The wavelength of two notes in air are `40/195 m` and `40/193 m`. Each note produces 9 beats per second separately with third note of fixed frequency. The velocity of sound in air in `m//s` is

Options

A. 360
B. 320
C. 330
D. 340

Answer

To find the velocity of sound in air based on the given wavelengths and the beat frequency, we can follow these steps:

### Step 1: Calculate the Frequencies of the Two Notes
The frequency \( f \) of a wave can be calculated using the formula:
\[ f = \frac{v}{\lambda} \]
where \( v \) is the velocity of sound and \( \lambda \) is the wavelength.

Given the wavelengths:
- \( \lambda_1 = \frac{40}{195} \, \text{m} \)
- \( \lambda_2 = \frac{40}{193} \, \text{m} \)

### Step 2: Express Frequencies in Terms of Velocity
Using the formula for frequency, we can express the frequencies of the two notes as:
\[ f_1 = \frac{v}{\lambda_1} = \frac{v}{\frac{40}{195}} = \frac{195v}{40} \]
\[ f_2 = \frac{v}{\lambda_2} = \frac{v}{\frac{40}{193}} = \frac{193v}{40} \]

### Step 3: Set Up the Beat Frequency Equations
The problem states that each note produces 9 beats per second with a third note of fixed frequency \( f_0 \). This gives us two equations:
1. \( f_0 - f_1 = 9 \)
2. \( f_2 - f_0 = 9 \)

From these, we can express \( f_0 \):
\[ f_0 = f_1 + 9 \]
\[ f_0 = f_2 - 9 \]

### Step 4: Equate the Two Expressions for \( f_0 \)
Setting the two expressions for \( f_0 \) equal gives:
\[ f_1 + 9 = f_2 - 9 \]
Rearranging this gives:
\[ f_2 - f_1 = 18 \]

### Step 5: Substitute the Frequencies
Substituting the expressions for \( f_1 \) and \( f_2 \):
\[ \frac{193v}{40} - \frac{195v}{40} = 18 \]
This simplifies to:
\[ \frac{-2v}{40} = 18 \]
Multiplying both sides by -40 gives:
\[ 2v = -720 \]
Thus:
\[ v = \frac{720}{2} = 360 \, \text{m/s} \]

### Final Answer
The velocity of sound in air is \( 360 \, \text{m/s} \).

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