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Two non-viscous, incompressible and immiscible liquids of densities p and 1.5pare 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 the figure. Each liquid occupies one fourth the circumference of the tube.(i)Find the angle theta that the radius to the interface makes with the vertical in equilibrium position. _________ (ii) If the whole liquid column is given a small displacement from its equilibrium position, show that the resulting oscillations are simple harmonic. Find the time period of these oscillations. ____________ |
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Answer» (i) The pressure due to liquid on left limb (at bottom) `P_(1)=R(1sintheta)1.5pg`...(i) The pressure due to liquids on right limb `P_(2)=(Rsintheta+Rcostheta)pg+R(1-costheta)1.5pg`...(ii) In EQUILIBRIUM, `P_(1)=P_(2)` Which gives, `tantheta=((1)/(5))impliestheta=tan^(-1)((1)/(5))` (ii) If the liquid is given a small angular displacement `alpha`, the pressure difference, `dP=P_(1)-P_(2)` `dP=[Rsintheta+alpha)+Rcos(theta+alpha)]pg+R[l-cos(theta+alpha)]1.5pg-R[l-sin(theta+alpha)]1.5pg` As `alpha` is small, `sinalpha=alpha`,`cosalpha=1` `dP=Rpg[2.5sintheta+2.5costhetaalpha-0.5costheta+0.5sinthetaalpha]` `tantheta=0.2`,`sintheta=(0.2)/(sqrt(1.04))` and `costheta=(1)/(sqrt(1.04))` `dP=2.55Rpgalpha=2.55pgy`(as `Ra=y`) Restoring force `F=dPx` area =`-2.55pgA` Mass of the liquid in tube `m=(2piR)/(4)Ap+(2pir)/(4)Axx1.5p=1.25piRAp` Hence acceleration `a=(F)/(m)=-(2.55pgyA)/(1.25piRAp)`,`a=-2.04((g)/(piR))y`,`a=-omega^(2)y` Hence, `omega=sqrt(2.04((g)/(piR)))`,TIME period, `T=(2pi)/(omega)=2.5sqrtRsec`  
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