Rain appears to fall vertically to a man walking at `3kmh^(-1)`, but when he changes his speed to double, the rain appears to fall at `45^@` with vertical. Study the following statements and find which of them are correct.
i. Velocity of rain is `2 (sqrt3) kmh^(-1)`.
ii. The angle of fall of rain (with vertical) is `theta = tan^(-1)(1/(sqrt2)).`
iii. The angle of fall of rain (with vertical) is `theta = sin^(-1) (1/(sqrt2))`.
iv. Velocity of rain is `3(sqrt2)kmh^(-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 will analyze the situation step by step.

### Step 1: Understanding the Problem
The man is walking at a speed of \(3 \, \text{km/h}\) and perceives the rain to be falling vertically. When he doubles his speed to \(6 \, \text{km/h}\), he perceives the rain to be falling at an angle of \(45^\circ\) with the vertical. We need to find the velocity of the rain and the angle of its fall with respect to the vertical.

### Step 2: Setting Up the Vectors
1. Let \( \vec{v_m} \) be the velocity of the man, which is \(3 \, \text{km/h}\) in the horizontal direction.
2. Let \( \vec{v_r} \) be the velocity of the rain with respect to the ground.
3. When the man is walking at \(3 \, \text{km/h}\), the rain appears to fall vertically, which means the horizontal component of the rain's velocity must equal the man's velocity.

### Step 3: Case 1 - Man Walking at \(3 \, \text{km/h}\)
- The horizontal component of the rain's velocity is \(v_{rh} = 3 \, \text{km/h}\).
- The vertical component of the rain's velocity is \(v_{rv}\), which we need to find.

### Step 4: Case 2 - Man Walking at \(6 \, \text{km/h}\)
- The horizontal component of the rain's velocity remains \(3 \, \text{km/h}\).
- Now, the man’s speed is \(6 \, \text{km/h}\), and the rain appears to fall at \(45^\circ\) with the vertical. This means:
\[ \tan(45^\circ) = 1 = \frac{v_{rh}}{v_{rv}} \]
Therefore, \(v_{rh} = v_{rv}\).

### Step 5: Finding the Velocity of Rain
From the above, we have:
- \(v_{rh} = 3 \, \text{km/h}\)
- \(v_{rv} = 3 \, \text{km/h}\)

Now, we can find the resultant velocity of the rain:
\[ v_r = \sqrt{v_{rh}^2 + v_{rv}^2} = \sqrt{(3)^2 + (3)^2} = \sqrt{18} = 3\sqrt{2} \, \text{km/h} \]

### Step 6: Finding the Angle of Fall of Rain
Using the relationship:
\[ \tan(\theta) = \frac{v_{rh}}{v_{rv}} = \frac{3}{3} = 1 \]
Thus,
\[ \theta = \tan^{-1}(1) = 45^\circ \]

### Step 7: Conclusion
Now we can evaluate the statements:
1. **Velocity of rain is \(2\sqrt{3} \, \text{km/h}\)** - **Incorrect** (we found \(3\sqrt{2} \, \text{km/h}\)).
2. **The angle of fall of rain (with vertical) is \(\theta = \tan^{-1}(1/\sqrt{2})\)** - **Incorrect** (it is \(45^\circ\)).
3. **The angle of fall of rain (with vertical) is \(\theta = \sin^{-1}(1/\sqrt{2})\)** - **Correct** (since \(\sin(45^\circ) = \frac{1}{\sqrt{2}}\)).
4. **Velocity of rain is \(3\sqrt{2} \, \text{km/h}\)** - **Correct**.

### Final Answer
The correct statements are:
- iii. The angle of fall of rain (with vertical) is \(\theta = \sin^{-1}(1/\sqrt{2})\).
- iv. Velocity of rain is \(3\sqrt{2} \, \text{km/h}\).