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Two non - viscous, incompressible and im...

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.

Text Solution

Verified by Experts

The correct Answer is:
(a) `tan^(-1)sqrt((1)/(5))` (b) `2pisqrt((R)/(6.11))`
(a) In equilibrium, pressure of same liquid at same level will be same.
Therefore `P_(1) = P_(2)`
or `P + (1.5 p g h_(1)) = P + (rho g h_(2))`
(`p =` pressure of gas in empty part of the tube)
`:. 1.5 h_(1) = h_(2)`
`1.5 [ R costheta - R sintheta] = rho (R costheta + R sintheta)`
or `3 costheta - 3 sintheta = 2 costheta + 2 sintheta`
or `5 tantheta = 1`
`theta = tan^(-1)((1)/(5))`

(b) When liquid are slightly disturbed by an angle `beta`. Net restoring pressure `DeltaP = 1.5 rhogh + rhogh` this pressure will be equal at all sections of the liquid. Therefore, net restoring torque on the whole liquid.

`h = R sin (theta + beta) - Rsintheta`
or, `tau = -(DeltaP) (A) (R)`
`tau = -2.5rhogh AR`
`= - 2.5rhog AR [R sin (theta + beta) - R sintheta]`
`= -2.5 rhog AR^(2) [ sintheta cosbeta + sinbeta costheta - sintheta]`
Assuming `cos beta = 1` and `sinbeta = beta` (given, `beta` is small)
`:. tau = -(2.5 rhoA gR^(2) costheta)beta`
or `Ialpha = -(2.5 rhoAgR^(2) costheta)beta ....(1)`
Here, `I = (m_(1) + m_(2))R^(2)`

`= [((piR)/(2).A)rho + ((piR)/(2)).A (1.5 rhop) ]R^(2)`
`= (1.25 piR^(3)rho)A`
and `costheta = (5)/(sqrt(26)) = 0.98`
Substituting in equation `(1)`, we have
`alpha = -((6.11)beta)/(R) rArr` angular acceleration `prop-`angular displacement
As angular acceleration is propotional to `-beta`, motion is simple harmonic in nature.
`T = 2pisqrt(|(beta)/(alpha)|) = 2pisqrt((R)/(6.11))`
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