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A cube of side 5cm is immersed in water and then in saturated salt solution. In

which case will it experience a greater buoyant force. If each side of the cube is
reduced to 4cm and then immersed in water compare the force experienced by the
cube, as compared to the first case. Give reasons for each case.

When a person is floating on the liquid the net buoyant force = weight of the liquid displaced( Vρg )

or 53 cm3

ρ is density of the liquid.

So, if the density water is more then the buoyant force is more. So, the buoyant force is more in the case of saturated salt solution due to greater density then water. 

If each side of cube is reduced and then immersed in water, buoyant force will also decrease by above given relation 

So, the net buoyant force becomes = V'ρg

where V' be the new volume, V = 64 cm3.

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 It will experience a greater buoyant force in salt water due to it's higher density. Salt water is more dense than fresh water because salt water has more mass than fresh water. According to the formula of density, it's D=m/V. D is the density, m is the mass, and V is the volume. The more mass and less volume, the more dense the "solution" is. The less mass and more volume, the less dense the "solution" is.

Cube A :Volume of cube of side 5cm is 5*5*5 = 125cm(cube)
Cube B:Volume of cube of side 4cm is 4*4*4 = 64cm(cube)
Due to greater volume, Cube A will experience greater buoyant force as it displaces more water.
Cube B will not due to lower volume.
Thank You.
 

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