Mercury is poured into a U-tube open at both ends until the total length of mercury is
$h$.
(a) If the level of mercury on one side of the tube is depressed and the mercury is allowed to oscillate with small amplitude, show that, neglecting friction, the period $\tau_{1}$ is given by
$$
\tau_{1}=2 \pi \sqrt{\frac{h}{2 g}}
$$
(b) One end of the U-tube is now closed so that the length of the entrapped air column is $L$, and again the mercury is caused to oscillate. Assuming friction to be negligible, the air to be ideal, and the changes of volume to be adiabatic, show that the period $\tau_{2}$ is now
$$
\tau_{2}=2 \pi \sqrt{\frac{h}{2 g+\gamma h_{0} g / L}}
$$
where $h_{0}$ is the height of the barometric column.
(c) Show that
$$
\gamma=\frac{2 L}{h_{0}}\left(\frac{\tau_{1}^{2}}{\tau_{2}^{2}}-1\right)
$$