Question

A cylindrical spray pump of tube Radius R has n fine holes each of r at one end. If the speed of liquid in the cylindrical tube is V the speed of ejection of the liquid through the holes is what?

Answer 1

given,
  area of cross section area (A) = πR²
  Number of holes = n
radius of hole is r , hence, cross section area of hole (a) = πr²
the speed of liquid in the cylindrical tube is V,
Let the speed of ejection of the liquid through the holes is v
        A/C to equation  of continuity,
          N₁ A₁V₁  = N₂A₂V₂
 where A is cross sectional area.
           V is the velocity is liquid.
           N is the number of path(hole )
      now, 
 AV = nav
 πR²V = nπr²v
  R²V/nr² = v
    v = R²V/nr²

hence, the speed of ejection of the liquid from the hole is v = R²V/nr²

Answer 2

Consider a cylindrical tube of spray pump having radius R,
one end having n holes.
 let r be the radius
 Let v be the speed of the liquid in tube.

cross section area of tube(A)=πR²

Cross sectional area of each hole=πr²

Therefore cross sectional area of n holes=a=πnr²

If v is the speed of ejection of fliquid hrough holes then
from continuity of flow we have

av=AV
v=AV/a=πR²V/πnr²=V/n (R/r)²

thus speed of ejection of liquid through holes is R²V/nr²