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 (in mm) is it buoyed up or down by surface tension? Surface of water = 0.07 Nm-1.

If surface tension is neglected the condition for floating gives800×10−3g=0.12xρg or x=0.08mρ=1000 kgm−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.∴ 800×10−3g+4×0.1×0.07=0.12x′ρg or  x′=0.08+0.028100=0.08+2.8×10−4Therefore, the additional distance = x'-x =2.8×10−4 m=0.28 mm