First slide

Fluid statics

Question

A tank with a square base of area 2 $m^{2}$ is divided into two compartments by a vertical partition in the middle. There is a small hinged door of face area 20 ${cm}^{2}$ at the bottom of the partition. Water is filled in one compartment and an acid of relative density 1.5 in the other, both to a height of 4 m. The force necessary to keep the door closed is (Take g = 10 ${ms}^{- 2}$ )

Moderate

A

10 N
B

20 N
C

40 N
D

80 N

Solution

The situation is as shown in the figure.
For compartment containing water,
H = 4m, $ρ_{w} = 10^{3} kg m^{- 3}$
Pressure exerted by the water at the door at the bottom is
$P_{w} = ρhg = 10^{3} kg m^{- 3} \times 4 m \times 10 {ms}^{- 2}$
= $4 \times 10^{4} N m^{- 2}$
For compartment containing acid,
$ρ$ _a = 1.5 x $10^{3} kg m^{- 3}$ , h = 4m
Pressure exerted by the acid at the door at the bottom is
$P_{a} = ρ_{a} hg = 1.5 \times 10^{3} kg m^{- 3} \times 4 m \times 10 {ms}^{- 3}$
= $(6 \times 10^{4}) {Nm}^{- 2}$
$∴ Net pressure on the door = P_{a} - P_{w}$
$= (6 \times 10^{4} - 4 \times 10^{4}) {Nm}^{- 2}$
$= 2 \times 10^{4} N m^{- 2}$
Area of the door = 20 ${cm}^{2}$ = 20 x $10^{- 4}$ $m^{2}$
Force = Pressure x Area = 40 N
Thus, to keep the door closed the force of 40 N must be applied horizontally from the water side.

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