__Question 10.21__:^{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.

__Solution__:*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 X 10

^{3}X 9.8 = 3.92 X 10

^{4}Pa

Pressure due to acid is given as:

P

_{2}= h

_{2}

*ρ*

_{2}g

= 4 X 1.7 X 10

^{3}X 9.8 = 6.664 X 10

^{4}Pa

Pressure difference between the water and acid columns:

Δ

*P*= P

_{2}- P

_{1}

= 6.664 X 10

^{4}- 3.92 X 10

^{4}

= 2.744 X 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.

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