Law of Floatation

A block of wo...

Question

A block of wood of mass $24$ kg floats on water. The volume of wood is $0.032$ $m^{3}$ . Find the volume of block below the surface of water. (Density of water $= 1000$ kg $m^{- 3}$ )

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Solution

Now the density of the wood given by

$\frac{m}{v} = \frac{24}{0.032}$

$= 750 \frac{k g}{m^{3}}$

By principle of floation
Mass of wood $=$ mass of liquid displaced.
mass of liquid displaced $= 24 k g$
Volume of liquid displaced (v)

$= \frac{m}{v} = \frac{24}{1000} = 0.024 m^{3}$

Therefore, volume of wood inside the water will be

$= \frac{0.024}{0.032} = 0.75 m^{3}$

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