A cylindrical vessel of radius 3 cm has at the bottom a horizontal capillary tube of length 20 cm and internal radius `0.4` mm. If the vessel is filled with water, find the time taken by it to empty one half of its contents. Given that the viscosity of water is `0.01` poise.

Explanation

At t= 0, H is height of water level in container
dh = fall of water level in time dt.
According to equation of continuity:
Rate of flow of liquid `=V =- (adh)/(dt)`
= area of container
According to Poiscuille.s formula: P=pgh `-a (dh)/(dt)=(pi Pr^(4))/(8 eta l)`
`(dt)=(-8 eta la)/(pi pgr^(4)) (dh)/(h)`
At t=0 height =H
at t=T height becomes, `=(2H)/3 [therefore 1/3"rd of water is moving out"]`
`underset(0)overset(T)int dt=(-8etala)/(pi pgr^(4) underset(H)overset((2H)/3)int (dh)/(h)`
`T=(-8 eta la)/(pi gr^(4)) [log ((2H)/(3)-log_(e) (H)]`
`=(-8eta la)/(pi pgr^(4)) [log_(e0 (2H)-loe_(e) 3-log_(e) H]`
`=(-8 etala)/(pi pgr^(4)) [log_(e) 2 xx 5-log_(e) 3 -log_(e)5]`
`=(-8 eta la)/(pi pgr^(4)) [2.30-1.09 -1.60]`
`=(8 eta la)/(pi pgr^(4)) xx 0.39`
`=(8 xx 0.01 xx 10 xx pi (5)^(2))/(3.14 xx 1 xx 980 xx (0.03)^(4)) xx 0.39`
=9826.1 s