Figure shows a large closed cylindrical tank containing water, Initially the air trapped above the water surface has a height `h_(0)`and pressure `2p_(0)` where `p_(0_`is the atmospheric pressure. There is a hole in the wall of tank at a depth `h_(1)`below the top from which water comes out. A long vertical tube is connected as shown. (a) Find the height `h_(2)` of the water in the long tube above the top initially. (b) Find the speed with which water comes out of hole.(c)Find the height of the water in the long tube above the top when the water stops coming out of the hole.

Explanation

(a) ` 2P_0 = (h_2 +h_0) rho g `
` [:. Since liquids at same level`
have same pressure
` rArr 2P_0 = (h_2) rho g + (h_0) rho g `
` rArr (h_2)rho g = 2(P_0)-(h_0)rho g `
` rArr (h_2) = (2(P_0)/rho g) - ((h_0) rho g/rho g)`
` = (2(P-0)/ rho g)- (h_0)`
` (b) K.E. of the water = pressure energy of `
the water at that layer
` rArr 1/2 m(V_2) = m xx (p/ rho)`
` rArr (V_2)= (2P/rho)= 2/rho [P-0 +rho g (h_1 - h_0)]`
` rArr V = ([2/rho {(P_0)+ rho g (h_1 -h_0)}]^1/2)`
(c) From question,
` 2(P_0)+ rho g (h_1 - h_0) = (P_0)+ rho g X`
` rArr X = (P_0/rho g )+(h_1 - h_0) = h_2+ h_1 `
` i.e. X is h_1 meter below the top `
` rArr X is -h_1 above the top. `