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In the above question, the resultant upward force exerted by the sides of the glass on the water is
A. 100 N
B. 102 N
C. 303 N
D. 211 N

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Correct Answer - D
Force exerted by atmosphere on water is
`F_(2)=(P_(a))A_(2)=(1.01xx10^(5))xx(3xx10^(-3)) =303 N` (downwards)
Force exerted by bottom on the water,
`F_(3)=-F_(1)=102 N` (downwards)
Let F be the force exerted by side walls on thwe water (upwards) there equilibrium of water : net upward force = net downward force
or `F+F_(3)=F_(2)+W`
`:. F=F_(2)+W-f_(3)=303+10-102 =211 N` (downwards)

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