Animated Solution for Physics - Properties of Solids and Liquids: A jar is filled with two non-mixing liquids 1 and 2 having densities ${\rho }_1$ and ${\rho }_2$, respectively. A solid ball, made of a material of density ${\rho }_3$, is dropped in the jar. It comes to equilibrium in the position shown in the figure. Which of the following is true for ${\rho }_1,{\rho }_2$ and ${\rho }_3$?

The problem of a ball floating between two immiscible liquids is a classic and elegant demonstration of Archimedes' principle and the concept of relative density. It might look like a complex system at first glance, but by breaking it down into simple, logical steps, the solution reveals itself beautifully.

Imagine you are looking at the jar. We have two liquids that refuse to mix: Liquid 1 is resting on top, and Liquid 2 has settled at the bottom. Right at the boundary between them, a solid ball is suspended in perfect equilibrium.

To solve this, we don't need complex equations; we just need to understand the fundamental rule of buoyancy: an object (or fluid) will float on another fluid if its density is lower, and it will sink if its density is higher.

First, let's decode the relationship between the two liquids. Why is Liquid 1 floating peacefully on top of Liquid 2?

In physics, a lighter fluid will always float above a denser fluid. Think of oil and water; oil floats because it is less dense. Since Liquid 2 has settled at the bottom of the jar, it must be the heavier of the two. This immediately gives us our first crucial piece of information: the density of Liquid 1 is strictly less than the density of Liquid 2.

Mathematically, we write this as:

Now, let's shift our focus to the solid ball and trace its journey. When dropped into the jar, it didn't float on the top surface of Liquid 1. Instead, it completely sank through it.

For an object to sink in a fluid, its density must be greater than the fluid's density. Because the ball sank through Liquid 1, we can confidently conclude that the density of the ball (${\rho }_3$) is greater than the density of Liquid 1 (${\rho }_1$).

or equivalently,

But the story doesn't end there. The ball didn't sink all the way to the bottom of the jar! It stopped and is now floating on Liquid 2.

Because it floats on Liquid 2 and doesn't sink into it, the ball must be lighter than Liquid 2. The upward buoyant force from Liquid 2 is strong enough to support the ball. Therefore, the density of the ball (${\rho }_3$) must be less than the density of Liquid 2 (${\rho }_2$).

Let's put all the pieces of the puzzle together. We have established three facts: 1. Liquid 1 is lighter than Liquid 2 (${\rho }_1<{\rho }_2$). 2. The ball is heavier than Liquid 1 (${\rho }_1<{\rho }_3$). 3. The ball is lighter than Liquid 2 (${\rho }_3<{\rho }_2$).

Combining these inequalities, we see that ${\rho }_1$ is the smallest, ${\rho }_2$ is the largest, and ${\rho }_3$ sits perfectly in the middle.

The correct order of densities is:

This simple visual puzzle beautifully demonstrates how relative densities govern the stratification of fluids and the buoyancy of objects within them.