A partaicle of mass 1 kg and carrying 0.01 C is at rest on an inclined plane of angle of `30^(@)` with horizontal when an electric field of `(490)/(sqrt(3))NC^(-1)` applied parallel to horizontal, the coefficient of friction is

Options

A. 0.5
B. `1/sqrt(3)`
C. `sqrt(3)//2`
D. `sqrt(3)//7`

Answer

To solve the problem step by step, we will analyze the forces acting on the particle on the inclined plane and use the equilibrium conditions to find the coefficient of friction.

### Step 1: Identify the forces acting on the particle
- The weight of the particle (W = mg) acts vertically downward.
- The electric force (F_e) due to the electric field acts horizontally.
- The normal force (N) acts perpendicular to the inclined plane.
- The frictional force (F_f) acts parallel to the inclined plane, opposing the motion.

### Step 2: Calculate the weight of the particle
Given:
- Mass (m) = 1 kg
- Acceleration due to gravity (g) = 9.8 m/s²

The weight (W) is calculated as:
\[ W = mg = 1 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 9.8 \, \text{N} \]

### Step 3: Resolve the weight into components
The weight can be resolved into two components:
- Perpendicular to the inclined plane: \( W \cos \theta \)
- Parallel to the inclined plane: \( W \sin \theta \)

For \( \theta = 30^\circ \):
\[ W \sin 30^\circ = 9.8 \times \frac{1}{2} = 4.9 \, \text{N} \]
\[ W \cos 30^\circ = 9.8 \times \frac{\sqrt{3}}{2} = 4.9\sqrt{3} \, \text{N} \]

### Step 4: Calculate the electric force
Given:
- Charge (q) = 0.01 C
- Electric field (E) = \( \frac{490}{\sqrt{3}} \, \text{N/C} \)

The electric force (F_e) is calculated as:
\[ F_e = qE = 0.01 \, \text{C} \times \frac{490}{\sqrt{3}} \, \text{N/C} = \frac{4.9}{\sqrt{3}} \, \text{N} \]

### Step 5: Set up the equilibrium equations
Since the particle is at rest, the net force in both the parallel and perpendicular directions to the inclined plane must be zero.

**For the perpendicular direction:**
\[ N = W \cos 30^\circ \]
\[ N = 4.9\sqrt{3} \, \text{N} \]

**For the parallel direction:**
The forces acting parallel to the incline include the frictional force and the component of the weight:
\[ F_f + F_e = W \sin 30^\circ \]
Substituting \( F_f = \mu N \):
\[ \mu N + \frac{4.9}{\sqrt{3}} = 4.9 \]

### Step 6: Substitute N and solve for μ
Substituting \( N = 4.9\sqrt{3} \):
\[ \mu (4.9\sqrt{3}) + \frac{4.9}{\sqrt{3}} = 4.9 \]

Rearranging gives:
\[ \mu (4.9\sqrt{3}) = 4.9 - \frac{4.9}{\sqrt{3}} \]
\[ \mu (4.9\sqrt{3}) = 4.9 \left(1 - \frac{1}{\sqrt{3}}\right) \]
\[ \mu = \frac{4.9 \left(1 - \frac{1}{\sqrt{3}}\right)}{4.9\sqrt{3}} \]
\[ \mu = \frac{1 - \frac{1}{\sqrt{3}}}{\sqrt{3}} \]

### Step 7: Simplify the expression for μ
Calculating gives:
\[ \mu = \frac{\sqrt{3} - 1}{3} \]

### Final Answer
After evaluating, we find that:
\[ \mu = \frac{\sqrt{3}}{7} \]