The terminal velocity of small sized spherical body of radius r falling veertically in a viscous liquid is given by a following proportionality

Options

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

Explanation

To solve the problem regarding the terminal velocity of a small-sized spherical body falling vertically in a viscous liquid, we will derive the relationship step by step.

### Step-by-Step Solution:

1. **Understanding Terminal Velocity**:
- Terminal velocity (Vt) is the constant speed that a freely falling object eventually reaches when the resistance of the medium prevents further acceleration. In this case, it occurs when the gravitational force is balanced by the drag force due to the viscosity of the liquid.

2. **Forces Acting on the Sphere**:
- The gravitational force acting on the sphere is given by \( F_g = mg \), where \( m \) is the mass of the sphere and \( g \) is the acceleration due to gravity.
- The drag force (or viscous force) opposing the motion is given by Stokes' law for small spheres, which can be expressed as \( F_d = 6 \pi \eta r v \), where \( \eta \) is the viscosity of the liquid, \( r \) is the radius of the sphere, and \( v \) is the velocity of the sphere.

3. **Setting Up the Equation**:
- At terminal velocity, the gravitational force is equal to the drag force:
\[ mg = 6 \pi \eta r V_t \]

4. **Expressing Mass in Terms of Density**:
- The mass \( m \) can be expressed in terms of the density of the sphere \( \rho_s \) and its volume \( V \):
\[ m = \rho_s \cdot V = \rho_s \cdot \frac{4}{3} \pi r^3 \]
- Substituting this into the force balance equation gives:
\[ \rho_s \cdot \frac{4}{3} \pi r^3 g = 6 \pi \eta r V_t \]

5. **Simplifying the Equation**:
- Canceling \( \pi \) from both sides and rearranging gives:
\[ \rho_s \cdot \frac{4}{3} r^2 g = 6 \eta V_t \]
- Rearranging for \( V_t \):
\[ V_t = \frac{\rho_s \cdot 4g r^2}{18 \eta} \]

6. **Identifying Proportionality**:
- From the equation \( V_t = \frac{4 \rho_s g}{18 \eta} r^2 \), we can see that \( V_t \) is directly proportional to \( r^2 \):
\[ V_t \propto r^2 \]

7. **Conclusion**:
- Therefore, the terminal velocity of a small-sized spherical body falling in a viscous liquid is proportional to the square of its radius \( r \). The correct answer is option D: \( r^2 \).