A tank with a square base of area `1.0m^(2)` is divided by a vertical partition in the middle The bottom of the partition has a small - hinged door of area `20cm^(2)` . The tank is filled with water in one compartement , and and an acid (of relative density 1.7) in the other , both to a height of 4.0 m . compute the force necessary to keep the door close.

Explanation

For tank filled with water ,
`h_(1)=4m,rho_(1)=10^(3)kgm^(-3)` (density of water)
Pressure due to water near door at bottom
`P_(1)=h_(1)rho_(1)g`
`=4xx10^(3)xx9.8`
`=39.2xx10^(3)Pa` …(1)
(b) For tnk filled with acid,
`h_(2)=4m,(rho_(2))/(rho_(1))=1.7`
`thereforerho_(2)=1.7xxrho_(1)=1.7xx10^(3)kgm^(-3)`
Pressure due to acid near door at bottom,
`P_(2)=h_(2)rho_(2)g`
`=4xx1.7xx10^(3)xx9.8`
`=66.64xx10^(3)Pa`...(2)
`thereforeP_(2)gtP_(1) " " [because"From result (1) and (2)"]`
`therefore` Pressure difference,
`DeltaP=P_(2)-P_(1)`
`=66.64xx10^(3)-39.2xx10^(3)`
`=27.44xx10^(3)Pa`
Area of the hinged door ,
`A=20cm^(2)=20xx10^(-4)m^(2)`
To keep door closed.
`therefore` The force necessary in horizontal direction from water to acid on door.
F= pressure difference `(DeltaP)xx` area A
`=27.44xx10^(3)xx20xx10^(-4)`
`=54.88N`
`thereforeF=55N`
Base area does not affect the magnitude of force .