Determine the period of oscillations of mercury of mass `m=200g` poured into a bent tube (figure) whose right arm forms an angle `theta=30^(@)` with vertical. The cross-sectional area of the tube is `S=0.50 cm^(2)`. The viscosity of mercury is to be neglected.

Explanation

If the mercury rises in the left arm by `x` it must fall by a slanting length equal to `x` in the other arm. Total pressure difference in the two arms wil then be
`rho gx+rho g x cos theta=rho g x (1+cos theta)`
This will give rise to a restoring force
`- rho g S x (1+ cos theta)`
This must equal mass times acceleration which can be obtained from work energy principle.
The K.E. of mercury in the tube is clearly `: (1)/(2) mdot(x ^(2))`
So mass times acceleration must be `: m ddot(x)`
Hence `m ddot (x) + rhog S(1+cos theta)x =0`
This is S.H.M. with a time period
`T=2pi sqrt((m)/(rho g S(1+cos theta)))`