The effect of making a hole exactly at $(1/3^{\mathrm{rd}})$ of the length of the pipe from its closed end is such that

Two sound waves one in air and the other in freshwater are equal in intensity.
$\left(a\right)$ Find the ratio of pressure amplitudes of the wave in water to that of the wave in air.

$\left(b\right)$ If the pressure amplitudes of the waves are equal in then what will be the ratio of the intensities of the waves.
$\left[v_{\mathrm{sound}}=340 \mathrm{m} {\mathrm{s}}^{-1}\right.$in air $\&$ density of air $=1.22 \mathrm{kg} {\mathrm{m}}^{-3}, v_{\mathrm{sound}}=1488 \mathrm{m} {\mathrm{s}}^{-1}$ in water]

If the number of possible natural oscillations of the air column in a pipe whose frequencies lie below $v_0=$$1250 \mathrm{Hz}.$ The length of the pipe is $L=85 \mathrm{cm}$. The velocity of sound is  $340 \mathrm{m} {\mathrm{s}}^{-1}$.
$\left(a\right)$ When the pipe is closed from one end is $n_1$

$\left(b\right)$ When the pipe is opened from both ends is $n_2$.

Then find the value of $n_1+n_2$

A student is performing an experiment using a resonance column and a tuning fork of frequency $244 {\mathrm{s}}^{-1}$. He told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is $\left(0.350\pm 0.005\right) \mathrm{m}$, the gas in the tube is (useful information: $\sqrt{167RT}=640 {\mathrm{J}}^{\frac{1}{2}} \mathrm{mole}{ }^{-\frac{1}{2}}, \sqrt{140RT}=590 {\mathrm{J}}^{\frac{1}{2}} \mathrm{mole}{ }^{-\frac{1}{2}}$. The molar mass $M$ in grams and given in the options. Take the values of $\sqrt{\frac{10}{M}}$ for each gas as given there).