Shown in the figure is a container whose top and botoom diameters are D and d respectively. At the bottom of the container, there is a capillary tube of outer radius b and inner radius a.
The volume flow rate in the capillary is Q. If the capillary is removed the liquid comes out with a velocity of `v_0`. The density of the liquid is given as `rho`. Calculate the coefficient of viscosity `eta`.

Answer

A, B, D

Explanation

KEY CONCEPT: When the tube is not there, using Bernoulli's theorem
`P+P_0+1/2rhov_1^2+rhogH=1/2rhov_0^2+P_0`
`impliesP+rhogH=1/2rho(v_0^2-v_1^2)`
But according to equation of continuity
`v_1=(A_2v_0)/(A_1)`
`:.P+rhogH=1/2rho[v_0^2-(A_2/A_1v_0)^2]`
`P+rhogH=1/2rhov_0^2[1-(A_2/A_1)^2]`
Here, `P+rhogH=DeltaP`
According to Poisseuille's equation
`Q=(pi(DeltaP)a^4)/(8etal)=eta=(pi(DeltaP)a^4)/(8Ql)`
`:. eta=(pi(P+rhogH)a^4)/(8Ql)=(pi)/(8Ql)xx1/2rhov_0^2[1-(A_2/A_1)^2]xxa^4`
Where `A_2/A_1=d^2/D^2`
`eta=(pi)/(8Ql)xx1/2rhov_0^2[1-d^4/D^4]xxa^4`