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 $54.88 \;N .$