An open container has dimensions of `4.0 m xx5.0 m ` and height of 3.0 m.
(i) Find the weight of the air in the container at `20^(@)C`.
(ii) What is the weight of an equal volume of water ? Also find pressure at the base of container due to this weight pof water.
(iii) What is the total downward force on the base of the container due to air pressure of 1.0 atm ?
Take the denities of air and water as `1.2 kg//m^(3)` and `10^(3) kg//m^(3)` respectively.

Explanation

It is given that density of air `rho_(air)=1.2 kg//m^(3)`
and density of water `rho_(water)=10^(3) kg//m^(3)`
(i) The volume of air in the container
`=(4.0)(5.0)(3.0)=60m^(3)`
`therefore ` The mass of the air is `m_(air )=rho_(air).V_(air)`
`=(1.2)(60)=72 kg`
The weight of the air is `w_(air)=m_(air)g=(72)(9.8)`
=705.6 N
(ii) The weight of water `w_(water) =m_(water)g=rho_(water). V_(water)g`
`=(10^(3))(60)(9.8)`
`=5.9xx10^(5) N`
Pressure at the base due to this weight
` rho=(F(=w))/(A)=(5.9xx10^(5))/(20)=2.95xx10^(4) Nm^(-2)`
(iii) The downward force on the base is
=(Air pressure) (surface area)
`=(1.013xx10^(5))(4.0xx5.0)=2.0xx10&^(6) N`