Animated Solution for Physics - Properties of Solids and Liquids: A hollow spherical shell at outer radius $R$ floats just submerged under the water surface. The inner radius of the shell is $r$. If the specific gravity of the shell material is $\frac{27}{8}$ with respect to water, value of $r$ is

The beauty of Archimedes' principle lies in its elegant simplicity: a floating object displaces its own weight in fluid. But what happens when the object isn't a solid block, but a hollow shell? Let's dive into this fascinating problem and unravel the physics step by step.

Imagine a hollow spherical shell floating in water. The problem states it is just submerged, which is a crucial piece of information. This means the water level is exactly tangent to the top of the sphere.

Because it is completely under the water line, the volume of water it displaces is equal to its entire outer volume. Let the outer radius be $R$ and the inner radius be $r$. The actual solid material of the shell only exists in the region between $r$ and $R$. Therefore, the volume of the shell's material is:

For the shell to float in equilibrium, the total downward gravitational force (its weight) must be perfectly balanced by the upward buoyant force exerted by the water.

The weight of the shell depends on the volume of its solid material and its density ${\rho }_o$:

The buoyant force, however, depends on the total volume of water displaced, which corresponds to the full outer volume of the shell, and the density of water ${\rho }_w$:

Equating the two forces, we get a beautiful symmetry where the $\frac{4}{3}\pi$ and $g$ terms cancel out immediately:

Now, let's rearrange this to isolate the ratio of the radii. Dividing both sides by $R^3{\rho }_o$, we obtain:

The problem gives us the specific gravity of the shell material, which is defined as the ratio of its density to the density of water: $\frac{{\rho }_o}{{\rho }_w}=\frac{27}{8}$. Notice that our equation requires the inverse of this ratio!

Let's solve for the ratio $\frac{r}{R}$:

Taking the cube root of both sides:

Here is where a bit of mathematical intuition comes in handy. We need to estimate the cube root of $19$ without a calculator. We know that $2^3=8$ and $3^3=27$. The number $19$ is closer to $27$. If we test a few decimals, $2.6^3\approx 17.5$ and $2.7^3\approx 19.68$. So, $(19{)}^{1/3}$ is approximately $2.66$.

Looking at our multiple-choice options, we can evaluate them to find the closest match. Option (a) is $\frac{8}{9}R$. Let's calculate $\frac{8}{9}$: it is exactly $\mathrm{0.888...}$, which perfectly matches our approximation!

Thus, the inner radius $r$ is $\frac{8}{9}R$.