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Question

# Some liquid is filled in a cylindrical vessel of radius R. Let F1 be the force applied by the liquid on the bottom of the cylinder. Now the same liquid is poured into a vessel of uniform square cross-section of side R. Let F2 be the force applied by the liquid on the bottom of this new vessel. Then:

A
F1=πF2
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B
F1=F2π
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C
F1=πF2
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D
F1=F2
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Solution

## The correct option is D F1=F2Given, Volume of liquid in cylindrical vessel = Volume of liquid in cubical vessel πR2×Hcylinder=R2×Hcube ⇒Hcube=πHcylinder ...(i) F1 = Force applied by the liquid on the bottom of the cylindrical vessel. F1=(Pbottom)× Base area or F1=(ρgHcyclinder)×πR2 F2 = Force applied by the liquid on the bottom of the cubical vessel. F2=(Pbottom)× Area or F2=(ρgHcube)×R2 Therefore, F1F2=ρgHcylinder×πR2ρgHcube×R2 Using eqn (i) F1F2=ρgHcylinder×πR2ρg(πHcylinder)×R2 ∴F1=F2 Alternate Solution: For those shapes whose cross sectional area is perpendicular to base area, force exerted by the liquid of same amount on the base area of different containers will be same. This is because of the reason that entire weight of liquid is supported by base of container for such shapes whose cross sectional area is perpendicular to base area

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