A particle undergoes three successive displacements given by `s_(1) =sqrt(2)` m north - east , `s_(2)=2` m due south and `s_(3) =4 m, 30^(@)` north of west , then magnitude of net displacement is

Options

A. `sqrt(14+4sqrt(3))`
B. `sqrt(14-4sqrt(3))`
C. `sqrt(4)`
D. None of these

Explanation

To solve the problem of finding the magnitude of the net displacement of a particle undergoing three successive displacements, we can follow these steps:

### Step 1: Define the Displacements
The displacements are given as:
- \( s_1 = \sqrt{2} \) m (North-East)
- \( s_2 = 2 \) m (South)
- \( s_3 = 4 \) m (30° North of West)

### Step 2: Break Down Each Displacement into Components
1. **Displacement \( s_1 \)**:
- North-East means it makes a 45° angle with both the North and East axes.
- Components:
\[ s_{1x} = s_1 \cos(45^\circ) = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \text{ m (East)} \]
\[ s_{1y} = s_1 \sin(45^\circ) = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1 \text{ m (North)} \]

2. **Displacement \( s_2 \)**:
- Due South means it only has a negative y-component.
- Components:
\[ s_{2x} = 0 \text{ m} \]
\[ s_{2y} = -2 \text{ m} \]

3. **Displacement \( s_3 \)**:
- 30° North of West means it has both negative x-component and positive y-component.
- Components:
\[ s_{3x} = -s_3 \cos(30^\circ) = -4 \cdot \frac{\sqrt{3}}{2} = -2\sqrt{3} \text{ m} \]
\[ s_{3y} = s_3 \sin(30^\circ) = 4 \cdot \frac{1}{2} = 2 \text{ m} \]

### Step 3: Sum the Components
Now, we can sum the x and y components of all three displacements to find the net displacement.

- **Net x-component**:
\[ s_{net,x} = s_{1x} + s_{2x} + s_{3x} = 1 + 0 - 2\sqrt{3} = 1 - 2\sqrt{3} \]

- **Net y-component**:
\[ s_{net,y} = s_{1y} + s_{2y} + s_{3y} = 1 - 2 + 2 = 1 \]

### Step 4: Calculate the Magnitude of the Net Displacement
The magnitude of the net displacement \( s_{net} \) can be calculated using the Pythagorean theorem:
\[ s_{net} = \sqrt{s_{net,x}^2 + s_{net,y}^2} \]
Substituting the values:
\[ s_{net} = \sqrt{(1 - 2\sqrt{3})^2 + (1)^2} \]

### Step 5: Simplify the Expression
Calculating the squares:
\[ (1 - 2\sqrt{3})^2 = 1 - 4\sqrt{3} + 12 = 13 - 4\sqrt{3} \]
So,
\[ s_{net} = \sqrt{(13 - 4\sqrt{3}) + 1} = \sqrt{14 - 4\sqrt{3}} \]

### Final Answer
The magnitude of the net displacement is:
\[ \boxed{\sqrt{14 - 4\sqrt{3}} \text{ m}} \]