Question

The terminal velocity v of a small spherical ball of radius r falling through a viscous liquid is directly proportional to:

• A. \( r \)
• B. \( r^2 \)
• C. \( \frac{1}{r} \)
• D. \( \frac{1}{r^2} \)

Hint:

In Stokes’ law problems: \[ \text{Driving force} \propto r^3,\quad \text{Drag force} \propto r \Rightarrow v_t \propto r^2 \]

Answer 1

The Correct Option is B

Step 1: Picture a ball falling in a thick liquid.
A small ball dropped into a viscous liquid speeds up at first, then settles to a steady speed. That steady speed is the terminal velocity. We want how it depends on the radius $r$.

Step 2: List the forces on the ball.
Three forces act: weight pulling down, buoyancy pushing up, and the viscous drag opposing motion. At terminal velocity these balance out.

Step 3: Write weight minus buoyancy.
Both weight and buoyancy depend on the ball's volume, which grows as $r^3$. Their difference is \[ \frac{4}{3}\pi r^3 (\rho - \sigma) g, \] where $\rho$ is ball density and $\sigma$ is liquid density.

Step 4: Write the drag force.
Stokes' law gives the drag as $6\pi \eta r v$, which grows only with the first power of $r$.

Step 5: Balance the forces.
Set the net downward force equal to drag: \[ \frac{4}{3}\pi r^3 (\rho - \sigma) g = 6\pi \eta r v. \] Cancel one factor of $r$ and solve for $v$: \[ v = \frac{2 r^2 (\rho - \sigma) g}{9 \eta}. \]

Step 6: State the answer.
All the other quantities are fixed, so the terminal velocity grows with the square of the radius. \[ \boxed{r^2} \]