A cylindrical container of cross-section area, `A` is filled up with water upto height `'h'` Water may exit through a tap of cross sectional area `'a'` in the bottom of container. Find out

(a) Velocity of water just after opening of tap.
(b) The area of cross-section of water stream coming out of tape at depth `h_(0)` below tap in terms of `'a'` just after opening of tap.
(c) Time in which conainer becomes empty.
(Given : `((a)/(A))^(1//2) = 0.22, h = 20 cm , h_(0) = 20 cm`

Explanation

Applying Bernoulli's between `(1)` and `(2)`.
`P_(a) + rhogh + (1)/(2)rhov_(1)^(2) = P_(a) + (1)/(2)rhov_(2)^(2)`

Through continuity equation :
`Av_(1) = av_(2), v_(1) = (av_(2))/(a) rhogh + (1)/(2)rhov_(1)^(2) = (1)/(2)rhov_(2)^(2)`
on solving - `v_(2) = sqrt((2gh)/(1-(a^(2))/(A^(2)))) = 2m//sec`. .........(1)
(b) Appling Bernoulli's equation between `(2)` and `(3)`
`(1)/(2)rhov_(2)^(2) + rhogh_(0) = (1)/(2)rhov_(3)^(2)`
Through continuity equation -
`av_(2) = a'v_(3) rArr v_(3) = (av_(2))/(a')`
`rArr (1)/(2)rhov_(2)^(2) + rhogh_(0) = (1)/(2) rho((av_(2))/(a'))^(2)`

`(1)/(2) xx 2 xx 2 + gh_(0) = (1)/(2) ((a)/(a'))^(2) xx 2 xx 2`
`((a)/(a))^(2) = 1 + (9.8 xx .20)/(2) rArr ((a)/(a'))^(2) = 1.98 rArr a' = (a)/(sqrt(1.98))`
(c) From `(1)` at any height `'h'` of liquid level in container, the velocity through tap,
`v = sqrt((2gh)/(0.98)) = sqrt(20 h)`
We know, volume of liquid coming out of tap `=` decrease in volume of liquid in container.
For any small ltime interval `'dt'`
`av_(2)dt = -A*dx`
`asqrt(20 x) dt = -A dx rArr overset(t)underset(0)(int)dt = -(A)/(a) overset(0)underset(h)(int)(dx)/(sqrt(20x))`

`t = (A)/(asqrt(20))[2 sqrt(x)]_(h)^(0) rArr t = (A)/(asqrt(20)) 2sqrt(h)`
`= (A)/(a) xx 2 xx sqrt((h)/(20)) = (2A)/(a) sqrt((0.20)/(20)) = (2A)/(a) xx 0.1`
Given `((a)/(A))^(1//2) = 0.2` or `(A)/(a) = (1)/(0.0004) = 2500`
Thus `t = 2 xx 2500 xx 0.1 = 500` second.