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Two non-viscous, incomoressible and immiscible liquids of densities rho and 1.5 rho are poured into the two limbs of a circules 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 the radius to the interface makes with the vertical in equilibrium position. (b) If the whole liquid column is given a small displacement from its equilibrium position, show that the resulting oscillations alre simple harmonic. Find the time period of these oscillations. |
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Answer» 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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