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.

Answer

A, B

Explanation

(a) Level will be maximum when

Rate of inflow of water=rate of outflow of water
i.e. `alpha =av`
`or, alpha =a sqrt(2gh_(max))`
`implies h_(max)=(alpha^(2))/(2ga^(2))`
(b) Let at time t, the level of water be h. then,
`A((dh)/(dt))=alpha-asqrt(2gh)` or `int_(0)^(h) (dh)/(alpha -asqrt(2gh)) =int_(0)^(t) (dt)/(A)`
Solving this, we get
`t=(A)/(ag)[(alpha)/(a)In{(alpha-a sqrt(2gh))/(alpha)}-sqrt(2gh)]`.