A man holds an umbrella at `30^@` with the vertical to keep himself dry. He, then, runs at a speed of `10ms^(-1)`, and find the rain drops to be hitting vertically. Study the following statement and find the correct options.
i. Velocity of rain w.r.t. Earth is `20 ms^(-1)`
ii. Velocity of rain w.r.t. man is `10 sqrt(3)ms^(-1)`.
iii. Velocity of rain w.r.t. Earth is `30 ms^(-1)`
iv. Velocity of rain w.r. t. man is `10sqrt(2) ms^(-1)` .

Options

A. Statements (i) and (ii) are correct.
B. Statements (i) and (iii) are correct.
C. Statements (iii) and (iv) are correct.
D. Statements (ii) and (iv) are correct.

Explanation

To solve the problem, we need to analyze the situation using vector components and relative motion. Here's a step-by-step breakdown:

### Step 1: Understand the Situation
The man is running at a speed of \(10 \, \text{m/s}\) while holding an umbrella at an angle of \(30^\circ\) with the vertical. The rain appears to be falling vertically to him.

### Step 2: Set Up the Coordinate System
Let's establish a coordinate system:
- Let the vertical direction (downward) be the y-axis.
- Let the horizontal direction (the direction the man is running) be the x-axis.

### Step 3: Define Velocities
- Let \(V_r\) be the velocity of rain with respect to the Earth.
- The man runs with a velocity \(V_m = 10 \, \text{m/s}\) in the horizontal direction (x-axis).

### Step 4: Resolve the Velocity of Rain
Since the rain appears to be falling vertically to the man, we can resolve the components of the rain's velocity:
- The horizontal component of the rain's velocity must equal the man's running speed to appear vertical.
- The vertical component of the rain's velocity can be found using trigonometric relations.

### Step 5: Use Trigonometry
From the umbrella angle:
- The horizontal component of the rain's velocity can be expressed as:
\[ V_{r_x} = V_r \sin(30^\circ) \]
- The vertical component of the rain's velocity can be expressed as:
\[ V_{r_y} = V_r \cos(30^\circ) \]

### Step 6: Set Up the Equations
Since the rain appears to be falling vertically:
\[ V_{r_x} = V_m \]
Thus,
\[ V_r \sin(30^\circ) = 10 \, \text{m/s} \]
Since \(\sin(30^\circ) = \frac{1}{2}\):
\[ V_r \cdot \frac{1}{2} = 10 \implies V_r = 20 \, \text{m/s} \]

### Step 7: Calculate the Velocity of Rain with Respect to the Man
Using the vertical component:
\[ V_{r_y} = V_r \cos(30^\circ) = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]

### Step 8: Summary of Results
1. **Velocity of rain with respect to Earth**: \(V_r = 20 \, \text{m/s}\) (True)
2. **Velocity of rain with respect to man**: \(10\sqrt{3} \, \text{m/s}\) (True)
3. **Velocity of rain with respect to Earth**: \(30 \, \text{m/s}\) (False)
4. **Velocity of rain with respect to man**: \(10\sqrt{2} \, \text{m/s}\) (False)

### Conclusion
The correct statements are:
- i. Velocity of rain w.r.t. Earth is \(20 \, \text{m/s}\) (True)
- ii. Velocity of rain w.r.t. man is \(10\sqrt{3} \, \text{m/s}\) (True)
- iii. Velocity of rain w.r.t. Earth is \(30 \, \text{m/s}\) (False)
- iv. Velocity of rain w.r.t. man is \(10\sqrt{2} \, \text{m/s}\) (False)