In a resonance tube experiment when the tube is filled with water up to a height of $$17.0\,\text{cm}$$, from bottom, it resonates with a given tuning fork. When the water level is raised the next resonance with the same tuning fork occurs at a height of $$24.5\,\text{cm}$$. If the velocity of sound in air is $$330\,\text{m s}^{-1}$$, the tuning fork frequency is:

We consider the resonance tube as a pipe that is closed at the water surface and open at the top. The air column above the water is therefore an open-closed column. For such a column the normal-mode formula is stated first:

For the first resonance (the fundamental) the length of the vibrating air column is

$$L_1 \;=\;\frac{\lambda}{4}\;+\;e,$$

where $$\lambda$$ is the wavelength of the sound and $$e$$ is the so-called end correction that accounts for the fact that the pressure node is a little outside the physical open end.

For the very next resonance in an open-closed column the mode number jumps from 1 to 3, so the length of the air column becomes

$$L_2 \;=\;\frac{3\lambda}{4}\;+\;e.$$

Now we turn to the data of the experiment. The water is initially $$17.0\,\text{cm}$$ high, so let us denote that height by $$h_1=17.0\ \text{cm}$$. The corresponding air-column length is the full length of the tube (call it $$L_0$$) minus this water height: $$L_1 = L_0 - h_1.$$

When the water is raised, its new height is $$h_2 = 24.5\ \text{cm}$$ and the new air-column length is $$L_2 = L_0 - h_2.$$

Taking the difference of the two air lengths we have

$$L_1 - L_2 \;=\;(L_0 - h_1)\;-\;(L_0 - h_2)\;=\;h_2 - h_1.$$

Substituting the numerical values,

$$h_2 - h_1 \;=\;24.5\ \text{cm} - 17.0\ \text{cm} \;=\;7.5\ \text{cm}.$$

On the other hand, subtracting the modal expressions gives

$$L_1 - L_2 \;=\;\Bigl(\frac{\lambda}{4}+e\Bigr) - \Bigl(\frac{3\lambda}{4}+e\Bigr) \;=\;-\frac{\lambda}{2}.$$

The end correction $$e$$ cancels out completely. We take the magnitude, so

$$\frac{\lambda}{2} \;=\;7.5\ \text{cm} \;=\;0.075\ \text{m}.$$

Hence,

$$\lambda \;=\;2 \times 0.075\ \text{m} \;=\;0.15\ \text{m}.$$

The speed-frequency-wavelength relation is stated next:

$$v \;=\;f\,\lambda.$$

We know the velocity of sound in air is given as $$v = 330\ \text{m s}^{-1}.$$ Substituting $$v$$ and $$\lambda$$ we solve for the frequency $$f$$:

$$f \;=\;\frac{v}{\lambda} \;=\;\frac{330\ \text{m s}^{-1}}{0.15\ \text{m}} \;=\; \frac{330}{0.15}\ \text{s}^{-1}.$$

To carry out the division step by step, observe

$$0.15 = \frac{15}{100}, \quad \frac{330}{\frac{15}{100}} = \frac{330 \times 100}{15} = \frac{33000}{15} = 2200.$$

So,

$$f = 2200\ \text{Hz}.$$

Hence, the correct answer is Option A.