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Q.

A glass full of water has a bottom of area20cm2 , top of area20cm2 , height 20 cm and volume half a liter. (i) Find the force exerted by the water on the bottom (ii) Considering the equilibrium of the water; Find the resultant force exerted by the sides of the glass on the water. [Atomsphericpressure=1.0×105Nm2,Densityofwater=1000kgm3andg=10ms2]

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a

(i) 404 N (ii) 5 N upward

b

(i) 204 N (ii) 1 N upward

c

(i) 104 N (ii) 5N downward

d

(i) 304 N (ii) 10 N downward

answer is C.

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Detailed Solution

P=P0+hρg

=105+0.2×1000×10

=105+2×103

=105+0.02×105

P=(1.02)×105Nm2

(i)F=P.A=1.02×105×20×104

=2.04×102=204N

(ii) To find out the resultant force exerted by the sides of the glass from the F.B.D of water inside the glass

Pa×A+mg=Ahρwg+Fs+PaA

mg=Ahρwg+Fs

Fs=mgAhρwg

=0.5×10-20×104×20×102×1000×10

=54×102×104×102×104

Fs=54=1Nupward

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A glass full of water has a bottom of area20 cm2 , top of area20 cm2 , height 20 cm and volume half a liter. (i) Find the force exerted by the water on the bottom (ii) Considering the equilibrium of the water; Find the resultant force exerted by the sides of the glass on the water. [Atomspheric pressure=1.0×105Nm−2,Density of water=1000 kgm−3 and g=10 ms−2]