A hemi-spherical tank of radius 2 m is initially full of water and has an outlet of `12c m^2` cross-sectional area at the bottom. The outlet is opened at some instant. The flow through the outlet is according to the law `v(t)=0.6sqrt(2gh(t)),` where `v(t)` and `h(t)` are, respectively, the velocity of the flow through the outlet and the height of water level above the outlet and the height of water level above the outlet at time `t ,` and `g` is the acceleration due to gravity. Find the time it takes to empty the tank.

Explanation

Let the water level be at a height h after time t, and water level falls by dh in time dt, and the corresponding volume of water gone out be dV.
`therefore dV=-pir^(2)dh`
`therefore (dV)/(dt) =-pir^(2)(dh)/(dt)` `therefore` as t increases, h decreases
Now, velocity of water, `v=3/5 sqrt(2gh)`
Rate of flow of water =Av `(A=12 cm^(2))`
`therefore (dV)/(dt) = (3/5sqrt(2gh)A)`
`=-pir^(2)(dh)/(dt)`

Also from the figure,
`R^(2)=(R-h)^(2)+r^(2)`
or `r^(2)=2hr-h^(2)`
So, `3/5sqrt(2g)sqrt(h)A=-pi(2hR-h^(2)) xx (dh)/(dt)`
or `(2hR-h^(2))/sqrt(h)dh=-3/(5pi)sqrt(2h)Adt`
Integrating, we get
`int_(R)^(0)(2Rsqrt(h)-h^(3//2))dh=-(3sqrt(2g))/(5pi) A. int_(0)^(t)dt`
or `T=-(5pi)/(3Asqrt(2g)) (2Rh^(3//2)/(3//2)-h^(5//2)/(5//2))_(R^(0)`
`=(5pi)/(3Asqrt(2g))((4R)/(3)R^(3//2)-2/5R^(5//2))`
`=(5pi)/(3Asqrt(2)g)14/15R^(5//2)`
For R=200cm,
`T=(56pi)/(9Asqrt(g))(10)^(5)`
`=(56pi)/(9 xx 12sqrt(g))(10)^(5)`
`=(14pi)/(27sqrt(g))(10)^(5)` units