Question

A cube of mass m =800 g floats on the surface of water. Water wets it completely. The cube is 10 cm on each edge. By what additional distance is it buoyed up or down by surface tension? Surface tension of water = 0.07 N $m^{-1}$.

Answer 1

If surface tension is neglected the condition for floating
gives $800\times {10}^{-3}g=\left({0.1}^{2}x\rho \right)g$
or x = 0.08 m ( p = 1000 kg $m^{-3}$ for water)
Since water wets the cube, the angle of contact is zero and force of surface tension acts vertically downwards. So it is buoyed down by surface tension.
$\therefore 800\times {10}^{-3}+4\times 0.1\times 0.07=\left({0.1}^2{x}^{\prime }p\right)g$
or $x^{\prime }=0.08+\frac{0.028}{98}=0.08+2.8\times {10}^{-4}$
Therefore, the additional distance = $2.8\times {10}^{-4}m$