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Question

A thin square plate of side 4 m is placed horizontally 2 m below the surface of water. Calculate the thrust on one of the surface of the plate.
(Given: Atmospheric pressure,Patm=1.013×105N/m2
Density of water, ρw = 1000 kg/m3,
Acceleration due to gravity, (g)=9.8 ms1)

A

19.344×105 N

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B

19.344×103 N

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C

17.344×105 N

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D

21.344×105 N

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Solution

The correct option is A

19.344×105 N


Given :
Depth of plate in the water from the surface, h=2 m
Density of water, ρw=1000 kg/m3,
Acceleration due to gravity, g=9.8 ms1
Atmospheric pressure, Patm=1.013×105N/m2

Area of plate,
A=4 m×4 m
A=16 m2

Total pressure at a depth h below the surface of water is given by:
P=Patm+hρwg
P=(1.013×105)+(2×103×9.8)
P=1.209×105 Pa

Thrust on surface of plate due to liquid above it is:
F=PA
F=1.209×105 Pa×16 m2
F=19.344×105 N

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