Animated Solution for Physics - Properties of Solids and Liquids: Consider a solid sphere of radius $R$ and mass density $\rho (r)={\rho }_0\left(1-\frac{r^2}{R^2}\right)$, $0<r\le R$. The min. density of a liquid in which it will float is

Imagine you are handed a solid sphere, but it's not an ordinary one. Its density isn't uniform; it's denser at the core and gets lighter as you move towards the surface. The question asks us to find the minimum density of a liquid in which this peculiar sphere will float.

This is a classic test of Archimedes' Principle, but with a calculus twist!

Why does the question specifically ask for the minimum density?

Think about how buoyancy works. The buoyant force pushing up on the sphere is equal to the weight of the liquid displaced. If the liquid is very dense, like mercury, the sphere might only need to displace a tiny volume to balance its weight, floating high above the surface.

But as we decrease the liquid's density, the sphere must sink deeper to displace more volume and generate enough buoyant force. The absolute limit—the minimum possible liquid density—occurs when the sphere is fully submerged but just barely touching the surface. At this point, it is displacing its maximum possible volume.

At this critical floating condition, the forces are perfectly balanced:

The buoyant force $F_B$ is straightforward. Since the sphere is fully submerged, the displaced volume is the total volume of the sphere.

The weight of the sphere, however, requires a bit more finesse. Because the density $\rho (r)$ changes with the radius $r$, we can't just multiply a single density value by the total volume. We must integrate!

We imagine the sphere as a collection of infinitely many thin, concentric spherical shells, like the layers of an onion. Let's take one such shell at a distance $r$ from the center, with a tiny thickness $dr$.

The volume of this shell is its surface area multiplied by its thickness: $4\pi r^{2}dr$. The mass of this shell $dm$ is its volume multiplied by the density at that specific radius:

To find the total mass, we integrate this expression from the center ($r=0$) to the surface ($r=R$):

Now, let's roll up our sleeves and solve the integral. We can pull the constants ${\rho }_0$ and $4\pi$ outside the integral:

Integrating term by term using the power rule:

Applying the upper limit $R$ (the lower limit $0$ just gives $0$):

Finding a common denominator to subtract the fractions:

We are almost there! Let's clean up the equation by canceling the common terms $\pi$ and $R^3$ from both sides:

Solving for our target, the liquid density ${\rho }_l$:

And there we have it! The minimum density of the liquid must be $\frac{2}{5}{\rho }_0$. This beautiful result shows how calculus and physical principles intertwine to solve complex, non-uniform problems.