Question

A cube with a mass '$m$' completely wettable by water floats on the surface of water. Each side of the cube is '$a$'. The distance $h$ between the lower face of cube and the surface of the water if surface tension is $S$ is given as $\frac{mg+xSa}{{\rho }_w-a^{2}g}$. Find $x$. Take density of water as ${\rho }_w$. Take angle of contact to be zero.

Answer 1

The force that act on the block are the following-

Weight of the block=$mg$ downwards

Surface tension force=$SL=S\times \left(4a\right)$ downwards($4a$ since length of cube in contact with water surface is $a$ for each side of the cube)

Buoyant force on the cube=$V_{immersed}{\rho }_{w}g$

$=\left(a^{2}h\right){\rho }_{w}g$ upwards

Hence under equilibrium of the cube,

$mg+4Sa={\rho }_w{a}^{2}gh$

$⟹h=\frac{mg+4Sa}{{\rho }_w{a}^{2}g}$

Hence x=4.