A glass full of water has a bottom of area `20 cm^(2)`, top of area `20 cm`, height `20 cm` and volume half a litre.

a. Find the force exerted by the water on the bottom.
b. Considering the equilibrium of the `v`, after. find the resultant force exerted by the side, of the glass on the water. Atmospheric pressure `= 1.0 xx 10^(5) N//m^(2)`. Density of water `= 1000 kg//m^(-3)` and `g = 10 m//s^(2)`

Explanation

The pressure intensity at the bottom of the container is `P=hrhog=0.20xx1000xx10=2xx10^(3)N//m^(2)`
Force at the bottome of container
`=Pxx"base area" = 2xx10^(3)xx(20xx10^(-4))`
`=4N` (force exerted by water only) Weight of the water in the container
`="Volume" xx "density of water" xxg`
`=0.5xx10^(3)xx1000xx10=5n`
`P_(2)gtP_(1)`

From the diagram, the force exerted by the container is in upward direction because `(P_(2)gtP_(1))`. Let force exerted by sides of container be `F` then
`4+F=5NimpliesF=1N`