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.
Find the angle (theta) that the radius to the interface makes with the verticles in equilibrium position.

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.