Surface tension and surface energy

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Question

A cube of mass m and side length ‘a’ wettable by water (contact angle ), floats on the surface of water. Distance between the lower face of cube and the surface of water is $\frac{a}{N}$ . Value of N is (Take $m = \frac{ρ a^{3}}{3}$ ; surface tension of water = $\frac{ρ a^{2} g}{24}$ where $ρ$ is density of water)

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Solution

$\begin{array}{l} ∴ a^{2} x ρ g = mg + 4 a σ \\ \Rightarrow x = \frac{mg + 4 a σ}{a^{2} ρ g} \\ \Rightarrow x = \frac{\frac{ρ a^{3} g}{3} + \frac{4 a \times ρ a^{2} g}{24}}{a^{2} ρ g} \\ \Rightarrow x = \frac{a}{3} + \frac{a}{6} = \frac{a}{2} \end{array}$

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Surface tension and surface energy

Remember concepts with our Masterclasses.

A cube of mass m and side length ‘a’ wettable by water (contact angle ), floats on the surface of water. Distance between the lower face of cube and the surface of water is aN. Value of N is (Take m=ρa33; surface tension of water =ρa2g24 where ρ is density of water)

∴a2xρg=mg+4aσ⇒x=mg+4aσa2ρg⇒x=ρa3g3+4a×ρa2g24a2ρg⇒x=a3+a6=a2

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$\begin{array}{l} ∴ a^{2} x ρ g = mg + 4 a σ \\ \Rightarrow x = \frac{mg + 4 a σ}{a^{2} ρ g} \\ \Rightarrow x = \frac{\frac{ρ a^{3} g}{3} + \frac{4 a \times ρ a^{2} g}{24}}{a^{2} ρ g} \\ \Rightarrow x = \frac{a}{3} + \frac{a}{6} = \frac{a}{2} \end{array}$