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.

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

Density of water = 1000 kg m^-3

By the principle of floatation,

$\dfrac{\text {Volume of immersed part}}{\text{Total volume}} = \dfrac{\text {Density of wood}}{\text {Density of water}} \\[1em] \therefore \dfrac{\dfrac{2}{3}V}{V} = \dfrac{\text {Density of wood}}{1000} \\[0.5em] \Rightarrow \text{Density of wood} = \dfrac{2}{3} \times 1000 = 666.6666 \text { kg m}^{-3} \approx 667 \text { kg m}^{-3}$

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

By the principle of floatation,

$\dfrac{\text {Volume of immersed part}}{\text{Total volume}} = \dfrac{\text {Density of wood}}{\text {Density of oil}} \\[1em] \therefore \dfrac{\dfrac{3}{4}V}{V} = \dfrac{\text {667}}{\text{Density of oil}} \\[0.5em] \Rightarrow \text{Density of oil} = \dfrac{4}{3} \times 667 = 889.33 \text { kg m}^{-3} \approx 889 \text { kg m}^{-3}$