Animated Solution for Physics - Properties of Solids and Liquids: Comprehension Passage

Imagine a problem that seamlessly bridges the gap between the macroscopic flow of fluids and the invisible world of electric fields. This JEE Advanced masterpiece does exactly that. We are tasked with tracking a dielectric liquid as it flows between two chambers and then calculating how this fluid movement alters the electrical capacitance of the entire system. It’s a beautiful symphony of mechanics and electromagnetism. Let’s break it down step by step.

We start with a container divided into two equal chambers. The left chamber is initially brimming with a dielectric liquid, while the right chamber is empty. When the small hole at the bottom is opened, the liquid begins to flow.

Because the liquid is incompressible, the total volume remains constant. Since both chambers have an identical base area of $1{\text{ m}}^2$, the sum of the liquid heights in both chambers must always equal the total initial height, which is $2\text{ m}$. If we let the height of the liquid in the right chamber be $h$, the height in the left chamber is simply $2-h$.

To find out how fast the liquid is flowing through the hole, we invoke Bernoulli's Theorem. We compare the energy at the top surface of the left chamber with the energy just outside the hole in the right chamber. Both surfaces are exposed to atmospheric pressure, so $p_0$ cancels out. The potential energy difference is what drives the kinetic energy of the efflux.

The velocity of efflux $v$ is given by Torricelli's law:

Substituting our height expressions, we get:

Tracking Height Over Time

Now, we need to link this velocity to the changing height in the right chamber. We use the Equation of Continuity, which states that the rate of volume increase in the right chamber must equal the volume flow rate through the hole.

Here, $A$ is the cross-sectional area of the chamber ($1{\text{ m}}^2$) and $a$ is the area of the hole ($\sqrt{10}\times 10^{-4}{\text{ m}}^2$). Substituting our expression for $v$, we get a differential equation:

To solve this, we separate the variables, bringing all $h$ terms to one side and $t$ to the other, and integrate:

The integration is straightforward. The left side integrates to $-2(\sqrt{1-h}-1)$. After substituting the given values ($a=\sqrt{10}\times 10^{-4}$, $g=10$, $t=500$), the right side simplifies beautifully to $1$.

Solving for $h$, we find that at $t=500\text{ s}$, the height in the right chamber is $h=0.75\text{ m}$. Consequently, the height in the left chamber is $2-0.75=1.25\text{ m}$.

Capacitance of a Dynamic System

With the fluid mechanics sorted, we shift our focus to the electrical aspect. The container is sandwiched between two large metal plates, forming a capacitor.

At $t=0$, the left chamber is completely filled with the dielectric liquid (${ϵ}_r=15$), and the right chamber is filled with air (${ϵ}_r=1$). These two chambers act as two capacitors connected in parallel.

At $t=500\text{ s}$, the situation is more complex. Each chamber now contains a layer of liquid and a layer of air. Because these layers are stacked vertically between the plates, they act as capacitors in series.

For the left chamber, we have $0.75\text{ m}$ of air and $1.25\text{ m}$ of liquid:

For the right chamber, we have $1.25\text{ m}$ of air and $0.75\text{ m}$ of liquid:

The total final capacitance is the sum of these two parallel branches:

The problem asks for the difference in capacitance, $C_i-C_f$, in the form $(8-n){ϵ}_0$.

By direct comparison, we find $n=\frac{128}{65}\approx 1.97$.

This problem is a fantastic reminder of how different branches of physics are deeply interconnected. By mastering the fundamental principles of both fluid dynamics and electrostatics, you can unravel even the most intimidating setups!