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

The Correct Option is C

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

Answer 2

Given:
Water height \(h = 4 \, \text{m}\), density \(\rho_{w} = 10^3 \, \text{kg/m}^3\)
Acid height \(h = 4 \, \text{m}\), density \(\rho_{a} = 1.5 \times 10^3 \, \text{kg/m}^3\)
Acceleration due to gravity \(g = 10 \, \text{m/s}^2\)
Area of the door \(A = 20 \, \text{cm}^2 = 20 \times 10^{-4} \, \text{m}^2\)

Calculate Pressures:
The pressure exerted by water \(P_{w} = \rho_{w} \cdot h \cdot g = 10^3 \times 4 \times 10 = 40,000 \, \text{N/m}^2\)
The pressure exerted by acid \(P_{a} = \rho_{a} \cdot h \cdot g = 1.5 \times 10^3 \times 4 \times 10 = 60,000 \, \text{N/m}^2\)

Calculate Net Pressure on the Door:
Net pressure \(P_{\text{net}} = P_{a} - P_{w} = 60,000 - 40,000 = 20,000 \, \text{N/m}^2\)

Calculate Force on the Door:
Force \(F = P_{\text{net}} \times A = 20,000 \times 20 \times 10^{-4} = 40 \, \text{N}\)

So, the correct option is (C): 40N