An inclined plane of length `5.60` m making an angle of `45^(@)` with the horizontal is placed in a uniform electric field `E=100 Vm^(-1)`. A particle of mass `1 kg` and charge `10^(-2)C` is allowed to slide down from rest position from maximum height of slope. If the coefficient of friction is 0.1, then the time taken by the particle to reach the bottom is1 s 1.41 s 2 s None of these

Options

A. 1 s
B. 1.41 s
C. 2 s
D. None of these

Answer

To solve the problem step by step, we will analyze the forces acting on the particle, calculate the net force, determine the acceleration, and finally find the time taken to slide down the inclined plane.

### Step 1: Analyze the forces acting on the particle
The forces acting on the particle are:
1. Gravitational force (\(mg\)) acting downwards.
2. Electric force (\(F_E = QE\)) acting horizontally.
3. Normal force (\(R\)) acting perpendicular to the inclined plane.
4. Frictional force (\(F_f = \mu R\)) acting opposite to the direction of motion.

### Step 2: Resolve the forces
Given:
- Mass \(m = 1 \, \text{kg}\)
- Charge \(Q = 10^{-2} \, \text{C}\)
- Electric field \(E = 100 \, \text{V/m}\)
- Angle of inclination \(\theta = 45^\circ\)
- Coefficient of friction \(\mu = 0.1\)

The gravitational force can be resolved into two components:
- Perpendicular to the incline: \(mg \cos \theta\)
- Parallel to the incline: \(mg \sin \theta\)

Calculating these components:
- \(mg = 1 \times 9.8 = 9.8 \, \text{N}\)
- \(mg \cos 45^\circ = 9.8 \times \frac{1}{\sqrt{2}} \approx 6.93 \, \text{N}\)
- \(mg \sin 45^\circ = 9.8 \times \frac{1}{\sqrt{2}} \approx 6.93 \, \text{N}\)

The electric force is:
- \(F_E = QE = 10^{-2} \times 100 = 1 \, \text{N}\)

### Step 3: Calculate the normal force
The normal force \(R\) can be calculated using the balance of forces in the perpendicular direction:
\[ R = mg \cos 45^\circ + QE \sin 45^\circ \]
Substituting the values:
\[ R = 6.93 + 1 \times \frac{1}{\sqrt{2}} \approx 6.93 + 0.707 \approx 7.64 \, \text{N} \]

### Step 4: Calculate the frictional force
The frictional force \(F_f\) is given by:
\[ F_f = \mu R = 0.1 \times 7.64 \approx 0.764 \, \text{N} \]

### Step 5: Calculate the net force acting down the incline
The net force \(F_{\text{net}}\) acting down the incline is given by:
\[ F_{\text{net}} = mg \sin 45^\circ + QE \cos 45^\circ - F_f \]
Substituting the values:
\[ F_{\text{net}} = 6.93 + 1 \times \frac{1}{\sqrt{2}} - 0.764 \approx 6.93 + 0.707 - 0.764 \approx 6.873 \, \text{N} \]

### Step 6: Calculate the acceleration
Using Newton's second law \(F = ma\):
\[ a = \frac{F_{\text{net}}}{m} = \frac{6.873}{1} \approx 6.873 \, \text{m/s}^2 \]

### Step 7: Calculate the time taken to slide down the incline
Using the kinematic equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \(s = 5.6 \, \text{m}\)
- \(u = 0 \, \text{m/s}\) (initial velocity)
- \(a = 6.873 \, \text{m/s}^2\)

Substituting the values:
\[ 5.6 = 0 + \frac{1}{2} \times 6.873 \times t^2 \]
\[ 5.6 = 3.4365 t^2 \]
\[ t^2 = \frac{5.6}{3.4365} \approx 1.63 \]
\[ t \approx \sqrt{1.63} \approx 1.28 \, \text{s} \]

### Conclusion
The time taken by the particle to reach the bottom is approximately \(1.28 \, \text{s}\), which does not match any of the provided options.