A cube of density $800 k g m^{- 3}$ is floating in water. Volume of the cube is $2 m^{3}$ . Calculate the force of buoyancy acting on the body.
Density of water: $ρ_{w a t e r} = 1000 k g m^{- 3}$

1600 N

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16000 N

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8000 N

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10000 N

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Solution

The correct option is B $16000 N$
Given:
Density of cube: $ρ = 800 k g m^{- 3}$
Total volume of cube: $V = 2 m^{3}$
Density of water: $ρ_{w} = 1000 k g m^{- 3}$

The buoyant force is equal to the weight of the body $W$ as it does not sinks.

$F_{b} = W$
$F_{b} = ρ V g$
$F_{b} = 800 \times 2 \times 10$
$F_{b} = 16000 N$

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A cube of density 800 kg m−3 is floating in water. Volume of the cube is 2 m3. Calculate the force of buoyancy acting on the body. Density of water: ρwater=1000 kg m−3

The correct option is B 16000 NGiven: Density of cube: ρ=800 kg m−3 Total volume of cube: V=2 m3 Density of water: ρw=1000 kg m−3 The buoyant force is equal to the weight of the body W as it does not sinks. Fb=W Fb=ρVg Fb=800×2×10 Fb=16000 N

A cube of density $800 k g m^{- 3}$ is floating in water. Volume of the cube is $2 m^{3}$ . Calculate the force of buoyancy acting on the body.
Density of water: $ρ_{w a t e r} = 1000 k g m^{- 3}$