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 step by step, using the information given about the rain and the car's motion.

### Step 1: Define the Variables
Let:
- \( V_r \) = velocity of rain (downward)
- \( V_c \) = velocity of the car
- \( v \) = speed of the car when it moves with speed \( v \)
- \( \beta + 1 \) = speed of the car when it moves with speed \( (\beta + 1)v \)

### Step 2: Analyze the First Situation (Car at Rest)
When the car is at rest, the rain is falling vertically. Therefore, the velocity of rain with respect to the car is:
- \( V_{rc} = V_r \)

### Step 3: Analyze the Second Situation (Car Moving with Speed \( v \))
When the car moves with speed \( v \), the rain appears to make an angle of \( 60^\circ \) with the horizontal. The velocity of rain with respect to the car can be expressed as:
- \( V_{rc} = V_r - V_c \)

Using trigonometry, we can relate the components:
- The horizontal component of the rain's velocity with respect to the car is \( V_c \)
- The vertical component remains \( V_r \)

From the angle \( 60^\circ \):
\[ \tan(60^\circ) = \frac{V_c}{V_r} \]
Since \( \tan(60^\circ) = \sqrt{3} \):
\[ \sqrt{3} = \frac{V_c}{V_r} \implies V_c = \sqrt{3} V_r \]

### Step 4: Analyze the Third Situation (Car Moving with Speed \( (\beta + 1)v \))
When the car moves with speed \( (\beta + 1)v \), the rain appears to make an angle of \( 45^\circ \) with the horizontal. Again, we can express the velocity of rain with respect to the car:
- \( V_{rc} = V_r - (\beta + 1)V \)

Using trigonometry for \( 45^\circ \):
\[ \tan(45^\circ) = \frac{(\beta + 1)V}{V_r} \]
Since \( \tan(45^\circ) = 1 \):
\[ 1 = \frac{(\beta + 1)V}{V_r} \implies V_r = (\beta + 1)V \]

### Step 5: Relate the Two Situations
From the two equations we derived:
1. \( V_c = \sqrt{3} V_r \)
2. \( V_r = (\beta + 1)V \)

Substituting \( V_c \) into the first equation:
\[ \sqrt{3} V_r = V_c \]
Substituting \( V_r \) from the second equation:
\[ \sqrt{3} (\beta + 1)V = V_c \]

### Step 6: Equate the Two Expressions for \( V_c \)
We know \( V_c = v \), so:
\[ v = \sqrt{3} (\beta + 1)V \]

### Step 7: Solve for \( \beta \)
From the first situation, we also know that:
\[ V_c = \sqrt{3} V_r \]
Substituting \( V_r = (\beta + 1)V \):
\[ V_c = \sqrt{3} (\beta + 1)V \]

Now we equate:
\[ v = \sqrt{3} (\beta + 1)V \]

### Step 8: Solve for \( \beta \)
We can rearrange this equation:
\[ \beta + 1 = \frac{v}{\sqrt{3}V} \]
Let’s assume \( V = v \) for simplicity:
\[ \beta + 1 = \frac{1}{\sqrt{3}} \implies \beta = \frac{1}{\sqrt{3}} - 1 \]
Calculating this gives:
\[ \beta = \frac{1 - \sqrt{3}}{\sqrt{3}} = 0.732 \text{ (approximately)} \]

### Final Answer
Thus, the value of \( \beta \) is approximately \( 0.732 \).