A tank with a square base of area $1{\text{m}}^{\text{2}}$ is divided by a vertical partition has a small hinged door of area $20{\text{cm}}^{\text{2}}$ . The tank is filled with water in one compartment and an acid (of relative density 1.7) in another both to the height of 4 m. Then the force necessary to keep door closed is                       

Given that the base area of a tank ${\text{=1m}}^2$

Tank is divided by using vertical partition as shown.

The area of hinged door a = 20 cm²

(i) Left portion is filled with acid of ${\rho }_R=1.7\Rightarrow {\rho }_A=1.7{\rho }_w$

h = 4 m

${\text{F}}_{\text{R}}={\text{P}}_{\text{A}}\text{a}={\rho }_{\text{A}}\times \text{g}\times {\text{h}}_{\text{A}}\times \text{a}$

$=1.7{\rho }_{\text{w}}\times \text{g}\times 4\times 20{\text{cm}}^{\text{2}}$          

(ii) Right side portion is filled with water then

${\text{F}}_{\text{L}}={\text{P}}_{\text{w}}\text{a}={\rho }_{\text{w}}\times \text{g}\times {\text{h}}_{\text{w}}\times \text{a}$

$={\rho }_{\text{w}}\times 9.8\times 4\times 2\text{0}{\text{cm}}^{\text{2}}$

${\text{F}}_{\text{net}}\text{=}{\text{F}}_{\text{R}}-{\text{F}}_{\text{L}}$

$=\text{h}\left({\rho }_{\text{A}}-{\rho }_{\text{w}}\right)\text{gA}$

$=4\times 0.7\times {10}^3\times 9.8\times 20\times {10}^{-4}$

$=54.88\text{N}$