InterviewSolution
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InterviewSolution
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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. . (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. |
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Answer» Correct Answer - A::B::C 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`). |
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