Rain is falling vertically when car is at rest. When car moves with speed v rain appears at `60 deg` with horizontal when car moves with speed (`beta + 1)v` rain appears at 45degrees with horizontal.Find value of `beta`

Options

A. 3
B. 0.732
C. 2
D. 0.5

Answer

To solve the problem, we need to analyze the situation using the concept of relative velocity. Let's break it down step by step.

### Step 1: Understand the scenario
When the car is at rest, the rain is falling vertically. When the car starts moving, the rain appears to come at an angle. We need to find the velocity of the rain in terms of the speed of the car.

### Step 2: Set up the first case
When the car moves with speed \( v \), the rain appears at an angle of \( 60^\circ \) with the horizontal.

- Let \( V_0 \) be the velocity of the rain (which is vertical).
- The horizontal component of the rain's velocity relative to the car is \( v \).
- The angle \( \theta = 60^\circ \), so we can use the tangent function:

\[ \tan(60^\circ) = \frac{V_0}{v} \]

### Step 3: Solve for \( V_0 \) in the first case
Using the value of \( \tan(60^\circ) = \sqrt{3} \):

\[ \sqrt{3} = \frac{V_0}{v} \]

This gives us:

\[ V_0 = v \sqrt{3} \]

### Step 4: Set up the second case
When the car moves with speed \( (\beta + 1)v \), the rain appears at an angle of \( 45^\circ \) with the horizontal.

- The horizontal component of the rain's velocity relative to the car is now \( (\beta + 1)v \).
- The angle \( \theta = 45^\circ \), so we can again use the tangent function:

\[ \tan(45^\circ) = \frac{V_0}{(\beta + 1)v} \]

### Step 5: Solve for \( V_0 \) in the second case
Using the value of \( \tan(45^\circ) = 1 \):

\[ 1 = \frac{V_0}{(\beta + 1)v} \]

This gives us:

\[ V_0 = (\beta + 1)v \]

### Step 6: Equate the two expressions for \( V_0 \)
From the two cases, we have:

1. \( V_0 = v \sqrt{3} \)
2. \( V_0 = (\beta + 1)v \)

Setting them equal to each other:

\[ v \sqrt{3} = (\beta + 1)v \]

### Step 7: Simplify the equation
Dividing both sides by \( v \) (assuming \( v \neq 0 \)):

\[ \sqrt{3} = \beta + 1 \]

### Step 8: Solve for \( \beta \)
Rearranging the equation gives:

\[ \beta = \sqrt{3} - 1 \]

### Final Answer
Thus, the value of \( \beta \) is:

\[ \beta \approx 0.732 \quad (\text{since } \sqrt{3} \approx 1.732) \]