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)=sqrt(0. 62gh(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.

Answer

((14pixx10^(5))/(27sqrt(g))unit)`

Explanation

Let O be the centre of hemispherical tank. Let at any
instant t, water level be `BAB_(1)` and t+dt, water level
is `B'A'B_(1).` Let `lt O_(1)OB_(1)=theta. `
`rArr AB_(1)=r cos theta and OA = r sin theta` decrease in the water
volume in time `di = pi AB_(1) ^(2). d (OA)`
[`pir^(2)` is surface area of water level and d (OA) is depth
of weter level]
`=pir^(2) cdot cos^(2)theta cdot r cos theta d theta`
`=pir^(3) cdot cos^(3)theta d theta`
Also, `h (t)=O_(2)A=r-rsin theta=r (1-sin theta)`
Now, outflow rate `Q=Acdot v(t)=Acdot 0.6sqrt(2gr(-sin theta))` ltbr. Where, A is the area of the outlet.
Thus, volume fliwing out in time dt.
`rArr Q_dt= A cdot (0.6) cdot sqrt (2gr) cdot sqrt (1-sin theta ) dt`
We have, `pir^(3) cos^(3)theta d theta = A (0.6) cdot sqrt (2gr) cdot sqrt (1-sin theta ) dt`
`rArr (pir^(3))/(A(0.6)sqrt(2gr))cdot (cos^(3) theta )/sqrt(1-sin theta ) d theta =dt `
Let the time taken to empty the tank be T.
Then, `T=int_(0)^(pi//2) (pir^(3))/(A(0.6)sqrt(2gr))cdot (cos^(3) theta )/sqrt(1-sin theta ) d theta `
`= (pir^(3))/(A(0.6)sqrt(2gr))int_(0)^(pi//2)(1-sin^(2)theta (-cos theta))/sqrt(1-sin theta )d theta`
Let `t_(1)= sqrt(1-sin theta )`
`rArr dt_(1)=(- cos theta)/(sqrt(1-sin theta ))d theta `
`therefore T=(-2pir^(3))/(A(0.6)sqrt(2gr))int_(1)^(0)[1-(1-t_(1)^(2))^(2)]dt_(1)`
`rArr T=(-2pir^(3))/(A(0.6)sqrt(2gr))int_(1)^(0)[1-(1+t_(1)^(4)-2t_(1)^(2))]dt_(1)`
`rArr T=(-2pir^(3))/(A(0.6)sqrt(2gr))int_(1)^(0)[1-(1-t_(1)^(4)+2t_(1)^(2)]dt_(1)`
`rArr T=(2pir^(3))/(A(0.6)sqrt(2gr))int_(1)^(0)(t_(1)^(4)+2t_(1)^(2))dt_(1)`
`rArr T=(2pir^(3))/(A(0.6)sqrt(2gr))int_(1)^(0)[t_(1)^(5)/5-(2t_(1)^(3))/3]_(1)^(0)`
`rArr T=(2pir cdot r^(5//2))/(A(6/10)sqrt(2gr)) cdot [0-1/5-0+2/3]`
`rArr T=(2pi cdot 2^(5//2)(10^(2))^(5//2))/(12cdot 3/5 cdot sqrt(2) cdot sqrt(g))[2/3-1/5]`
`=(2pixx10^(5) cdot 4 cdot 5)/((12xx3)sqrt(g))[(10-3)/15]`
`=(2pixx10^(5) xx7)/(3 cdot 3 cdot sqrt(g) cdot3 )=(14pixx10^(5))/(27sqrt(g)) unit`