For compartment containing water, $h_{1}=4\ \text{m},\ \rho_{1}=10^{3}\ \text{kg}\ \text{m}^{-3}$ The pressure exerted by water at the door provided at bottom, $P_{1}=h_{1}\rho_{1} g$$=4 \times 10^{3} \times 9.8$$=3.92 \times 10^{4}\ \text{Pa}$For compartment containing acid,$h_{2}=4\ \text{m},\ \rho_{2}=1.7 \times 10^{3}\ \text{kg}/\text{m}^{3}$The pressure exerted by acid at the door provided at bottom,$P_{2}=h_{2}\rho_{2} g$$=4 \times 1.7 \times 10^{3} \times 9.8$$=6.664 \times 10^{4}\ \text{Pa}$∴ Difference of pressure $= P_{2} - P_{1}$$=6.664 \times 10^{4} - 3.92 \times 10^{4}$$=2.744 \times 10^{4}\ \text{Pa}$Given, area of door, $A=20\ \text{cm}^{2}=20 \times 10^{-4}\ \text{m}^{2}$Force on the door = difference in pressure × area$= (P_{2} - P_{1}) \times A$$= (2.744 \times 10^{4}) \times (20 \times 10^{-4})$$=54.88\ \text{N}=55\ \text{N}$To keep the door closed, the force equal to 55 N should be applied horizontally on the door from compartment containing water to that containing acid.