Animated Solution for Physics - Properties of Solids and Liquids: If the terminal speed of a sphere of gold (density = $19.5{\text{ kgm}}^{-3}$) is $0.2{\text{ ms}}^{-1}$ in a viscous liquid (density = $1.5{\text{ kgm}}^{-3}$), find the terminal speed of a sphere of silver (density = $10.5{\text{ kg/m}}^3$) of the same size in the same liquid.

Imagine you are standing in a laboratory, holding two identical spheres in your hands. One is made of gleaming gold, and the other of shiny silver. You drop them simultaneously into two tall glass cylinders filled with the exact same viscous liquid. As they fall, they accelerate initially, but soon, they reach a constant speed. This constant speed is what physicists call terminal velocity.

In this problem, we are given the terminal velocity of the gold sphere and asked to find the terminal velocity of the silver sphere. To do this, we need to understand the delicate dance of forces acting on these spheres as they plunge through the liquid.

When a sphere falls through a viscous fluid, it experiences three primary forces. First, there is gravity pulling it downwards, which depends on the sphere's mass (and thus its density, $\rho$). Second, there is the buoyant force pushing it upwards, which depends on the density of the displaced liquid, $\sigma$. Finally, there is the viscous drag force, which opposes the motion and increases as the sphere falls faster.

Terminal velocity is achieved when these three forces perfectly balance each other out. The net force becomes zero, and the sphere stops accelerating. The mathematical expression for the terminal velocity $v_T$ of a sphere of radius $r$ falling through a fluid of viscosity $\eta$ is given by Stokes' Law:

Let's look closely at this equation. For our two spheres, they are of the same size, meaning their radius $r$ is identical. They are falling through the same liquid, meaning the liquid density $\sigma$ and the viscosity $\eta$ are identical. And, of course, the acceleration due to gravity $g$ is the same.

Since $r$, $g$, $\eta$, and $\sigma$ are all constants in this scenario, we can see that the terminal velocity is directly proportional to the difference between the density of the sphere and the density of the liquid:

This term $(\rho -\sigma )$ represents the effective density of the sphere in the fluid. It's the driving factor that determines how fast the sphere will ultimately fall.

Because of this direct proportionality, we can easily compare the two spheres by setting up a ratio. The ratio of the terminal velocity of the silver sphere to that of the gold sphere will be equal to the ratio of their effective densities:

Now, we simply plug in the values provided in the problem. The density of silver ${\rho }_{\text{Ag}}$ is $10.5{\text{ kgm}}^{-3}$, the density of gold ${\rho }_{\text{Au}}$ is $19.5{\text{ kgm}}^{-3}$, and the density of the liquid $\sigma$ is $1.5{\text{ kgm}}^{-3}$. The terminal velocity of the gold sphere $v_{T,\text{Au}}$ is given as $0.2{\text{ ms}}^{-1}$.

Let's simplify the numerator and the denominator. The effective density of silver is $10.5-1.5=9.0$. The effective density of gold is $19.5-1.5=18.0$.

This fraction simplifies beautifully to exactly $\frac{1}{2}$. This means the driving force for the silver sphere is exactly half of that for the gold sphere.

And there we have it! The silver sphere, being less dense than the gold sphere, experiences a relatively stronger buoyant force compared to its weight. Consequently, it falls at exactly half the terminal speed of the gold sphere. Physics is beautifully consistent!