Animated Solution for Physics - Properties of Solids and Liquids: Two small drops of mercury each of radius $R$ coalesce to form a single large drop. The ratio of total surface energy before and after the change is

Imagine you are in a laboratory, observing two tiny, perfectly spherical drops of liquid mercury resting on a smooth surface. Let's say each of these small drops has a radius of $R$. Now, what happens when they touch? They instantly merge, or coalesce, to form one single, larger drop. Let's call the radius of this new big drop $R^{\prime }$.

The most fundamental principle governing this phenomenon is that the total amount of mercury hasn't changed. Matter is strictly conserved, which means the total volume before they merge must be exactly equal to the total volume after they merge.

Let's translate this physical reality into a mathematical equation. We know from basic geometry that the volume of a sphere is $\frac{4}{3}\pi R^3$. Since we start with two identical small drops, their combined initial volume is simply $2\times \frac{4}{3}\pi R^3$.

On the other side of our equation, this must equal the volume of our newly formed large drop, which is $\frac{4}{3}\pi R^{\prime 3}$. This equation is our master key to finding the new radius:

Now, let's simplify our master equation. Look at both sides—the constant factor $\frac{4}{3}\pi$ is present on both the left and the right. We can elegantly cancel it out. This leaves us with a much simpler relation:

To isolate $R^{\prime }$, we need to take the cube root of both sides. The cube root of $R^3$ is just $R$, and the cube root of $2$ is $2^{1/3}$. So, we find that the new radius $R^{\prime }$ is exactly $2^{1/3}$ times the original radius $R$:

With the geometry sorted, let's shift our focus to the physics of surface energy. Every liquid surface acts like a stretched elastic membrane, possessing potential energy. This surface energy, denoted by $U$, is directly proportional to the surface area of the liquid. The constant of proportionality is the surface tension, $T$. So, the formula is simply:

Since we are dealing with mercury throughout the process, the surface tension $T$ remains a constant value. Let's calculate the total surface energy we started with, which we'll call $U_i$. Initially, we had two separate small drops. The surface area of a single sphere is $4\pi R^2$. Since there are two of them, the total initial surface area is $2\times 4\pi R^2$. To find the initial surface energy, we just multiply this total area by the surface tension $T$:

Now, what about the final state? After the drops coalesce, we are left with just one single, larger drop of radius $R^{\prime }$. The surface area of this new big drop is $4\pi R^{\prime 2}$. Therefore, the final surface energy, $U_f$, is simply the surface tension $T$ multiplied by this new area:

Notice how the total surface area has actually decreased, which is why small drops naturally want to merge to minimize their energy!

The question asks for the ratio of the initial surface energy to the final surface energy. Let's set up this ratio by dividing $U_i$ by $U_f$:

Look at all those common terms! The surface tension $T$ and the $4\pi$ factor cancel out beautifully from top and bottom. We are left with a very clean expression:

We are almost there. We have our ratio in terms of $R$ and $R^{\prime }$, but we need a numerical answer. Remember that crucial relationship we derived earlier? We found that $R^{\prime }=2^{1/3}R$. Let's substitute this into our denominator. We need to square the entire term, so $(2^{1/3}{)}^2$ becomes $2^{2/3}$, and $R^2$ is just $R^2$. Our ratio is now:

For the final step, let's simplify this expression. The $R^2$ terms in the numerator and denominator cancel each other out completely. We are left with $2$ divided by $2^{2/3}$. Using the laws of exponents, when we divide terms with the same base, we subtract their powers. So, this is $2^{1-2/3}$.

One minus two-thirds is exactly one-third. Therefore, our final ratio is $2^{1/3}$. Written as a ratio, it is $2^{1/3}:1$. This matches option (a) perfectly!