Three liquids of equal masses are taken in three identical cubical vessels A,B and C. Their densities are `rho_(A),rho_(B)` and `rho_(C)` respectively. But `rho_(A)ltrho_(B)ltrho_(C)`. The force exerted by liquid on the base of the cubical vessel is

Options

A. maximum in vessel C
B. minimum in vessel C
C. the same in all the vessels
D. maximum in vessel A

Answer

C

Explanation

Pressure exerted by the liquid on the base of the vessel is
`P=rhoxxgxxh` . . . (i)
where
`rho=` density of the liquid
`g=` acceleration due to gravity
h=hright of the liquid column in the vessel
`"Density"=("Mass")/("Volume")=("Mass")/("Area"xx"Height")`
`rho=(m)/(axxh),h=(m)/(rhoxxa)`
Substituting the value of h in equation (i), we get
`P=rhoxxgxx(m)/(rhoxxa)=("g m")/(a)`
F=Pressure `xx` Area `=Pxxa=(" gm")/(a)xxa="g m"`
`becausem_(A)=m_(B)=m_(C):.F_(A)=F_(B)=F_(C)`