A circular tube of uniform cross section is filled with two liquids densities `rho_(1)` and `rho_(2)` such that half of each liquid occupies a quarter to volume of the tube. If the line joining the free surfaces of the liquid makes an angle `theta` with horizontal find the value of `theta.`

Options

A. `2pi sqrt((R )/(6.1))`
B. `2pi sqrt((2R)/(3))`
C. `2pisqrt((3R)/(2))`
D. `2pisqrt((R )/(9.8))`

Answer

A

Explanation

When liquids are slightly disturbed by an angle `beta`. Net restoring pressure `dP = 1.5 rho gh +rho gh =2.5 rho gh`. This pressure will be equal at all sections of the liquid. Therefore, et restoring torque on the whole liquid.

`tau =- (dP)(A)(R) ` or `tau =- 2.5 rho gh AR`
`=- 2.5 rho gAR [R sin (theta +beta) -R sin theta]`
`=- 2.5 rho gAR^(2)[sin theta cos beta +sin beta cos theta -sin theta]`
Assuming `cos beta =1` and `sin beta =beta` (as `beta` is small) `:. tau =-(2.5 rho AgR^(2) cos theta) beta`
or `I alpha =- (2.5 rho AgR^(2) cos theta) beta ...(1)`

Here, `I = (m_(1)+m_(2))R^(2) = [((piR)/(2).A)rho+((piR)/(2)).A(1.5rho)]`
`R^(2)alpha = (1.25 pi R^(3) rho)A` and `cos theta =(5)/(sqrt(26)) =0.98`
Substituting in equation (1), we have `alpha = -((6.11)beta)/(R )`
As angular acceleration is proportional to `-b`, motion is simple harmonic in nature.
`T = 2pi sqrt(|(beta)/(alpha)|)=2pisqrt((R)/(6.1))`