Question

While cleaning his flat a physics research student knocks over a bucket containing 5.4 litre of water, which then forms itself into one continuous puddle. The contact angle between floor and water is $60^\circ$. we can assume that the area of water-air interface and water-floor interface to be almost same and it is $A$.

The area (exposed to air) of the puddle formed is A. Take $g = 10 \ m/s^2$, density of water = $10^3 \ kg/m^3$ and surface tension of water = $72.9 \times 10^{-3} \ N/m$. The value of $A$ is

Answer 1

2

Answer 2

The volume of water is $V = 5.4 \text{ litre} = 5.4 \times 10^{-3} \ m^3$. The density of water is $\rho = 10^3 \ kg/m^3$.

Let the area of the puddle be $A$. We assume the puddle forms a thin film of uniform thickness $h$. The volume of the puddle is $V = A \times h$, so $h = V/A$.

The potential energy of the water is the sum of gravitational potential energy and surface energy. The gravitational potential energy, taking the floor as the reference level, is $U_g = m g (h/2)$. Since the thickness is assumed uniform, the center of mass is at $h/2$. $U_g = \rho V g \left(\frac{V}{2A}\right) = \frac{\rho g V^2}{2A}$.

The surface energy arises from the interfaces. We consider the water-air interface (area $A$) and the water-floor interface (area $A$). When the puddle forms, these interfaces replace the floor-air interface over area $A$. The change in surface energy compared to the bare floor is $\Delta U_s = \gamma_{wa} A + \gamma_{wf} A - \gamma_{fa} A$, where $\gamma_{wa}$ is the surface tension of the water-air interface (given as $\gamma$), $\gamma_{wf}$ is the surface tension of the water-floor interface, and $\gamma_{fa}$ is the surface tension of the floor-air interface. According to Young's equation, at the contact line, the surface tensions are related by $\gamma_{fa} = \gamma_{wf} + \gamma \cos \theta$, where $\theta$ is the contact angle. So, $\gamma_{wf} - \gamma_{fa} = -\gamma \cos \theta$. The change in surface energy is $\Delta U_s = \gamma A + (\gamma_{wf} - \gamma_{fa}) A = \gamma A - \gamma A \cos \theta = \gamma A (1 - \cos \theta)$.

The total potential energy of the puddle is $U = U_g + \Delta U_s = \frac{\rho g V^2}{2A} + \gamma A (1 - \cos \theta)$. The puddle spreads to minimize this total energy. We find the minimum by differentiating $U$ with respect to $A$ and setting the derivative to zero. $\frac{dU}{dA} = \frac{d}{dA}\left(\frac{\rho g V^2}{2A}\right) + \frac{d}{dA}\left(\gamma A (1 - \cos \theta)\right) = 0$. $-\frac{\rho g V^2}{2A^2} + \gamma (1 - \cos \theta) = 0$.

Solving for $A$: $\frac{\rho g V^2}{2A^2} = \gamma (1 - \cos \theta)$. $A^2 = \frac{\rho g V^2}{2 \gamma (1 - \cos \theta)}$. $A = \sqrt{\frac{\rho g V^2}{2 \gamma (1 - \cos \theta)}} = V \sqrt{\frac{\rho g}{2 \gamma (1 - \cos \theta)}}$.

Now, substitute the given values: $V = 5.4 \times 10^{-3} \ m^3$. $\rho = 10^3 \ kg/m^3$. $g = 10 \ m/s^2$. $\gamma = 72.9 \times 10^{-3} \ N/m$. $\theta = 60^\circ$, so $\cos \theta = \cos 60^\circ = 0.5$. $1 - \cos \theta = 1 - 0.5 = 0.5$.

$A = (5.4 \times 10^{-3}) \sqrt{\frac{10^3 \times 10}{2 \times (72.9 \times 10^{-3}) \times 0.5}}$. $A = (5.4 \times 10^{-3}) \sqrt{\frac{10^4}{72.9 \times 10^{-3}}}$. $A = (5.4 \times 10^{-3}) \sqrt{\frac{10^4 \times 10^3}{72.9}}$. $A = (5.4 \times 10^{-3}) \sqrt{\frac{10^7}{72.9}}$. To simplify $\sqrt{\frac{10^7}{72.9}}$, we can write $\frac{10^7}{72.9} = \frac{10^7}{729/10} = \frac{10^8}{729}$. $729 = 27^2$. So, $\sqrt{\frac{10^8}{729}} = \frac{\sqrt{10^8}}{\sqrt{729}} = \frac{10^4}{27}$.

$A = (5.4 \times 10^{-3}) \times \frac{10^4}{27}$. $A = \frac{5.4 \times 10^{-3} \times 10^4}{27} = \frac{5.4 \times 10}{27} = \frac{54}{27} = 2$.

The area of the puddle is $2 \ m^2$.