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ICSE Class 9 Physics Question 16 of 16

Upthrust in Fluids, Archimedes' Principle and Floatation — Question 7

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Question 7

A wooden block floats in water with two-third of it's volume submerged. (a) Calculate the density of wood. (b) When the same block is placed in oil, three-quarter of it's volume is immersed in oil. Calculate the density of oil.

Answer

(a) Let volume of wooden block be V.
Volume of block submerged in water = 23V\dfrac{2}{3}\text V

Density of water = 1000 kg m-3

By the principle of floatation,

Volume of immersed partTotal volume=Density of woodDensity of water∴23VV=Density of wood1000⇒Density of wood=23×1000=666.6666 kg m−3≈667 kg m−3\dfrac{\text {Volume of immersed part}}{\text{Total volume}} = \dfrac{\text {Density of wood}}{\text {Density of water}} \\[1em] \therefore \dfrac{\dfrac{2}{3}\text V}{\text V} = \dfrac{\text {Density of wood}}{1000} \\[1 em] \Rightarrow \text{Density of wood} = \dfrac{2}{3} \times 1000 = 666.6666 \text { kg m}^{-3} \approx 667 \text { kg m}^{-3}

(b) Volume of block submerged in oil = 34V\dfrac{3}{4}\text V

By the principle of floatation,

Volume of immersed partTotal volume=Density of woodDensity of oil∴34VV=667Density of oil⇒Density of oil=43×667=889.33 kg m−3≈889 kg m−3\dfrac{\text {Volume of immersed part}}{\text{Total volume}} = \dfrac{\text {Density of wood}}{\text {Density of oil}} \\[1em] \therefore \dfrac{\dfrac{3}{4}\text V}{\text V} = \dfrac{\text {667}}{\text{Density of oil}} \\[0.5em] \Rightarrow \text{Density of oil} = \dfrac{4}{3} \times 667 = 889.33 \text { kg m}^{-3} \approx 889 \text { kg m}^{-3}