Coefficient of Friction:**
- At \( \theta = 30^\circ \):
\[ \sin 30^\circ = \frac{1}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2} \]
- Plugging these values into the equation gives:
\[ \frac{1}{2} = \mu \cdot \frac{\sqrt{3}}{2} \]
- Solving for \( \mu \):
\[ \mu = \frac{1}{\sqrt{3}} \approx 0.577 \]

5. **Change in Surface Area:**
- The problem states that the surface area of the block is doubled to \( 4 \, \text{m}^2 \). However, the coefficient of friction \( \mu \) does not depend on the surface area of the block. Therefore, the condition for sliding remains the same.

6. **Determine the New Angle of Inclination:**
- Since the coefficient of friction remains unchanged, the angle at which the block will start sliding down the incline will still satisfy the same equation:
\[ \sin \theta' = \mu \cos \theta' \]
- This means that the angle \( \theta' \) at which the block begins to slide will still be \( 30^\circ \).

### Final Answer:
The angle at which the block starts sliding down the incline when the surface area is doubled remains \( 30^\circ \).