A cylindrical vessel of area of cross-section A and filled with liquid to a height of `h_(1)` has a capillary tube of length l and radius r protuding horizontally at its bottom. If the viscosity of liquyid is `eta` and density `rho`. Find the time in which the level of water in the vessel falls to `h_(2)`.

Options

A. `(8etalA)/(pirhogr^(4))` ln `(h_(1))/(h_(2))`
B. `(8etalA)/(pirhogr^(4))`
C. `(etaA)/(g)(sqrt(h_(1))-sqrt(h_(2)))`
D. `(8etalA)/(pirhogr^(4))` ln `(h_(2))/(h_(1))`

Answer

A

Explanation

Let `h` be the height of water level in the vessel at instant `t` which decreases by dh in time dt.
`:.` rate of flow of water through the capillary tube,
`V=-A((dh)/(dt))` ….(1)
Further, the rate of flow from poiseuille formula
`V=(piPr^(4))/(8etal)` ....(2)
The hyddrostatic pressure at depth h is `P=rhogh`
from eqns (1) and (2) we have
`-A(dh)/(dt)=(pirhohr^(4))/(8etal)`
`dt=-(8etalA)/(pirhor^(4))(dh)/(h),t=(-8etalA)/(pirhogr^(4))int_(h_(1))^(h_(2))(dh)/(h)`