An engineer was given a task to measure the rate of increase in pressure at the bottom of an empty cylindrical tank which is filled with water through hose pipe. If the speed of water coming out of the hose pipe is `"10 m s"^(-1)`, diameter and radius of the cylinder and hose pipe are 5 m and 25 cm, respectively, find the result shown by the engineer. (Take g = `10" m s"^(-2))`

Explanation

Speed of water coming out of the hose pipe,
`"V=10 m s"^(-1)`.
`:.` the length of the water column coming out of the hose pipe in one second of time, = 10 m.
Radius of the hose = 25 cm = `(1)/(4)m`.
`:.` area of cross section of the hosse pipe,
`a=pi((1)/(4))^(2)m^(2)`.
`:.` volume of water coming out of the hose pipe per second `=v=la=10xxpi((1)/(4))^(2)m^(3)`
Hence, the volume of water filled in the tank in one second of time `10xxpi((1)/(4))^(2)m^(3)`
Given, diameter of the tank = 5 m
`:.` radius of the tank = 2.5 m
`:.` area of cross-section of the tank `=pi(2.5)^(2)m^(2)`
Let 'h' be the increase in height of the water column, due to water filled in the tank in one second of time.
`:.` volume of the water filled in the tank in one second of `=hpi(2.5)^(2)m^(3)` (2)
From equation (1) an (2), we get
`10xxpi((1)/(4))^(2)=hxx(2.5)^(2)pirArrh=0.1m`
`:.` Increase in the height of water column in the tank in one second, h=0.1 m
`:.` rate of increase in pressure at the bottom the tank. = hpg
Given, `"g=10 m s"^(-2)` and density of water,
`p=10^(3)"kg m"^(-3)`
`:.` rate of increase in pressure at the bottom of the tank.
`=(0.1)xx(10^(3))xx(10)"Pa s"^(-1)=10^(3)"Pa s"^(-1)`