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A cube of side 5 cm is immersed in water and then in a saturated salt solution. In which case will it experience a greater buoyant force. If each side of the cube is reduced to 4 cm and then immersed in water, what will be the effect on the buoyant force experienced by the cube as compared to the first case for water. Give the reason for each case.


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Solution

Step 1: Given data

Side of the cube, s=5cm

Side of the cube, s=0.05m

Volume of the cube, v=0.000125m3

Reduced side of the cube, sr=4cm

Reduced side of the cube, sr=0.04cm

Volume of the reduced cube, vr=0.000064m3

Step 2: Assumptions

Density of water, ρw=1000kg/m3

Density of saturated salt solution=ρs

Buoyant force exerted on the cube of side 5cm immersed in water=fbw

Buoyant force exerted on the cube of side 5cm immersed in saturated salt solution=fbs

Buoyant force exerted on the cube of side 4cm immersed in water=f4

Acceleration due to gravity, g=10m/s2

Step 3: Predicting the relative magnitude of the buoyant force exerted on the cube of side 5cm immersed in water and saturated salt solution

Buoyant force exerted on the cube of side 5cm immersed in water, fbw=ρw×g×V ………………….(a)

Buoyant force exerted on the cube of side 5cm immersed in the saturated salt solution, fbs=ρs×g×V …………………….(b)

From equations (a) and (b), it is clear that"the buoyant force exerted on the cube is directly proportional to the density of the liquid in which the cube is immersed.

Since acceleration due to gravity and the volume of the cube remains constant, therefore the “density” is the deciding parameter to depict whether the cube experiences more buoyant force in water or saturated salt solution.

But the density of the salt solution is always greater than water i.e ρs>ρw

Therefore, fbs>fbw

Hence, the buoyant force exerted on the cube will be more when it is immersed in a saturated salt solution.

Step 4: Estimating the buoyant force exerted on the cubes of the side 5cm and 4cm

Buoyant force exerted on the cube of the side 5cm immersed in water, fbw=ρw×g×V

Substituting the given values in the above equation, we get

fbw=1000kg/m3×10m/s2×0.000125m3

fbw=1.25N

Buoyant force exerted on the cube of the side 4cm immersed in water, f4=ρw×g×vr

Substituting the given values in the above equation, we get

f4=1000kg/m3×10m/s2×0.000064m3

f4=0.64N

Hence, the buoyant force will be more on the cube of the side 5cm as compared to the cube of the side 4cm.


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