A pipe of length 85 cm 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 340 m/s.

We have a pipe that is closed at one end and open at the other. For such a pipe, the standing-wave modes of the air column obey a special harmonic condition. First, we recall the formula for the resonant frequencies of a pipe closed at one end:

$$f_n = \dfrac{n \, v}{4L}$$

where $$f_n$$ is the frequency of the $$n^{\text{th}}$$ mode, $$v$$ is the velocity of sound in air and $$L$$ is the length of the pipe. Because one end is closed, only odd values of $$n$$ appear; that is $$n = 1, 3, 5, 7, \dots$$.

Now we substitute the given data. The length of the pipe is 85 cm, so

$$L = 85 \text{ cm} = 0.85 \text{ m}.$$

The velocity of sound is

$$v = 340 \text{ m/s}.$$

Substituting $$v$$ and $$L$$ into the formula, we find a handy constant factor:

$$ \dfrac{v}{4L} \;=\; \dfrac{340}{4 \times 0.85} \;=\; \dfrac{340}{3.4} \;=\; 100 \text{ Hz}. $$

So each resonant frequency can be written compactly as

$$f_n = n \times 100 \text{ Hz}, \qquad n = 1, 3, 5, 7, \dots$$

We are asked to keep only those natural oscillations whose frequencies lie below 1250 Hz. Therefore we impose

$$ n \times 100 \lt 1250 \quad\Longrightarrow\quad n \lt \dfrac{1250}{100} \quad\Longrightarrow\quad n \lt 12.5. $$

Remembering that $$n$$ must be odd, we list all positive odd integers less than 12.5:

$$n = 1, \; 3, \; 5, \; 7, \; 9, \; 11.$$

Counting them gives

$$\text{Number of possible oscillations} = 6.$$

Hence, the correct answer is Option C.