Question

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 $20c{m}^2$ at the bottom of the partition. Water is filled in one compartment and an acid of relative density $1.53\times 10kg{m}^{-3}$ in the other, both to a height of $4m$. The force necessary to keep the door closed is (Take $g=10m{s}^{-2})$

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

Answer 1

Answer: (C)

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
$P_w={\rho }_{w}hg=10^{3}kg{m}^{-3}\times 4m\times 10m{s}^{-2}$
$=4\times 10^{4}N{m}^{-2}$
For compartment containing acid,
${\rho }_a=1.5\times 10^{3}kg{m}^{-3},h=4m$
Pressure exerted by the acid at the door at the bottom is
$P_a={\rho }_{a}hg=1.5\times 10^{3}kg{m}^{-3}\times 4m\times 10m{s}^{-2}$
$=6\times 10^{4}N{m}^{-2}$
$\therefore$ Net pressure on the door$=P_a-P_w$
$=(6\times 10^4-4\times 10^4)N{m}^{-2}=2\times 10^{4}N{m}^{-2}$
Area of the door $=20c{m}^2=20\times 10^{-4}{m}^2$
$\therefore$ Force on the door $=2\times 10^{4}N{m}^{-2}\times 20\times 10^{-4}{m}^2=40N$
Thus, to keep the door closed the force of $40N$ must be applied horizontally from the water side