Solve the following problem:
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 τ1 is given by
T1 = 2π √h/2g
(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 τ2 is now
T2 = 2π √h/2g + γh0g/L
Where h0 is the height of the barometric column.
(c) Show that
γ = 2L/h0(T21/T22 - 1) .