Question

Terminal velocity of a small sphere falling in viscous liquid varies with radius as:

• A. v_t r
• B. v_t r^2
• C. v_t 1r
• D. v_t 1r^2

Hint:

Terminal velocity increases with square of radius because gravitational force grows as r^3 while drag grows as r.

Answer 1

The Correct Option is B

Step 1: Understand terminal velocity.
When a small sphere falls through a thick liquid, it speeds up at first, then reaches a steady speed. This steady speed is called the terminal velocity. We want to see how it depends on the radius.

Step 2: Balance the forces.
At terminal velocity the downward weight is balanced by the upward buoyancy and the viscous drag. So the net force is zero and the sphere falls steadily.

Step 3: Use Stokes' law.
Solving the force balance using Stokes' law for drag gives the terminal velocity: \[ v_t = \frac{2 r^2 (\rho - \sigma) g}{9 \eta} \] Here $r$ is the radius, $\rho$ is the density of the sphere, $\sigma$ is the density of the liquid and $\eta$ is the viscosity.

Step 4: Spot the radius dependence.
Look at the formula. Everything except $r$ is a constant for a given sphere and liquid. The only place radius appears is as $r^2$.

Step 5: Write the proportionality.
Since the rest is constant, the terminal velocity is proportional to the square of the radius: \[ v_t \propto r^2 \]

Step 6: Choose the correct option.
A bigger sphere falls faster, and the dependence is on the square of the radius. So the correct relation is $v_t \propto r^2$. \[ \boxed{v_t \propto r^2} \]