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

Explanation

Force on the door `="Pressure difference "xx" area of door "`
For acid `h_(1)=4m, p_(1)=1.7 xx 10^(3) "kg/m"^(3)`
`p_(1)=4 xx 1.7 xx 10^(3) xx 9.8 `
`p_(1)=66.64 xx 10^(3) Pa`
For water, `h_(2)=4m, p_(2)=10^(3) kg//m^(3)`
Pressure exerted by the acid,
`=4 xx 1000 xx 9.8`
Difference of pressure `trianlgep=p_(1)-p_(2)=(66.64-39.2) xx 10^(3)`
`=27.44 xx 10^(3) Pa`
Area of door, `A=20 xx 10^(-4) m^(2)`
Force applied on the door `=trianglep. A`
`=27.44 xx 10^(3) xx 20 xx 10^(-4)`
=54.88 N
To keep the door closed a force equal to 54.88 N should be applied from water compartment to acid compartment