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 $$=$$ $$0.07$$ $$Nm-1$$.

Question

Infinity Learn · Accepted Answer

If surface tension is neglected the condition for floating gives

$\begin{array}{l} 800 \times 10^{- 3} g = (0 {.1}^{2} xρ) g or x = 0 .08 m \\ (ρ = 1000 {kgm}^{- 3} for water) \end{array}$

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.

$∴ 800 \times 10^{- 3} g + 4 \times 0 .1 \times 0 .07 = (0 {.1}^{2} x^{'} ρ) g$

$or x^{'} = 0 .08 + \frac{0 .028}{100} = 0 .08 + 2 .8 \times 10^{- 4}$

Therefore, the additional distance = x'-x $= 2.8 \times 10^{- 4} m = 0.28 mm$

Surface tension and surface energy

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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.

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Surface tension and surface energy

Remember concepts with our Masterclasses.

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.

Ready to Test Your Skills?

free study material

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.