d
Base area of the given \(\operatorname{tank}, A=1.0 m ^{2}\)

Area of the hinged door, \(a=20 cm ^{2}=20 \times 10^{-4} m ^{2}\)

Density of water, \(\rho_{1}=10^{3} kg / m ^{3}\)

Density of acid, \(\rho_{2}=1.7 \times 10^{3} kg / m ^{3}\)

Height of the water column, \(h_{1}=4 m\)

Height of the acid column, \(h_{2}=4 m\)

Acceleration due to gravity, \(g=9.8\)

Pressure due to water is given as:

\(P_{1}=h_{1} \rho_{1} g\)

\(=4 \times 10^{3} \times 9.8\)

\(=3.92 \times 10^{4} Pa\)

Pressure due to acid is given as:

\(P_{2}=h_{2} \rho_{2} g\)

\(=4 \times 1.7 \times 10^{3} \times 9.8\)

\(=6.664 \times 10^{4} Pa\)

Pressure difference between the water and acid columns:

\(\Delta P=P_{2}-P_{1}\)

\(=6.664 \times 10^{4}-3.92 \times 10^{4}\)

\(=2.744 \times 10^{4} Pa\)

Hence, the force exerted on the door \(=\Delta P \times a\) \(=2.744 \times 10^{4} \times 20 \times 10^{-4}\)

\(=54.88 N\)

Therefore, the force necessary to keep the door closed is \(54.88 \;N .\)