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ICSE Class 9 Physics Question 13 of 16

Upthrust in Fluids, Archimedes' Principle and Floatation — Question 2

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Question 2

A block of wood of mass 24 kg floats on water. The volume of wood is 0.032 m3.

Find —

(a) the volume of block below the surface of water,

(b) the density of wood.

(Density of water = 1000 kg m-3)

Answer

(a) Given,

Mass = 24 kg

Volume = 0.032 m3

Upthrust = Volume of block below the water x density of liquid x acceleration due to gravity

Hence,

24kgf=v×1000×gv=241000⇒v=0.024 m324 \text {kgf} = \text v \times 1000 \times \text g \\[0.5em] \text v = \dfrac{24}{1000} \\[0.5em] \Rightarrow \text v = 0.024 \text{ m}^{3} \\[0.5em]

Hence, volume of block below the surface of water = 0.024 m3

(b) By the principle of floatation,

Volume of immersed part Total volume=Density of woodDensity of water\dfrac{\text {Volume of immersed part }}{\text{Total volume}} = \dfrac{\text {Density of wood}}{\text {Density of water}}

and

Density of water = 1000 kg m-3

Substituting the values in the formula we get,

0.0240.032=Density of wood1000⇒Density of wood=0.0240.032×1000=7.5×102 kg m−3\dfrac{0.024}{0.032} = \dfrac{\text {Density of wood}}{1000} \\[0.5em] \Rightarrow \text{Density of wood} = \dfrac{0.024}{0.032} \times 1000 \\[0.5em] = 7.5 \times 10^{2} \text{ kg m}^{-3} \\[0.5em]