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.
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(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

(###JMA_CHMO_C10_027_S01##)
At equilibrium taking of liquid about `O`
(Torque due to liquid densith `rho` ) = )Torque due to loquid of density `1.5 rho`)
`m_(2)g xx QM =m_(1)g xx PN`
`:. M_(2)g sin (45^(@)) =m_(1)g R sin(45^(@)-theta)`
`V rho g R sin (45^(@)+theta) = 1.5 V rho g R sin(45^(@)-theta)` ...(i) ltbr `implies (SIN(45+theta))/(sin(45-theta)) =1.5`
`implies (sin45^(@)cos theta + cos 45^(@) sin theta)/(sin45^(@)cos theta - cos 45^(@) sin theta)=3/2`
`implies tan theta = 1/5`
`tau = m_(2)g R sin(45^(@)+theta+alpha)-m_(1)g R sin gR sin(45^(@)-theta-alpha)`
`=V rho g R sin(45^(@)-theta - alpha)-1.5 V rho g R(45^(@)-theta-alpha)`
`=V rho g Rsin(theta+45^(@)) cos alpha+V rho g R cos(45^(@)+theta)`
`sin alpha-1.5 V rho g R sin(45^(@)-theta) cos alpha + 1.5 V rho R cos (45^(@)-theta) sin alpha`
Using eq. (i) we get
`tau = V rho g R[cos (45^(@)+theta)sin alpha+1.5 cos (45^(@)-theta)sin alpha]`
`tau = V rho g R[cos (45^(@)+theta)+1.5 cos (45^(@)-theta)sin alpha]
When `alpha` is small (given)
`:. sin alpha ~~ alpha`
`tau = V rho g R[cos(45^(@)+theta)+1.5 cos (45^(@)-theta)]alpha`
Since, `tau` and `alpha` are proportional and directed towards mean position.
`:. The motion is simple harmonic.
Moment of interia about `O` is
`I = V rhoR^(2) +1.5V rho R^(2)`
`T = 2 pi sqrt((1)/(C)`
`=2 pi (sqrt((V rho xx 2.5 R^(2))))/(sqrt([cos(45+theta)+1.5 cos (45-theta)]Vrho gR))`
`=2pi (sqrt(1.803R))/(sqrt(g))` (using the value ` tan theta = 1/5`).