A wooden cube of side 10 cm has mass 700 g. What part of it remains above the water surface while floating vertically on the water surface?

Given,

Side of wooden cube = 10 cm

Hence,

Volume of wooden cube = 10 cm x 10 cm x 10 cm = 1000 cm³

Mass = 700 g

$\text{Density} = \dfrac{\text{mass}}{\text {volume}} \\[0.5em] \therefore \text{Density of wooden cube} = \dfrac{700}{1000} \\[0.5em] = 0.7 \text{ g cm}^{-3}$

By the principle of floatation,

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

Density of water = 1 g cm^-3

Density of wooden cube = 0.7 g cm^-3

$\therefore \dfrac{\text {Volume of immersed part}}{\text{Total volume}} = \dfrac{0.7}{1}$

Hence, fraction submerged = 0.7

Height of wooden cube = 10 cm

Part of woden cube which is submerged = 10 x 0.7 = 7 cm

Therefore, part above water = 10 - 7 = 3 cm

Hence, 3 cm of height of wooden cube remains above water while floating.