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  • (a) A cube of side 5 cm 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 4 cm and then immersed in water, what will be the effect on the buoyant force

(a) A cube of side 5 cm 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 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 reason for each case.

(b) A ball weighing 4 kg of density 4000 kgm-3 is completely immersed in water of density 1000 kgm-3 .Find the force of buoyancy on it.

Answers (1)

(a) Force of buoyancy in any liquid is equal to weight of displaced fluid.

F_{b}=dVg

As the cube is completely immersed, displaced volume will be equal to volume of cube.

In both cases, volume and acceleration due to gravity is same.

Force of buoyancy will be only affected by density.

Density of saline solution will be more than density of water, hence force in water will be less.

If we change the side length of cube, its volume will be changed.

The force of buoyancy will be changed in same ratio as of volume.

Old volume = (5cm)3 = 125cm3

New volume = (4cm)3= 64cm3

Hence, the new force will be 64/125 times of old buoyancy force.

(b)     Force of buoyancy in any liquid is equal to weight of displaced fluid.

F_{b}=dVg

As the density of solid is more than density of water, it will drown in it. Therefore, the displaced volume will be equal to the volume of solid.

We know that mass is product of density and volume.

M=d \times V0\\ V=\frac{M}{d}=\frac{4}{4000}m^{3}

Now, we know the displaced volume, we can calculate force of buoyancy.

F_{b}=dVg\\ where\; \; V=\frac{4}{4000}m^{3}\\ \Rightarrow F_{b}=1000 \times \frac{4}{4000}\times 10=10N

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