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.

A tank with a square base of area 1.0m^(2) is divided by a vertical pa

1.	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.
Answer» Solution :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 .