Terminal velocity `(V)` of a spherical object varies with a radius of object `(r)` -

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Answer

To solve the problem regarding the relationship between terminal velocity (V) and the radius (r) of a spherical object, we can follow these steps:

### 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 spherical object falling through a viscous medium, the forces acting on it are the gravitational force (downward) and the drag force (upward).

2. **Forces Acting on the Object**:
- The gravitational force acting on the object is given by \( F_g = mg \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.
- The drag force acting on the object can be expressed as \( F_d = 6 \pi \eta r V \), where \( \eta \) is the coefficient of viscosity, \( r \) is the radius of the spherical object, and \( V \) is the terminal velocity.

3. **Setting Up the Equation**:
At terminal velocity, the drag force equals the gravitational force:
\[ mg = 6 \pi \eta r V \]

4. **Expressing Mass in Terms of Density**:
The mass \( m \) can be expressed in terms of the density of the object \( \rho \) and its volume. The volume \( V \) of a sphere is given by \( V = \frac{4}{3} \pi r^3 \), so:
\[ m = \rho \cdot V = \rho \cdot \frac{4}{3} \pi r^3 \]

5. **Substituting Mass into the Equation**:
Substituting \( m \) into the force balance equation gives:
\[ \rho \cdot \frac{4}{3} \pi r^3 g = 6 \pi \eta r V \]

6. **Simplifying the Equation**:
Canceling \( \pi \) from both sides and rearranging gives:
\[ \frac{4}{3} \rho r^2 g = 6 \eta V \]
\[ V = \frac{2}{9} \frac{\rho g r^2}{\eta} \]

7. **Identifying the Relationship**:
From the final equation, we can see that terminal velocity \( V \) is proportional to the square of the radius \( r \):
\[ V \propto r^2 \]

8. **Graphical Representation**:
The relationship \( V \propto r^2 \) indicates that if we were to plot terminal velocity \( V \) against radius \( r \), the graph would be a parabola opening upwards. However, since the radius cannot be negative, the graph will only exist in the first quadrant.

### Conclusion:
The terminal velocity \( V \) of a spherical object varies with the square of its radius \( r \) as \( V \propto r^2 \).