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A cubical body of mass $2 kg$ and side $20 cm$ is completely immersed in water. Density of water is $1000 kg / m^{3}$ . The buoyant force exerted by water on the body is

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78.4 N

68.4 N

66.6 N

46.5 N

answer is A.

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Detailed Solution

A cubical body of mass

2 kg

and the side

20 cm

is completely immersed in water. Density of water is

1000 kg / m^{3}

. The buoyant force exerted by water on the body is

78.4 N

.
Given that,

density of water, ρ = 1000 kg / m^{3}

Side of cube = 20 cm

\Rightarrow Side of cube = 20 \times 10^{- 2} m

Volume of body immersed in water,

V = {(side)}^{3}

\Rightarrow V = {(20 \times 10^{- 2} m)}^{3}

\Rightarrow V = 8 \times 10^{- 3} m^{3}

This is the volume of displaced water.

Weight of water displaced = volume \times density of water \times g

Where,

g = 9.8 m / s^{2}

We know the formula for buoyant force

F_{b} = ρ \times V \times g

where,

ρ

is fluid density,

V

is fluid volume,

g

is acceleration due to gravity,

F_{b}

is buoyant force.
On putting all the values in the above equation, we get,

{\Rightarrow F}_{b} = 8 \times 10^{- 3} \times 1000 \times 9.8

{\Rightarrow F}_{b} = 78.4 N

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