Animated Solution for Physics - Properties of Solids and Liquids: Two mercury drops (each of radius $r$) merge to form a bigger drop. The surface energy of the bigger drop, if $T$ is the surface tension, is

The phenomenon of liquid drops merging is not just a beautiful visual; it is a profound demonstration of energy minimization in physics. Let's dive into the mechanics of what happens when two identical mercury drops decide to become one.

Imagine two identical tiny drops of liquid mercury, each with a radius $r$, floating in space. When they come into contact, they rapidly merge to form a single, larger drop. Let's denote the radius of this new, bigger drop as $R$.

While the shape and surface area change dramatically during this process, one fundamental property remains absolutely constant: the total amount of matter. Because mercury is an incompressible liquid, the total volume before merging must exactly equal the total volume after merging.

Since the volume of a sphere is given by $\frac{4}{3}\pi r^3$, we can set up our master equation:

Notice how the geometric constants, $\frac{4}{3}\pi$, beautifully cancel out from both sides of the equation. This leaves us with a simple, elegant relationship between the radii:

Taking the cube root of both sides, we find the exact radius of our new giant drop:

This is a crucial intermediate result. The new radius is not simply $2r$; it scales by the cube root of $2$.

Now, let's shift our focus to the core of the question: the surface energy. The surface energy ($E$) of any liquid drop is the energy required to maintain its surface against the inward pull of surface tension ($T$). It is simply the product of the surface tension and the total surface area ($A$).

For our newly formed big drop, which is a perfect sphere, the surface area is $4\pi R^2$. Therefore, the surface energy is:

Now, we bring back our crucial intermediate result. We substitute $R=2^{1/3}r$ into our surface energy equation:

This is where we must be careful with our exponents. When we square the term $(2^{1/3}r)$, we must square both the numeric factor and the variable:

Substituting these back into our equation gives:

To match the options provided in the question, we need to combine the numeric constants. Remember that $4$ can be written as $2^2$.

When multiplying terms with the same base, we simply add their exponents. So, we add $2$ (which is $\frac{6}{3}$) and $\frac{2}{3}$:

This brings us to our final, elegant expression for the surface energy of the merged drop:

Why do drops merge in the first place? If you calculate the initial surface energy of the two separate drops, you will find it is greater than the final surface energy of the single merged drop.

Nature always seeks the lowest possible energy state. By merging, the liquid reduces its total surface area, thereby releasing surface energy. This released energy is what drives the spontaneous merging process, often causing the newly formed drop to briefly vibrate or heat up!