By increasing the volume of water coming out per second the water pump to n times , what is the power of motor has to be increases ?

Explanation

Power `P = W/t =(1/2mv^(2))/t`
Where m = mass of water coming out from pump .
` :. P =(mv^(2))/(2t)`
` =(Vrhov^(2))/(2t) " " [ :. m = rhoV] `
` :. P = (Alrhot^(2))/(2t) " " [ :. V = Al] `
V = volume of water coming from the pump in time t
A = cross sectional area of pipe ( is constant )
l = length of flow of water
`v = l/t` velocity of water thrown by pump
t = density of water
` :. P = 1/2 A rhov^(2) xx1/t`
` =1/2 Arhov^(3) " " [ :. l/t =v] `
` :. P = 1/2 Arho v^(3)` , here A and `rho ` are constant
` :. P prop v^(3) " "....(1)`
` rArr` Volume of water coming from cross section of pipe of length l per second
`V = V/t =(Al)/t`
` :. V = Av - to [l/t = v] `
If by increasing the value of V to n times then V = Av (here A is constant )
` :. ` The value of v has to done to n times so from equation (1) ,
` :. ` The value of v has to done t n times so from equation (1) , New power of pump ,
`P prop (nv)^(3)`
` :. P prop n^(3) v^(3)`
` :. P prop n^(3) P `