The velocity of the liquid coming out of a small hole of a vessel containing two different liquids of densities `2rho` and `rho` as shown in the figure is

Answer

`tan^(-1)(1/5)`, (ii)`2.5sqrt( R)`

Explanation

(i) `tan^(-1)((1)/(5))` (ii) `2.5sqrtR`
(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`