Q.A metal block of area $0.10\ \text{m}^{2}$ is connected to a $0.010\ \text{kg}$ mass via a string that passes over an ideal pulley (considered massless and frictionless). A liquid with a film thickness of $0.30\ \text{mm}$ is placed between the block and the table. When released the block moves to the right with a constant speed of $0.085\ \text{m s}^{-1}$. Find the coefficient of viscosity of the liquid.

Step 1 — Why the ball is at terminal velocity.

A sphere falling through a viscous liquid quickly reaches a steady speed at which the three forces balance:

where $\rho$ is the ball's density, $\sigma$ the liquid's, and $6\pi\eta r v$ is Stokes' drag. Solving for $v$ gives the standard terminal-velocity result:

Notice that only the difference of densities $(\rho - \sigma)$ appears — which is exactly what the question gives us ($7000\ \mathrm{kg\,m^{-3}}$). Since the ball falls with uniform velocity over the measured distance, using $v = h/t$ is legitimate.

Step 2 — Find the mean time from the three trials.

Step 3 — Terminal velocity.

Step 4 — Square the radius.

(The awkward $\sqrt{3}$ in the data is there precisely so that $r^2$ comes out clean.)

Step 5 — Substitute into the formula for $\eta$. …

Concept & Intuition

Terminal velocity in a viscous fluid (like air) is reached when the downward gravitational force is balanced by the upward viscous drag (Stokes’ law) and buoyancy. For small spherical droplets, Stokes’ law gives drag $\propto r v$, while weight minus buoyancy $\propto r^3$. Equating them shows $v \propto r^2$. When droplets coalesce, volume is conserved, so the new radius is larger by the cube root of the number of drops. That squared gives the factor by which terminal velocity increases.

Step-by-step reasoning

1. Volume conservation upon coalescence Let each original drop have radius $r$. The volume of one drop is $\frac{4}{3}\pi r^3$. 64 drops have total volume $64 \times \frac{4}{3}\pi r^3$. After coalescence, the single big drop has radius $R$ and volume $\frac{4}{3}\pi R^3$. Equating:

2. Terminal velocity dependence on radius For a sphere moving slowly through a viscous fluid, Stokes’ law gives drag force $F_d = 6\pi \eta r v$. The net downward force (weight minus buoyancy) is $\frac{4}{3}\pi r^3 (\rho - \sigma)g$, where $\rho$ is droplet density and $\sigma$ is air density. At terminal velocity $v_t$:

Solving for $v_t$:

Hence $v_t \propto r^2$.

3. Apply the scaling …

Step 1 — The two physical situations.

• Field ON: the drop is balanced (stationary), so the electric force exactly supports the weight:

• Field OFF: the drop falls at terminal velocity, so the viscous drag exactly supports the weight (Stokes' law):

The weight $mg$ is the common bridge between the two — that is the key to the whole problem.

Step 2 — Find the radius from the terminal-velocity condition.

Write the mass in terms of the drop's density and radius, $m = \rho \cdot \tfrac{4}{3}\pi r^3$, and substitute into (2):

Substituting $\eta = 1.8\times10^{-5}$, $v = 2\times10^{-3}$, $\rho = 900$, $g = 9.8$:

Step 3 — Get the weight from Stokes' law, equation (2).

Step 4 — Evaluate the field.

$E = \frac{81}{7}\pi \times 10^{5} = (11.571)(3.1416)\times10^{5} = 3.636\times10^{6}\ \mathrm{V/m}$ …

Terminal velocity in a viscous medium is

For identical glass balls in air, only the radius differs, so …