Animated Solution for Physics - Properties of Solids and Liquids: A drop of liquid of radius $R=10^{-2}\text{ m}$ having surface tension $S=\frac{0.1}{4\pi }{\text{ Nm}}^{-1}$ divides itself into $K$ identical drops. In this process the total change in the surface energy $\Delta U=10^{-3}\text{ J}$. If $K=10^{\alpha }$, then the value of $\alpha$ is

Imagine holding a tiny droplet of water in your hand. To us, it looks like a simple, peaceful sphere of liquid. But at the microscopic level, there is a silent, energetic tug-of-war happening on its surface!

Every single molecule inside the liquid is surrounded by its neighbors, happily pulled in all directions by cohesive forces. But the molecules on the very edge—the surface—have no neighbors above them. They feel a relentless, net inward pull. This molecular tension acts like an elastic stretched membrane, trying to minimize the surface area. This beautiful phenomenon is what we call Surface Tension.

In this problem from JEE Advanced 2017, we are asked to explore what happens when a single large liquid drop of radius $R$ is shattered into $K$ identical smaller droplets. Let's embark on this thrilling thermodynamic journey to find the value of $\alpha$!

The Law of Volume Conservation

Before we dive into the energy calculations, we must address a fundamental physical constraint: matter cannot be created or destroyed.

When our parent drop of radius $R$ splits into $K$ smaller droplets of radius $r$, the total mass of the liquid remains absolutely constant. Assuming the density of the liquid does not change, this means the total volume must be conserved.

Let's write down the volume of the initial large drop:

After the split, we have $K$ identical spherical droplets. The total volume of these $K$ droplets is:

Equating the two volumes due to conservation:

Notice how beautifully the factor of $\frac{4}{3}\pi$ cancels out from both sides! This leaves us with a clean geometric relationship:

Taking the cube root on both sides, we can express the radius of the smaller droplets, $r$, in terms of the parent radius $R$ and the number of drops $K$:

This is our first major milestone. Keep this relation close, as we will need it very soon!

The Energy Tug-of-War

Why does it take work to split a drop?

As we discussed, surface tension always tries to minimize surface area. When a single large drop splits into many smaller ones, the total surface area increases dramatically. Creating new surface area requires pulling molecules from the interior of the liquid to the surface, which requires doing work against the cohesive forces.

This work is stored as potential energy, known as Surface Energy ($U$). The surface energy of any spherical liquid drop is given by the product of its surface tension ($S$) and its surface area ($A$):

Let's calculate the change in surface energy, $\Delta U$, during the splitting process. The change is simply the final surface energy minus the initial surface energy:

We can factor out $4\pi$ to make the equation look much cleaner:

Now, let's bring back our volume conservation relation, $r=R\cdot K^{-1/3}$, and substitute it into our energy equation:

Simplifying the exponent of $K$:

Substituting this back, we get:

This is the master equation of our problem! It beautifully connects the change in surface energy to the initial radius, surface tension, and the number of droplets.

Cracking the Numbers

Now comes the exciting part—substituting the given values to find the unknown exponent $\alpha$.

We are given: - Initial radius, $R=10^{-2}\text{ m}$ - Surface tension, $S=\frac{0.1}{4\pi }{\text{ Nm}}^{-1}$ - Change in surface energy, $\Delta U=10^{-3}\text{ J}$

Let's plug these values into our master equation:

Look at that! The $4\pi$ in the numerator and denominator cancel out perfectly. This is why physics is so elegant—complex constants often vanish when we set up the equations correctly.

Let's simplify the remaining terms:

To isolate the term with $K$, we divide both sides by $10^{-5}$:

Since $101$ is extremely close to $100$, we can make a standard approximation for integer-type questions:

Now, to find $K$, we cube both sides of the equation:

We are given that $K=10^{\alpha }$. Comparing the exponents:

And there we have it! The value of $\alpha$ is 6.

What makes this problem a classic is its connection to thermodynamics. If this process were reversed—meaning $K$ tiny droplets coalesced to form a single large drop—the surface area would decrease.

This decrease in surface area would release the excess surface energy as heat! If the process is adiabatic, this heat would raise the temperature of the liquid. Can you calculate this temperature rise? Try using the formula $\Delta U=mc\Delta T$, where $m$ is the mass of the drop and $c$ is its specific heat capacity. This is a fantastic way to deepen your understanding of fluid mechanics and thermodynamics!