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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 ms−1)

A

19.344×105 N

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B

19.344×103 N

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C

21.344×105 N

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D

17.344×105 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 ms−1 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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