A glass tube of length 1.0 is completely filled with water. A vibrating tuning fork of frequency 500Hz is kept over the mouth of the tube and the water is drained out slowly at the bottom of the tube. If the velocity of sound in air is $330\dfrac{m}{s}$ , then the total number of resonances that occur will
A. 2
B. 3
C. 1
D. 5

Hint: Let us have a discussion about the resonant frequency of a tube with one closed end. The expression we consider will relate the frequency to the number of harmonics, the speed of sound and the length of the tube. Rearranging the equation to express resonant length in terms of all the other parameters in the expression.

Formula used:
Resonance frequency $f = \dfrac{{nv}}{{4l}}$

Complete answer:
We know that resonance tubes form only odd harmonics, we can rewrite the obtained equations in such a way that we can equate them to consequently obtain the speed of sound.

We know that acoustic resonance is a phenomenon wherein an acoustic system amplifies the sound waves whose frequency matches its own natural frequency of vibration which is resonant frequency.

A resonance tube with the lower end closed by water surface will have resonances at frequencies
$f = \dfrac{{nv}}{{4l}}$
Where n is an odd positive integer, v is the speed of sound, and l is the length of the resonant tube. This tube produces only odd harmonics.
Hence,
$ \Rightarrow l = \dfrac{{nv}}{{4f}}$

Let the successive lengths at which resonances are observed be at ${l_1},{l_2},{l_3},{l_4},$ respectively. Then, we have:
$ \Rightarrow {l_1} = \dfrac{v}{{4{f_1}}} = \dfrac{{330}}{{4 \times 500}} = 0.165m$
Similarly
$ \Rightarrow {l_2} = \dfrac{{3v}}{{4{f_1}}} = \dfrac{{3 \times 330}}{{4 \times 500}} = 0.495m$
$ \Rightarrow {l_3} = \dfrac{{5v}}{{4{f_1}}} = \dfrac{{5 \times 330}}{{4 \times 500}} = 0.825m$
$ \Rightarrow {l_4} = \dfrac{{7v}}{{4{f_1}}} = \dfrac{{7 \times 330}}{{4 \times 500}} = 1.155 > 1m$
Since ${4^{th}}$ length is exceeding glass tube length we cannot consider it to produce resonant frequency.

So, Total number of resonances is equal to 3. Therefore, the correct answer is “Option B”.

Note: Remember that the expression for resonances of the resonance tube varies depending on the structure of the tube. The expression we used is only applicable for tubes that are closed on one end, and for open tubes. Also, the resonance properties can be understood by considering the behaviour of a sound wave in air, which propagates via the compressions and rarefactions of the molecules.