Question:
The terminal velocity v of a small spherical ball of radius r falling through a viscous liquid is directly proportional to:
The terminal velocity v of a small spherical ball of radius r falling through a viscous liquid is directly proportional to:
Show Hint
In Stokes’ law problems:
\[
\text{Driving force} \propto r^3,\quad \text{Drag force} \propto r
\Rightarrow v_t \propto r^2
\]
Updated On: Jun 10, 2026
- \( r \)
- \( r^2 \)
- \( \frac{1}{r} \)
- \( \frac{1}{r^2} \)
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The Correct Option is B
Solution and Explanation
Concept:
When a small spherical body moves through a viscous fluid, it experiences a resistive force given by Stokes' law:
\[
F_v = 6 \pi \eta r v
\]
where \( \eta \) is viscosity, \( r \) is radius, and \( v \) is velocity.
At terminal velocity, net force becomes zero:
\[
\text{Weight} - \text{Buoyant force} = \text{Viscous force}
\]
Step 1: Write forces Weight of sphere: \[ W = \frac{4}{3}\pi r^3 \rho g \] Buoyant force: \[ B = \frac{4}{3}\pi r^3 \sigma g \] Net downward force: \[ F = \frac{4}{3}\pi r^3 (\rho - \sigma) g \]
Step 2: Balance at terminal velocity \[ \frac{4}{3}\pi r^3 (\rho - \sigma) g = 6 \pi \eta r v_t \]
Step 3: Solve for \( v_t \) Cancel \( \pi \) and simplify: \[ \frac{4}{3} r^3 (\rho - \sigma) g = 6 \eta r v_t \] \[ v_t = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta} \]
Step 4: Proportionality Since all other terms are constant for a given system: \[ v_t \propto r^2 \] Thus, terminal velocity is proportional to \( r^2 \).
Step 1: Write forces Weight of sphere: \[ W = \frac{4}{3}\pi r^3 \rho g \] Buoyant force: \[ B = \frac{4}{3}\pi r^3 \sigma g \] Net downward force: \[ F = \frac{4}{3}\pi r^3 (\rho - \sigma) g \]
Step 2: Balance at terminal velocity \[ \frac{4}{3}\pi r^3 (\rho - \sigma) g = 6 \pi \eta r v_t \]
Step 3: Solve for \( v_t \) Cancel \( \pi \) and simplify: \[ \frac{4}{3} r^3 (\rho - \sigma) g = 6 \eta r v_t \] \[ v_t = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta} \]
Step 4: Proportionality Since all other terms are constant for a given system: \[ v_t \propto r^2 \] Thus, terminal velocity is proportional to \( r^2 \).
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