State Archimedes' Principle. A ball weighing 4 kg of density $4000 k g m^{- 3}$ is completely immersed in water of density $1000 k g m^{- 3}$ . Find the apparent weight of the ball in water. $(g = 10 m s^{- 2})$

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Solution

According to Archimedes' principle, when an object is wholly or partially immersed in a liquid, it experiences a buoyant force which is equal to the weight of liquid diplaced by the object.
The buoyant force acting on the ball, F = Volume of the ball $\times$ density of water $\times$ acceleration due to gravity
Volume of the ball = Mass of the ball/Density of theball
$= k g / 4000 k g / m^{3}$
$= 0.001 m^{3}$
The buoyant force acting on the ball, $F = 0.001 m^{3} \times 1000 k g / m^{3} \times 10 m / s^{2}$
F = 10 N
The buoyant force acting on the ball = 10 N
The weight of the ball $= 4 k g \times 10 m / s^{2} = 40 N$
The apparent weight of the ball in water = 40 N - 10 N

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State Archimedes' Principle. A ball weighing 4 kg of density 4000kgm−3 is completely immersed in water of density 1000kgm−3. Find the apparent weight of the ball in water. (g=10ms−2)

State Archimedes' Principle. A ball weighing 4 kg of density $4000 k g m^{- 3}$ is completely immersed in water of density $1000 k g m^{- 3}$ . Find the apparent weight of the ball in water. $(g = 10 m s^{- 2})$