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ICSE Class 9 Physics Question 15 of 16

Upthrust in Fluids, Archimedes' Principle and Floatation — Question 6

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Question 6

A piece of wood of uniform cross section and height 15 cm floats vertically with it's height 10 cm in water and 12 cm in spirit. Find the density of (i) wood and (ii) spirit.

Answer

Given,

Height of the wood = 15 cm

Height of the wood in water = 10 cm

Height of the wood in spirit = 12 cm

As the block of wood is of uniform cross-sectional area, the height of the block is proportional to the volume

By the principle of floatation,

Volume of immersed partTotal volume=Density of woodDensity of water∴1015=Density of wood1⇒Density of wood=1015=0.667g cm−3\dfrac{\text {Volume of immersed part}}{\text{Total volume}} = \dfrac{\text {Density of wood}}{\text {Density of water}} \\[1em] \therefore \dfrac{10}{15} = \dfrac{\text {Density of wood}}{1} \\[1 em] \Rightarrow \text{Density of wood} = \dfrac{10}{15} = 0.667 \text {g cm}^{-3}

In the case of spirit, substituting the values in the formula above, we get,

Volume of immersed partTotal volume=Density of woodDensity of spirit1215=0.667Density of spiritDensity of spirit=1512×0.667=0.833 g cm−3\dfrac{\text {Volume of immersed part}}{\text{Total volume}} = \dfrac{\text {Density of wood}}{\text {Density of spirit}} \\[0.5em] \dfrac{12}{15} = \dfrac{0.667}{\text{Density of spirit}} \\[0.5em] \text{Density of spirit} = \dfrac{15}{12} \times 0.667 = 0.833 \text { g cm}^{-3} \\[0.5em]