A gardening pipe having an internal radius R is connected to a water sprinkler havmg n holes each of radius r. The water in the pipe has a speed v. What is the speed of water leaving the sprinkler ?

Options

A. `((R^2)/(r^2))nv`
B. `(R^2v)/(nr^2)`
C. `((r^2)/(R^2))v`
D. `((nR^2)/(r^2))v`

Answer

A

Explanation

From the equation of continuity,
`Av=A'v'`, where v and v' are the speeds of water in the pipe while entering and leaving the sprinkler respectively.
Area of pipe`=A=piR62`
`:.` Area of each hole in the sprinkler`=pir^2`
`:.` Area of n holes A'`=npir^2`
`:'" "Av=A'v'" " :." "piR^2xxv=npir^2v'`
`:." "v'=((R^2)/(nr^2))v`