A cylindrical tank `1 m` in radius rests on a platform `5 m` high. Initially the tank is filled with water to a height of `5 m`. A plug whose area is `10 ^-4 m^2`, is removed from an orifice on the side of the tank at the bottom. Calculate the following :
(a) Initial speed with which the water flows from the orifice.
(b) Initial speed with which the water strikes the ground,
( c) Time taken to empty the tank to half its original value.

Explanation

Area of orifice : `a = 10^-4 m^2`
Area of tank : `A = pi (1)^2 = pi m^2`
(a) Velocity of efflux : `v_1 = sqrt(2 gH) = sqrt(1 xx 10 xx 5) = 10 m//s`
(b) When water reaches to ground, its speed is `v_2`.
Applying energy conservation between `A` and `B`
`(1)/(2) mv_1^2 + mg H' = (1)/(2) mv_2^2`
`v_2 = sqrt(v_1^2 + 2gh) = sqrt((10)^2 + 2 xx 10 xx 5) = 10 sqrt(2) m//s`
( c) Let at instant height of water is `h` and in time `dt`, level of liquid falls by `dh`
Velocity of efflux : `v = sqrt(2 gh)`
Volume of water coming out of tank per second
`Adh = -a sqrt(2 gh) dt`
(As time increases, height of water decreases)
`A dh = a = sqrt(2 gh) dt`
`dt = -(A)/(a sqrt(2g)) (dh)/(h^(1//2))`
`int dt = -(A)/(a sqrt(2 g)) int h^(-1//2)//dh`
`t = -(A)/(a sqrt(2g)) (|h^(1//2)|_H^(H//2))/(1//2)`
=`-sqrt((2)/(g)) (A)/(a) (sqrt((H)/(2)) - sqrt(H))`
=`sqrt((2)/(g)) (A)/(a) (sqrt(H) - sqrt((H)/(2))) = (A)/(2) (sqrt((2 H)/(g)) - sqrt((H)/(g)))`
=`(pi)/(10^-4) (1 - (1)/(sqrt(2)))`
=`pi ((sqrt(2) -1)/(sqrt(2))) xx 10^4 sec`.

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