A pipe of length `85cm` is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below `1250 Hz`. The velocity of sound in air is `34m//s`.

Options

A. (a) `12`
B. ( b) `8`
C. ( c ) `6`
D. (d) `4`

Explanation

To solve the problem, we need to determine the number of possible natural oscillations of the air column in a closed pipe of length 85 cm, with the frequencies below 1250 Hz. The velocity of sound in air is given as 340 m/s.

### Step-by-Step Solution:

1. **Convert Length to Meters**:
The length of the pipe is given as 85 cm. We convert this to meters:
\[ L = 85 \, \text{cm} = 0.85 \, \text{m} \]

2. **Use the Formula for Frequencies in Closed Pipes**:
For a pipe closed at one end, the frequency of the natural oscillations can be given by the formula:
\[ f_n = \frac{(2N - 1)V}{4L} \]
where \( N \) is the mode number (1, 2, 3,...), \( V \) is the velocity of sound, and \( L \) is the length of the pipe.

3. **Set Up the Inequality**:
We need to find the maximum value of \( N \) such that the frequency \( f_n \) is less than 1250 Hz:
\[ \frac{(2N - 1)V}{4L} < 1250 \]

4. **Substitute Known Values**:
Substitute \( V = 340 \, \text{m/s} \) and \( L = 0.85 \, \text{m} \) into the inequality:
\[ \frac{(2N - 1) \cdot 340}{4 \cdot 0.85} < 1250 \]

5. **Simplify the Inequality**:
Calculate \( 4L \):
\[ 4L = 4 \cdot 0.85 = 3.4 \]
Now substitute this back into the inequality:
\[ \frac{(2N - 1) \cdot 340}{3.4} < 1250 \]

6. **Multiply Both Sides by 3.4**:
\[ (2N - 1) \cdot 340 < 1250 \cdot 3.4 \]
Calculate \( 1250 \cdot 3.4 \):
\[ 1250 \cdot 3.4 = 4250 \]
Thus, we have:
\[ (2N - 1) \cdot 340 < 4250 \]

7. **Divide Both Sides by 340**:
\[ 2N - 1 < \frac{4250}{340} \]
Calculate \( \frac{4250}{340} \):
\[ \frac{4250}{340} \approx 12.5 \]
Therefore:
\[ 2N - 1 < 12.5 \]

8. **Solve for \( N \)**:
Add 1 to both sides:
\[ 2N < 12.5 + 1 \]
\[ 2N < 13.5 \]
Divide by 2:
\[ N < \frac{13.5}{2} = 6.75 \]

9. **Determine the Maximum Integer Value of \( N \)**:
Since \( N \) must be a whole number, the maximum integer value for \( N \) is:
\[ N = 6 \]

### Conclusion:
The number of possible natural oscillations of the air column in the pipe whose frequencies lie below 1250 Hz is **6**.