A pipe open at both ends has a fundamental frequency $$f$$ in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now:

Let the total length of the pipe be $$L$$. Because both ends are originally open, the air column supports a fundamental (first harmonic) whose antinodes are at each open end and whose single node is midway between. For an open-open pipe, the well-known relation between the fundamental wavelength $$\lambda_1$$ and the length $$L$$ is stated first:

$$\text{For an open-open pipe : }\; \lambda_1 = 2L$$

If the speed of sound in air is $$v$$, the fundamental frequency $$f$$ is therefore

$$f = \dfrac{v}{\lambda_1} = \dfrac{v}{2L} \qquad (1)$$

Now the pipe is dipped vertically into water until exactly one half of its length is submerged. The water surface behaves like a rigid boundary for sound waves and therefore acts as a closed end, while the top of the pipe remains an open end. Consequently, only the upper half of the pipe, of length $$L/2$$, is now filled with vibrating air, and the pipe has effectively changed from an open-open organ pipe to an open-closed organ pipe of length $$L/2$$.

For an open-closed pipe, the fundamental standing wave has a node at the closed end and an antinode at the open end. The standard relation for such a pipe is first recalled:

$$\text{For an open-closed pipe : }\; \lambda'_1 = 4\left(\text{length of the air column}\right)$$

Here the vibrating column length is $$L/2$$, so the new fundamental wavelength $$\lambda'_1$$ becomes

$$\lambda'_1 = 4\left(\dfrac{L}{2}\right) = 2L$$

We notice that $$\lambda'_1$$ is numerically equal to $$\lambda_1$$ obtained earlier. Proceeding, the new fundamental frequency $$f'$$ is

$$f' = \dfrac{v}{\lambda'_1} = \dfrac{v}{2L} \qquad (2)$$

Comparing expressions (1) and (2), we have

$$f' = f$$

Thus, immersing half the pipe in water does not change the fundamental frequency of the vibrating air column.

Hence, the correct answer is Option B.