A cylinderical tank 1 m in radius rests on a plaform 5 m high. Initially the tank is filled with upto a height of 5m a plug whose area is `10^(-4)cm^(2)` is removed from an orifice on the side of the tank at the bottom.
Calculate (a). Initial speed with which the water flows from the orifice
(b). Initial speed with which the water strikes the ground.

Explanation

(a) Speed of efflux, `u = sqrt(2gh)`
`= sqrt(2 xx 10 xx 5) = 10 ms^(-1)`
(b) the initial vertical velocity of water flowing out of orifice is zero. Its vertical velocity when it hits the ground is:
`upsilon_(0) = sqrt(u^(2)+upsilon^(2)) = sqrt(10^(2)+10^(2)) = 10sqrt(2)`
`= 14.14 ms^(-1)`.
(c ) Let A, `A_(0)` be the area of cross-section of the cylindrical tank an plug fitted to orifice respectivley. Then `A = pixx (1)^(2)=pi m^(2)` and `A_(0)= 10^(-4)m^(2)`. When the height of water level above the hole is y, the velocity of flow through the orifice is `upsilon=sqrt(2gy)`. If dV volume of water flows out in time dt and dy. be the decreases. in height of water in tank the rate of flow of water is :
`(dV)/(dt)=A_(0)upsilon = A_(0)sqrt(2gy) or (-Ady)/(dt) =A_(0) sqrt(2gy)`
or `dt = (-A)/(A_0) (dy)/(sqrt(2gy)) [ :' dV=-A dy]`
on intergration it within the conditions of flow of water
`int_(0)^(t) dt = int_(h)^(h//2) (-A)/(A_(0)) (dy)/(sqrt(2gy)`
`t - (A)/((A_0) sqrt((2)/(g)) (sqrt(h)-sqrt(h//2))`
`=(pi xx (1)^(2))/(10^(-4)) sqrt(2/10) (sqrt(5)-sqrt(5//2))`
`= 9.2 xx 10^(3) s ~~2.5 h`.