A siren placed at a railway platform is emitting sound of frequency `5 kHz`. A passenger sitting in a moving train `A` records a frequency of `5.5kHz` while the train approaches the siren. During his return journey in a different train `B` he records a frequency of `6.0 kHz` while approaching the same siren. the ratio the velocity of train `B` to that of train `A` is

Options

A. `242//252`
B. `2`
C. `5//6`
D. `11//6`

Explanation

To solve the problem, we need to use the Doppler effect formula for sound waves. The frequency observed by a moving observer can be calculated using the formula:

\[ f' = f \frac{v + v_o}{v - v_s} \]

Where:
- \( f' \) is the observed frequency,
- \( f \) is the source frequency,
- \( v \) is the speed of sound in air,
- \( v_o \) is the speed of the observer (train),
- \( v_s \) is the speed of the source (which is stationary in this case).

### Step 1: Set up the equations for both trains

1. For train A (approaching the siren):
\[ f_A = f \frac{v + v_A}{v} \]
Given:
- \( f_A = 5.5 \, \text{kHz} \)
- \( f = 5 \, \text{kHz} \)

Plugging in the values:
\[ 5.5 = 5 \frac{v + v_A}{v} \]

2. For train B (also approaching the siren):
\[ f_B = f \frac{v + v_B}{v} \]
Given:
- \( f_B = 6.0 \, \text{kHz} \)

Plugging in the values:
\[ 6.0 = 5 \frac{v + v_B}{v} \]

### Step 2: Simplify the equations

1. For train A:
\[ 5.5 = 5 \frac{v + v_A}{v} \]
Rearranging gives:
\[ \frac{5.5}{5} = \frac{v + v_A}{v} \]
\[ 1.1 = \frac{v + v_A}{v} \]
\[ 1.1v = v + v_A \]
\[ v_A = 0.1v \quad \text{(Equation 1)} \]

2. For train B:
\[ 6.0 = 5 \frac{v + v_B}{v} \]
Rearranging gives:
\[ \frac{6.0}{5} = \frac{v + v_B}{v} \]
\[ 1.2 = \frac{v + v_B}{v} \]
\[ 1.2v = v + v_B \]
\[ v_B = 0.2v \quad \text{(Equation 2)} \]

### Step 3: Find the ratio of velocities

Now, we need to find the ratio of the velocities of train B to train A:
\[ \frac{v_B}{v_A} = \frac{0.2v}{0.1v} = \frac{0.2}{0.1} = 2 \]

### Final Answer

The ratio of the velocity of train B to that of train A is:
\[ \frac{v_B}{v_A} = 2 \]