Both neon `[ M_(Ne) = 20 xx 10^(-3) kg]` and helium `[M_(He) = 4 xx 10^(-3) kg]` are monoatomic gases and can be assumed to be ideal gases. The fundamental frequency of a tube (open at both ends ) of neon is `300 Hz at 270 K (R = ( 25//3) J//K mol)`
The fundamental frequency of the tube if the tube is filled with helium , all other factors remaining the same is

Options

A. `300 Hz`
B. `sqrt(2) xx 300 Hz`
C. `sqrt(3) xx 300 Hz`
D. `sqrt(5) xx 300 Hz`

Explanation

To find the fundamental frequency of a tube filled with helium, we can use the relationship between frequency and molecular weight for an ideal gas. The fundamental frequency \( f \) of a tube open at both ends is given by the formula:

\[ f = \frac{1}{2L} \sqrt{\frac{\gamma RT}{M}} \]

Where:
- \( L \) is the length of the tube (constant in this case),
- \( \gamma \) is the adiabatic index (constant for monoatomic gases),
- \( R \) is the universal gas constant (constant in this case),
- \( T \) is the temperature (constant in this case),
- \( M \) is the molar mass of the gas.

### Step 1: Understand the relationship
Since the length of the tube, temperature, and gas constant are the same for both gases, the fundamental frequency is inversely proportional to the square root of the molar mass:

\[ f \propto \frac{1}{\sqrt{M}} \]

### Step 2: Set up the ratio of frequencies
Let \( f_{Ne} \) be the frequency of neon and \( f_{He} \) be the frequency of helium. We can express the ratio of the frequencies as follows:

\[ \frac{f_{Ne}}{f_{He}} = \sqrt{\frac{M_{He}}{M_{Ne}}} \]

### Step 3: Substitute known values
We know:
- \( f_{Ne} = 300 \, \text{Hz} \)
- \( M_{Ne} = 20 \times 10^{-3} \, \text{kg} \)
- \( M_{He} = 4 \times 10^{-3} \, \text{kg} \)

Substituting these values into the equation gives:

\[ \frac{300}{f_{He}} = \sqrt{\frac{4 \times 10^{-3}}{20 \times 10^{-3}}} \]

### Step 4: Simplify the ratio
Now, simplify the right-hand side:

\[ \frac{4 \times 10^{-3}}{20 \times 10^{-3}} = \frac{4}{20} = \frac{1}{5} \]

Thus, we have:

\[ \frac{300}{f_{He}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \]

### Step 5: Cross-multiply to find \( f_{He} \)
Cross-multiplying gives:

\[ 300 = f_{He} \cdot \frac{1}{\sqrt{5}} \]

So,

\[ f_{He} = 300 \sqrt{5} \]

### Step 6: Calculate the value of \( f_{He} \)
Now, we can calculate \( f_{He} \):

\[ f_{He} = 300 \cdot \sqrt{5} \approx 300 \cdot 2.236 \approx 670.8 \, \text{Hz} \]

Thus, the fundamental frequency of the tube when filled with helium is approximately \( 670.8 \, \text{Hz} \).

### Final Answer
The fundamental frequency of the tube filled with helium is approximately \( 670.8 \, \text{Hz} \).

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