Mercury Splash

First, recognize that the surface tension $σ=0.5\frac{N}{m}=0.5\frac{J}{m^{2}}$ , which better represents the provided definition of surface tension. When the original droplet splits into $900$ different droplets, it is divided $900$ times by volume. Thus, we can find the total surface area of the $900$ droplets, and subtract it by the surface area of the original droplet to get the increase in surface area. The relation so far can be written as so: $σ=\frac{E}{SA_{900}-SA_{1mm}}$ , where $E$ is the energy needed to create the surface area change. Since we are asked for the height, we can input it directly into this equation by rewriting $E$ as gravitational potential energy, $mgh$ : $σ=\frac{mgh}{SA_{900}-SA_{1mm}}$ .

First, we establish what we need:

1. Surface area of the original droplet, $SA_{1mm}$ . This can be found using the formula for the surface area of a sphere, $SA=4\pi r^{2}$ .

2. Combined surface area of the 900 droplets. Equating the volume of the original droplet and that of the 900 droplets combined yields the radius of each of the 900 droplets, and thus their surface area.

3. Mass of the original droplet. This can be easily calculated since we can find its volume, and its density is given. We use $m=ρV$ .

Now to make calculations:

1. Using $R=1 mm$ , $SA_{1mm}=4\pi (0.001 m)^{2}=1.257\times 10^{-5} m^{2}$ .

2. We start by equating two volumes: $\frac{4}{3}\pi (0.001 m)^{3}=900\times \frac{4}{3}\pi r_{900}^{3}$ . Solving gives $r_{900}=1.036\times 10^{-4} m$ . The combined surface area of all 900 droplets is $SA_{900}=900\times 4\pi r_{900}^{2}=900\times 4\pi \times (1.036\times 10^{-4} m)^{2}=1.213\times 10^{-4} m^{2}$ . Now, we can calculate the difference in surface area, $SA_{900}-SA_{1mm}=1.213\times 10^{-4} m^{2}-1.257\times 10^{-5} m^{2}=1.088\times 10^{-4} m^{2}$ .

3. Plugging in the density and volume give the mass $m=ρV=(13.5\frac{g}{cm^{3}}\times \frac{1kg}{1000g}\times \frac{1m^{3}}{(100cm)^{3}})\times \frac{4/3}\pi(0.001 m)^{3}=5.655\times 10^{-5} kg$ .

Now, we can finally plug in the values back into our original equation. Solving for $h$ , we get $h=\frac{σ(SA_{900}-SA_{1mm})}{mg}=\frac{(0.5\frac{J}{m^{2}})(1.088\times 10^{-4} m^{2})}{(5.655\times 10^{-5}kg)(9.8\frac{m}{s^{2}})}=0.0981m=\fbox{9.81 cm}$ .

The original volume of the droplet is $V_0 = \dfrac {4}{3} \pi \cdot \left( 10^{-3} \, \text{m} \right)^3.$ The surface area of this droplet is $S_0 = 4 \pi \cdot \left( 10^{-3} \, \text{m} \right)^2 = 1.2566 \times 10^{-5} \, \text{m}^2.$ The volume of each of the mini-droplets is $\dfrac {V_0}{900}$ , so, if the radius of each of these droplets is $r$ , we have $\dfrac {4}{3} \pi r^3 = \dfrac {V_0}{900} \implies \dfrac {4}{3} \pi r^3 = \dfrac {1}{900} \cdot \dfrac {4}{3} \pi \cdot \left( 10^{-3} \, \text{m} \right)^3.$ We solve for $r$ : $r = \dfrac {10^{-3} \, \text{m}}{\sqrt[3]{900}} = 1.0357 \times 10^{-4} \, \text{m}.$ Clearly, the new total surface area of all the mini-droplets is $S = 900 \cdot 4 \pi r^2 = 3600 \pi \cdot r^2 = 1.2133 \times 10^{-4} \, \text{m}^2,$ while the original surface area was $1.2566 \times 10^{-5} \, \text{m}^2$ . This is a change in surface area of $1.0876 \times 10^{-4} \, \text{m}^2$ . We are given the surface tension $\sigma$ , so we can find the energy required: $\left( 0.5 \, \dfrac {\text {J}}{\text {m}^2} \right) \cdot \left( 1.0876 \times 10^{-4} \, \text{m}^2 \right) = 5.4382 \times 10^{-5} \, \text{J},$ which we will call $E$ . But we know that this was gained in the transfer of potential energy into kinetic energy. If $h$ is the height from which the droplet was dropped and $m$ is the mass of the droplet, we have $E = mgh = \rho V_0 g h \implies h = \dfrac {E}{g\rho V_0}.$ But, from our very first step, we have $V_0 = \dfrac {4}{3} \pi \cdot \left( 10^{-3} \, \text {m} \right)^3 = 4.1888 \times 10^{-9} \, \text{m}^3.$ Substituting $E = 5.4382 \times 10^{-5} \, \text{J}$ , $\rho = 1.35 \times 10^{4} \, \text{kg/m}^3$ , $g = 9.81 \, \text{m/s}^2$ , and $V_0 = 4.1888 \times 10^{-9} \, \text{m}^3$ , we get $h = 9.81 \times 10^{-2} \, \text{m} = \boxed {9.81} \, \text{cm},$ which is our answer.

$\blacksquare$

In this solution, I assume each mercury droplet is being considered as a perfect sphere.

Note that the volume of the mercury droplet before the fall is $V= \frac{4}{3} \pi R^3$ , and the surface area is $A_1= 4 \pi R^2$ . The initial mass of the droplet will then be $m= V \rho= \frac{4}{3} \pi R^3 \rho$ . Before the fall, the only contributions to the total energy of the system are the gravitational potential energy and the energy due to surface tension in the droplet. Thus, the initial energy of the system is: $E_{initial}= mgh_{min} + \sigma A_1= \frac{4}{3} \pi R^3 \rho h_{min} + \sigma 4 \pi R^2$ After the fall, the droplet breaks into $n$ droplets, each with the same volume. Since the radius of a sphere is directly proportional to the cube root of its volume, after the fall, the radius of each droplet will be: $R_{small}= \frac{R}{\sqrt[3]{n}}$ The total surface area of the droplets after the fall will then be: $A_{2}= n \times 4\pi R_{small}^2 = 4 n \pi \frac{R^2}{n^{\frac{2}{3}}}= 4 \pi n^{\frac{1}{3}} R^2$ After the fall, there is gravitational potential energy stored in the droplets. The only contribution to the total energy of the system is the surface tension of the mercury droplets. Hence, the total energy in the system after the fall is: $E_{final}= \sigma A_2= 4 \pi n^{\frac{1}{3}} R^2 \sigma$ Since the process is isothermic, there is no exchange of heat with the outer surroundings. Hence the total energy of the system must be constant. Equating, we obtain: $E_{initial}= E_{final}$ $\implies \frac{4}{3} \pi R^3 \rho h_{min} + \sigma 4 \pi R^2= 4 \pi n^{\frac{1}{3}} R^2 \sigma$ $\implies h_{min}= 3 \sigma \frac{n^{\frac{1}{3}}-1}{R d g}$ Plugging in the values, we obtain $\boxed{h_{min}= 9.8 \ cm }$ .

mgh = σA, where A = 4(3.14)(900 * ((1/900)^(1/3)* 10^-3)^2 - (10^-3)^2) = 1.088 * 10^-4 . So h = 1.088 * 10^-4 / (13500 * 4/3 * 3.14 (10^-3)^3 9.8) = 0.09818 m = 9.82 cm