A cylindrical tank has a small hole at bottom. At `t = 0`, a tap starts to supply water into the tank at a constant rate `beta m^3//s`.
(a) Find the maximum level of water `H_(max)` in the tank ?
(b) At what time level of water becomes `h(h lt H_(max))` Given `a` , area of hole, `A` : area of tank.

Explanation

(a) Level will be maximum when
Rate of inflow of water = rate of outflow of water
`beta = av = a sqrt(2 g H_(max))`
`H_(max) = (beta^2)/(2 a^2 g)`
(b) Let at time `t`, level of water is `h`, level falls by `dh` in time `dt`
`A dh = (beta - av) dt`
=`(beta - a sqrt(2 gh)) dt`
`int_0^t dt = A int_0^h (dh)/(beta - a sqrt(2 gh))`
`t = A int_0^h (dh)/(beta - a sqrt(2 gh)) = (A)/(ag) [(beta)/(a) 1n {(beta - a sqrt(2 gh))/(beta)} - sqrt(2 gh) ]`
The Solution : of this integral is not required for medical aspirant.
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