Terminal velocity of sphere falling through viscous liquid — derive

Question

Derive the expression for the terminal velocity of a small sphere of radius $r$ and density $\rho$ falling through a viscous liquid of density $\sigma$ and coefficient of viscosity $\eta$ .

Solution — Step by Step

As the sphere falls through the liquid, three forces act on it:

Weight (downward): $W = \frac{4}{3}\pi r^3 \rho g$
Buoyant force (upward): $F_B = \frac{4}{3}\pi r^3 \sigma g$
Viscous drag (upward, opposing motion): $F_v = 6\pi\eta r v$ (Stokes’ law)

Here $v$ is the instantaneous velocity of the sphere.

Initially, the sphere accelerates because $W > F_B + F_v$ . As velocity increases, the viscous drag $F_v$ increases (proportional to $v$ ).

Terminal velocity $v_T$ is reached when the net force on the sphere becomes zero — the upward forces exactly balance the downward weight. After this point, the sphere moves at constant velocity.

At terminal velocity, net force = 0:

W = F_B + F_v

\frac{4}{3}\pi r^3 \rho g = \frac{4}{3}\pi r^3 \sigma g + 6\pi\eta r v_T

\frac{4}{3}\pi r^3 (\rho - \sigma) g = 6\pi\eta r v_T

v_T = \frac{\frac{4}{3}\pi r^3 (\rho - \sigma) g}{6\pi\eta r}

v_T = \frac{4\pi r^3 (\rho - \sigma) g}{3 \times 6\pi\eta r}

\boxed{v_T = \frac{2r^2(\rho - \sigma)g}{9\eta}}

Why This Works

Stokes’ law ( $F = 6\pi\eta r v$ ) applies when:

The sphere is small and moves slowly (laminar/streamline flow)
The liquid is infinite in extent compared to the sphere
The sphere is rigid and smooth

The terminal velocity formula shows several important physical insights:

$v_T \propto r^2$ — larger spheres reach higher terminal velocity (rain drops fall faster than mist)
$v_T \propto (\rho - \sigma)$ — if the sphere’s density equals the liquid’s density, $v_T = 0$ (neutral buoyancy)
$v_T \propto \frac{1}{\eta}$ — more viscous liquids give lower terminal velocity (honey vs. water)

Common Mistake

A frequent error is using $\rho$ (sphere density) in the buoyancy term instead of $\sigma$ (liquid density). Buoyancy depends on the density of the fluid displaced, not the sphere. The net downward effective weight is $\frac{4}{3}\pi r^3 (\rho - \sigma)g$ , not $\frac{4}{3}\pi r^3 \rho g$ . If you forget the buoyancy term, your answer overestimates $v_T$ .

For JEE: This derivation appears as a 3-mark question. Note that $v_T \propto r^2$ — if the radius doubles, terminal velocity becomes 4 times. This relation between $v_T$ and $r$ is tested frequently in multiple-choice questions. Also remember the units: $[\eta] = \text{Pa·s} = \text{N·s/m}^2 = \text{kg/(m·s)}$ .

Question

Solution — Step by Step

Why This Works

Common Mistake

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