Two non - viscous, incompressible and immiscible liquids of densities (rho) and (1.5 rho) are poured into the two limbs of a circular tube of radius ( R) and small cross section kept fixed in a vertical plane as shown in fig. Each liquid occupies one fourth the cirumference of the tube.
(##JMA_CHMO_C10_027_Q01##).
(a) Find the angle (theta) that the radius to the interface makes with the verticles in equilibrium position.
(b) If the whole is given a small displacement from its equilibrium position, show that the resulting oscillations are simple harmonic. Find the period of these oscillations.

Answer

A, B, C

Explanation

(a) Let the liqud of density `1.5 rho` occupy the lect portion AB and the liquid of density `rho`occupy the right portion BC of the tube. The pressure at the lowest point D due to the liquid on the left is
`P_(1) = (R-Rsintheta) 1.5rhog`

The pressure due to the liquid on the right is
`P_(2)(Rsintheta + Rcostheta) rhog + (R-Rcostheta)1.5 rhog`
Since the liquids ar in equilibrium
`P_(1) = P_(2)` or `(R - Rsintheta) 1.5 rhog = (R-Rcostheta)rhog + (R-Rcostheta) 1.5rhog`
Solving, we get
`tantheta = 0.2` or `theta = tan^(-1)(0.2)`
(b) Let the whole liquid lbe given a small angular displacement `alpha` towards right. Then the pressure difference between the right and the left limbs is
`dP = P_(2) - P_(1)`
`= [Rsin(theta+alpha)+Rcos(theta+alpha)rhog+[R-Rcos(theta+alpha)1.5rhog-[R-Rsin(theta+alpha)]1.5rhog`
`= Rrhog[2.5sin(theta+alpha)-0.5cos(theta+alpha)]`
`= Rrhog[2.5(sinthetacosalpha+costheta sinalpha)-0.5(costheta cosalpha-sinthetasinalpha)]`
For small `alpha`
`sinalpha = alpha,cosalpha ~= 1 :. dP = Rrhog`
`[2.5 sintheta + 2.5alpha costheta - 0.5 costheta+0.5 alpha sin theta]`
As `tan theta = 0.2, sintheta (0.2)/(sqrt(1.04)) ~~0.2 , costheta = (1)/(sqrt(1.04)) ~~ 1`
`:. dP = Rrhog[2.5 xx 0.2 + 2.5 alpha - 0.5 + 0.5 xx 0.2 alpha]`
`= Rrhog[2.6 alpha]`
`= 236 rhog`
where `y = Ralpha`, the linear displacement
`:.` Restoring force `F = 2.6 rhogAy`
This shows that `F prop y`
Hence the motion is simple harmonic with force constant
`k = 2.6 rhogA`
Now, total mass of the liquid
`m = (2piR)/(4)Arho + (2piR)/(4)A(1.5p) = (5piRArho)/(4)`
`:.` Time period
`T = 2pisqrt((m)/(k)) = 2pisqrt((5pirArho)/(4xx236rhogA)) = pisqrt((1.93piR)/(9.8))`
`= 2.47 sqrt(R)` seconds.