The terminal velocity of a sphere moving through a viscous medium is :

Options

A. directly proportional to the radius of the sphere
B. inversely proportional to the radius of the sphere
C. directly propotional to the square of the radius of sphere
D. inversely propotional to the square of the radius of sphere

Answer

To find the relationship between the terminal velocity of a sphere moving through a viscous medium and its radius, we can derive the formula for terminal velocity and analyze the dependencies.

### Step-by-Step Solution:

1. **Understanding Terminal Velocity**:
Terminal velocity is the constant speed that a freely falling object eventually reaches when the resistance of the medium prevents further acceleration. For a sphere falling through a viscous fluid, the forces acting on it are gravity and drag force.

2. **Forces Acting on the Sphere**:
- **Weight of the Sphere (W)**: The weight can be expressed as \( W = V \cdot \rho \cdot g \), where \( V \) is the volume of the sphere, \( \rho \) is the density of the fluid, and \( g \) is the acceleration due to gravity.
- **Drag Force (F_d)**: The drag force acting on the sphere can be expressed using Stokes' law as \( F_d = 6 \pi \eta r v \), where \( \eta \) is the viscosity of the fluid, \( r \) is the radius of the sphere, and \( v \) is the velocity of the sphere.

3. **Setting Up the Equation**:
At terminal velocity, the drag force equals the weight of the sphere:
\[ V \cdot \rho \cdot g = 6 \pi \eta r v_t \]
where \( v_t \) is the terminal velocity.

4. **Volume of the Sphere**:
The volume \( V \) of the sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting this into the weight equation gives:
\[ \frac{4}{3} \pi r^3 \cdot \rho \cdot g = 6 \pi \eta r v_t \]

5. **Simplifying the Equation**:
Canceling \( \pi \) from both sides and rearranging gives:
\[ \frac{4}{3} r^3 \cdot \rho \cdot g = 6 \eta r v_t \]
Dividing both sides by \( r \) (assuming \( r \neq 0 \)):
\[ \frac{4}{3} r^2 \cdot \rho \cdot g = 6 \eta v_t \]

6. **Solving for Terminal Velocity**:
Rearranging for \( v_t \):
\[ v_t = \frac{4}{18} \cdot \frac{r^2 \cdot \rho \cdot g}{\eta} \]
Simplifying further:
\[ v_t = \frac{2}{9} \cdot \frac{r^2 \cdot \rho \cdot g}{\eta} \]

7. **Analyzing the Proportionality**:
From the final equation, we can see that:
\[ v_t \propto r^2 \]
This indicates that the terminal velocity is directly proportional to the square of the radius of the sphere.

### Conclusion:
Thus, the terminal velocity of a sphere moving through a viscous medium is directly proportional to the square of the radius of the sphere.