### Q.21:- A tank with a square base of area 1.0 m^{2} is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm^{2}. The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. compute the force necessary to keep the door close.

**Answer:-**Base area of the given tank, *A* = 1.0 m^{2}

Area of the hinged door, *a* = 20 cm^{2 }= 20 × 10^{–4} m^{2}

Density of water, *ρ*_{1} = 10^{3} kg/m^{3}

Density of acid, *ρ*_{2} = 1.7 × 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}*ρ*_{1}*g*

= 4 × 10^{3} × 9.8 = 3.92 × 10^{4} Pa

Pressure due to acid is given as:

*P*_{2} = *h*_{2}*ρ*_{2}*g*

= 4 × 1.7 × 10^{3} × 9.8 = 6.664 × 10^{4} Pa

Pressure difference between the water and acid columns:

Δ*P *= *P*_{2} – *P*_{1}

= 6.664 × 10^{4} – 3.92 × 10^{4}

= 2.744 × 10^{4} Pa

Hence, the force exerted on the door = Δ*P* × *a*

= 2.744 × 10^{4} × 20 × 10^{–4}

= 54.88 N

Therefore, the force necessary to keep the door closed is 54.88 N.