Question

A tank with a square base of area 2 m${}^2$ 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 1.53 x 10 kg m${}^{-3}$ in the other, both to a height of 4 m. The force necessary to keep the door closed is (Take g = 10 m s${}^{-2}$)

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

Answer 1

Answer: (C)

40 N

The situation is as shown in the figure.
For compartment contain 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 dosed the force of $40N$ must be applied horizontally from the water side.