A tank with a square base of area `2m^2` is divided into two compartments by a vertical partition in the middle. There is a small hinged door of face area `20 cm^2` at the bottom of the partition. Water is filled in one compartment and an acid of relative density 1.5 in the other, both to a height of 4m. The force necessary to keep the door closed is `(Take g = 10 m s^(-2))`

Options

A. 10N
B. 20 N
C. 40 N
D. 80 N

Answer

C

Explanation

The situation is as shown in the figure.

For compartment containing water,
`h=4m, rho_(w)=10^(3) kg m^(-3)`
Pressure exerted by the water at the door at the bottom is
`P_(w)=rho_(w)hg=10^(3) kg m^(-3) xx 4 m xx 10 m s^(-2)`
`=4xx10^(4) N m^(-2)`
For compartment containing acid,
`rho_(a)=1.5xx10^(3) kg m^(-3) , h=4 m`
Pressure exerted by the acid at the door at the bottom is
`P_(a)=rho_(a)hg=1.5xx10^(3) kg m^(3) xx 4 m xx 10 m s^(-2)`
`=6xx10^(4) N m^(-2)`
`=6xx10^(4) N m^(-2)`
`therefore` Net pressure on the door `=P_(a)-P_(w)`
`=(6xx10^(4)-4xx10^(4))N m^(-2) =2xx10^(4) N m^(-2)`
Area of the door `=20 cm^(2)= 20xx 10^(-4) m^(2)`
`therefore` Force on the door `= 2 xx 10^(4) N m^(-2) xx 10^(-4) m^(2)`
=40 N
Thus, to keep the door closed the force fo 40 N must be applied horizontally from the water side.