What is the buoyant force experienced by a cube of side $10 c m$ which is completely immersed in water?
(Take $g = 10 m s^{- 2}$ and density of water $= 1000 k g m^{- 3}$ )

$100 N$

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$10 N$

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$1000 N$

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$1 N$

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Solution

The correct option is B
$10 N$

Given,
Length of each side of the cube, $l = 10 c m = 0.1 m$
Density of water, $ρ = 1000 k g m^{- 3}$
Acceleration due to gravity, $g = 10 m s^{- 2}$
Volume of the cube, $V = l^{3} = (0.1)^{3} = 0.001 m^{3}$

Archimedes' principle states that the buoyant force acting on an object is equal to the weight of the fluid displaced by it.
In the given case, volume of cube is equal to the volume of water displaced.

Mass of water displaced (m) = Density of water $\times$ Volume of water displaced
$\Rightarrow m = 1000 k g m^{- 3} \times 0.001 m^{3}$
$= 1 k g$

Weight of water displaced, W = mg
$\Rightarrow W = 1 \times 10$ = $10 N$
Hence, the buoyant force is $10 N$ .

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What is the buoyant force experienced by a cube of side 10 cm which is completely immersed in water? (Take g=10 m s−2 and density of water =1000 kg m−3)

What is the buoyant force experienced by a cube of side $10 c m$ which is completely immersed in water?
(Take $g = 10 m s^{- 2}$ and density of water $= 1000 k g m^{- 3}$ )