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 $1.53 \times 10\, kg\, m^{-3}$ 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)

$40 \, N$

Answer 2

The situation is as shown in the figure For compartment containing water, $h=4 \, m$, $\rho_{w}=10^{3} \, kg\, m^{-3}$ Pressure exerted by the water at the door at the bottom is $P_{w}=\rho_{w} \, hg = 10^{3}\, kg \, m^{-3}\times4 m \times10\,m \,s^{-2}$ $=4\times10^{4}\, N\,m^{-2}$ For compartment containing acid, $\rho_{a}=1.5 \times10^{3} \, kg \,m^{-3}, h=4\,m$ Pressure exerted by the acid at the door at the bottom is $P_{a} = \rho_{a}hg =1.5 \times 10^{3}\,kg \,m^{-3} \times 4 \,m \times 10\,m \, s^{-2}$ $=6 \times 10^{4} \, N \, m^{-2}$ $\therefore$ 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 $= 20 cm^{2} = 20 \times 10^{-4} m^{2}$ $\therefore$ Force on the door $= 2 \times 10^{4} N \,m^{-2} \times 20\times 10^{-4} m^{2} = 40 \, N$ Thus, to keep the door closed the force of $40\, N$ must be applied horizontally from the water side