The viscous force acting on a solid ball of surface area. A moving with terminal velocity v is proportional to

Options

A. `A`
B. `A^(1//2)`
C. `V`
D. `V^(1//2)`

Answer

To solve the problem, we need to analyze the relationship between the viscous force acting on a solid ball moving with terminal velocity and its surface area. We will use Stokes' law, which relates the viscous force to the characteristics of the object.

### Step-by-Step Solution:

1. **Understanding Stokes' Law**:
Stokes' law states that the viscous force \( F \) acting on a sphere moving through a viscous fluid is given by:
\[ F = 6 \pi \eta r v \]
where:
- \( F \) is the viscous force,
- \( \eta \) is the dynamic viscosity of the fluid,
- \( r \) is the radius of the sphere,
- \( v \) is the terminal velocity of the sphere.

2. **Relating Surface Area to Radius**:
The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi r^2 \]
From this equation, we can express the radius \( r \) in terms of the surface area \( A \):
\[ r = \sqrt{\frac{A}{4\pi}} \]

3. **Substituting Radius into Stokes' Law**:
We can substitute the expression for \( r \) into Stokes' law:
\[ F = 6 \pi \eta \left(\sqrt{\frac{A}{4\pi}}\right) v \]
Simplifying this, we get:
\[ F = 6 \pi \eta \left(\frac{1}{2} \sqrt{\frac{A}{\pi}}\right) v \]
\[ F = \frac{3 \eta \sqrt{A}}{\sqrt{\pi}} v \]

4. **Identifying Proportional Relationships**:
From the final equation, we can see that the viscous force \( F \) is proportional to both the square root of the surface area \( \sqrt{A} \) and the terminal velocity \( v \):
\[ F \propto \sqrt{A} \cdot v \]

### Conclusion:
Thus, the viscous force acting on a solid ball of surface area \( A \) moving with terminal velocity \( v \) is proportional to \( \sqrt{A} \cdot v \).