To a man running at a speed of 20 m/hr, the rain drops appear to be falling at an angle of `30^(@)` from the vertical. If the rain drops are actually falling vertically downwards, their velocity in km.hr is

Options

A. `10 sqrt"" 3`
B. 10
C. `20 sqrt"" 3`
D. 40

Answer

To solve the problem, we need to analyze the situation using the concept of relative velocity. The man is running at a speed of 20 m/hr, and the rain appears to be falling at an angle of 30 degrees from the vertical. We need to find the actual velocity of the rain in km/hr.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Speed of the man (Vm) = 20 m/hr
- Angle of apparent rain (θ) = 30 degrees
- The actual velocity of the rain (Vr) is vertical.

2. **Convert the Speed of the Man**:
- Since we need to find the rain's velocity in km/hr, we convert the man's speed:
\[ Vm = 20 \text{ m/hr} = \frac{20}{1000} \text{ km/hr} = 0.02 \text{ km/hr} \]

3. **Setting Up the Relative Velocity**:
- The velocity of the rain with respect to the man (Vr,m) can be expressed using trigonometry:
\[ \tan(θ) = \frac{Vm}{Vr} \]
- Here, θ = 30 degrees, and we know that:
\[ \tan(30) = \frac{1}{\sqrt{3}} \]

4. **Substituting Values into the Equation**:
- We can rearrange the equation to find Vr:
\[ \frac{1}{\sqrt{3}} = \frac{0.02}{Vr} \]
- Cross-multiplying gives us:
\[ Vr = 0.02 \cdot \sqrt{3} \]

5. **Calculating the Value of Vr**:
- We know that \(\sqrt{3} \approx 1.732\):
\[ Vr = 0.02 \cdot 1.732 \approx 0.03464 \text{ km/hr} \]

6. **Final Conversion to km/hr**:
- Since we need to express the final answer in km/hr, we can simply state:
\[ Vr \approx 0.03464 \text{ km/hr} \]

### Final Answer:
The actual velocity of the rain is approximately **0.03464 km/hr**.