IIT JEE , AIEEE Forum - Physics , Mechanics

konichiwa2x

Hi,

A metal cube is placed in an empty vessel. When water is filled in the vessel so that the cube is completely immersed in the water, the force on the bottom of the vessel in contact with the cube

(A) will increase

(B) will decrease

(C) will remain the same

(D) will become zero.

[Solution: C]

please explain.

deepak_1990

i think the force acting against cube is negligible compared to the force due to weight of the cube
let the cube b of any size.....................

deepak_1990

if u think am correct rate me plz..

raman_shadow

yaar in my view the buoyant force acting on metal cube decreases pressure at pt. of contact.
but the fluid itself exerts pressure at bottom.
so this pressure compensates the loss of pressure due to buoyancy.
and the pressure remains same.

but i m not 100% sure.
pls rate if u think i m crrect

nick

let the density of cube be 'd'
force exerted when in vessel = dVg where V is the volume....1

now density of water be 'w'
buoyant force=wgV.........2

pressure inside the depth 'l' ( here l is the length of the edge as it is "just" immersed)=wgl
AREA=l^2 as a cubical block

therefore force experi. = wgl*l^2=wgV as l^3=V for a cube.....3

2 and 3 act opposite to each other and cancel each other out hence net FORCE REMAINS THE SAMEEEEEEEEE.

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