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**Hint:**The tube partly filled water may be considered as one end closed. The glass tube shall behave as an organ pipe. I will behave as a closed end organ pipe and the length of the pipe is varied by varying the water level.

Formula used: $V=f\lambda $ Where $V$is velocity, $f$ if frequency.

$\lambda $ wavelength.

**Complete step by step answer:**

We are given the following data;

Frequency $f=686Hz$

Assume speed of sound to be ${{V}_{s}}=343\text{ m/s}$

Let $L$ be the length of the air column.

So, we have $h=1-L$

We know that, the resonance condition in tube is given by;

$\Rightarrow 4L=n\lambda $

Simplifying the above equation we get;

$\Rightarrow L=\dfrac{n\lambda }{4}$

And by the formula we have wavelength

$\Rightarrow \lambda =\dfrac{V}{f}=\dfrac{343}{686}=\dfrac{1}{2}=0.5\,\text{m}\text{.}$

for $n=1,{{L}_{1}}=\dfrac{\lambda }{4}=\dfrac{0.5}{4}=0.125\text{ m}$

for $n=2,{{L}_{2}}=\dfrac{3\lambda }{4}=\dfrac{3\times 0.5}{4}=0.375\text{ m}$

for $n=3,{{L}_{3}}=\dfrac{5\lambda }{4}=\dfrac{5\times 0.5}{4}=0.625\text{ m}$

$\because $ one end is fixed $n=1,3,5,7,9....$

(a) There are $4$ values of $n$ i.e. $n=1,3,5,7$ which satisfies $h>0$

(b) Smallest height of water for resonance to correspond to $n=7$ with $h=0.125\text{ m}\text{.}$

for $n=7,{{L}_{7}}=7\times 0.125=0.875$

Thus we have;

$h=1-{{L}_{7}}=1-875-0.125\text{ m}$

(c) The second smallest water height $h$ is at $n=5$

for $n=5,{{L}_{5}}=5\times 0.125=0.625\text{ m}$

thus we have;

$h=1-{{L}_{5}}=1-0.625=0.375\text{ m}$

**Note:**Two sound waves in opposite directions interfere with each other to create resonance. Waves produced are longitudinal stationary. Resonance tube is a partially filled hollow cylinder with water and forced into vibration by a running fork. The running fork is the object that forced the air inside of the resonance tube into resonance.

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