A cylinder of length `1m` is divided by a thin perfectly flexible diaphragm in the middle. It is closed by similar flexible diaphragams at the ends. The two chambers into which it is divided contain hydrogen and oxygen. The two diaphragms are set in vibrations of same frequency. What is the minimum frequency of these diaphragms for which the middle diaphragm will be motionless? Velocity of sound in hydrogen is `1100 m//s` and that in oxygen is `300 m//s`.

Explanation

To solve the problem, we need to find the minimum frequency of the diaphragms such that the middle diaphragm remains motionless. We will use the properties of sound waves in the two gases (hydrogen and oxygen) and the relationship between frequency, velocity, and wavelength.

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a cylinder of length 1 meter divided into two equal chambers (0.5 m each) by a diaphragm. One chamber contains hydrogen, and the other contains oxygen.

2. **Identify the Velocities**:
- The velocity of sound in hydrogen (V_H2) = 1100 m/s.
- The velocity of sound in oxygen (V_O2) = 300 m/s.

3. **Determine the Frequencies**:
- The frequency of the sound wave in a medium is given by the formula:
\[ f = \frac{n \cdot V}{4L} \]
where \( n \) is the harmonic number, \( V \) is the velocity of sound in the medium, and \( L \) is the length of the chamber.

4. **Calculate the Frequencies for Each Gas**:
- For hydrogen:
\[ f_{H2} = \frac{n_1 \cdot V_{H2}}{4 \cdot L} = \frac{n_1 \cdot 1100}{4 \cdot 0.5} = \frac{n_1 \cdot 1100}{2} \]
- For oxygen:
\[ f_{O2} = \frac{n_2 \cdot V_{O2}}{4 \cdot L} = \frac{n_2 \cdot 300}{4 \cdot 0.5} = \frac{n_2 \cdot 300}{2} \]

5. **Set the Frequencies Equal**:
- Since the diaphragms vibrate at the same frequency, we set \( f_{H2} = f_{O2} \):
\[ \frac{n_1 \cdot 1100}{2} = \frac{n_2 \cdot 300}{2} \]
- Simplifying this gives:
\[ n_1 \cdot 1100 = n_2 \cdot 300 \]
- Rearranging gives:
\[ \frac{n_1}{n_2} = \frac{300}{1100} = \frac{3}{11} \]

6. **Express \( n_1 \) in Terms of \( n_2 \)**:
- Let \( n_1 = 3k \) and \( n_2 = 11k \) for some integer \( k \).

7. **Substitute into Frequency Equation**:
- Using \( n_1 \) in the frequency equation for hydrogen:
\[ f_{H2} = \frac{3k \cdot 1100}{2} \]
- This simplifies to:
\[ f_{H2} = 1650k \]

8. **Determine Minimum Frequency**:
- For the minimum frequency, we take \( k = 1 \):
\[ f_{min} = 1650 \text{ Hz} \]

### Final Answer:
The minimum frequency of the diaphragms for which the middle diaphragm will be motionless is **1650 Hz**.