A student is performing an experiment using a resonance column and a tuning fork of frequency `244s^(-1)`. He is told that the air in the tube has been replaced by another gas (assuming that the air column remains filled with the gas). If the minimum height at which resonance occurs is `(0.350+- 0.005)m`, the gas in the tube is (Useful information : `sqrt(167RT) = 640J^(1//2)mol^(-1//2)`,
`sqrt(140RT) = 590J^(1//2)mol^(-1//2)`. The molar masses `M` in grams are given in the options. take the values of `sqrt((10)/(M))` for each gas as given there.)

Options

A. Neon(M=20,`sqrt((10)/(20))=(7)/(10)`)
B. Nitrogen(M=28,`sqrt((10)/(28))=(3)/(5)`)
C. Oxygen(M=32,`sqrt((10)/(32))=(9)/(19)`)
D. Argon(M=36,`sqrt((10)/(36))=(17)/(32)`)

Explanation

To solve the problem, we need to determine which gas is in the resonance column based on the given frequency of the tuning fork and the minimum height at which resonance occurs. We will use the relationship between the speed of sound in a gas, the frequency of the tuning fork, and the length of the gas column.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Frequency of tuning fork, \( f = 244 \, \text{s}^{-1} \)
- Minimum height for resonance, \( L_{\text{min}} = 0.350 \pm 0.005 \, \text{m} \)
- Useful information:
- \( \sqrt{167RT} = 640 \, \text{J}^{1/2} \text{mol}^{-1/2} \)
- \( \sqrt{140RT} = 590 \, \text{J}^{1/2} \text{mol}^{-1/2} \)

2. **Understand Resonance Condition:**
The minimum length for resonance in a tube is given by:
\[ L_{\text{min}} = \frac{V}{4f} \]
where \( V \) is the speed of sound in the gas.

3. **Speed of Sound in a Gas:**
The speed of sound in a gas is given by:
\[ V = \sqrt{\frac{\gamma RT}{M}} \]
where \( \gamma \) is the adiabatic index, \( R \) is the universal gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas.

4. **Substituting for \( V \):**
We can substitute \( V \) into the resonance condition:
\[ L_{\text{min}} = \frac{1}{4f} \sqrt{\frac{\gamma RT}{M}} \]

5. **Rearranging for Molar Mass \( M \):**
Rearranging gives:
\[ M = \frac{\gamma RT}{(4f L_{\text{min}})^2} \]

6. **Calculate for Each Gas:**
We will calculate \( L_{\text{min}} \) for each gas using the provided values of \( \sqrt{167RT} \) and \( \sqrt{140RT} \).

- **For Neon (\( \gamma = 1.67 \)):**
\[ M = \frac{167RT}{(4 \cdot 244 \cdot 0.350)^2} \]
Using \( \sqrt{167RT} = 640 \):
\[ L_{\text{min}} = \frac{1}{4 \cdot 244} \cdot 640 \cdot \frac{1}{\sqrt{M}} \Rightarrow L_{\text{min}} \approx 0.459 \, \text{m} \quad \text{(not suitable)} \]

- **For Nitrogen (\( \gamma = 1.4 \)):**
\[ M = \frac{140RT}{(4 \cdot 244 \cdot 0.350)^2} \]
Using \( \sqrt{140RT} = 590 \):
\[ L_{\text{min}} \approx 0.363 \, \text{m} \quad \text{(not suitable)} \]

- **For Oxygen (\( \gamma = 1.4 \)):**
\[ L_{\text{min}} \approx 0.340 \, \text{m} \quad \text{(not suitable)} \]

- **For Argon (\( \gamma = 1.67 \)):**
\[ L_{\text{min}} \approx 0.348 \, \text{m} \quad \text{(suitable)} \]

7. **Conclusion:**
The gas in the tube is Argon, as it is the only gas that produces resonance within the specified height range.