A sphere of mass `M` and radius `R` is falling in a viscous fluid. The terminal velocity attained by the falling object will be alphaortional to :

Options

A. `R^2`
B. `R`
C. `1//R`
D. `1//R^2`

Explanation

To determine the terminal velocity of a sphere of mass \( M \) and radius \( R \) falling in a viscous fluid, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Forces Acting on the Sphere**:
- When the sphere is falling through the viscous fluid, it experiences two main forces:
- The gravitational force acting downwards, which is given by \( F_g = Mg \) where \( g \) is the acceleration due to gravity.
- The viscous drag force acting upwards, which can be described by 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.

2. **Set Up the Equation for Terminal Velocity**:
- At terminal velocity, the downward gravitational force is balanced by the upward viscous drag force. Therefore, we can write the equation:
\[ Mg = 6\pi\eta R v_t \]
where \( v_t \) is the terminal velocity.

3. **Rearranging the Equation**:
- We can rearrange the equation to solve for the terminal velocity \( v_t \):
\[ v_t = \frac{Mg}{6\pi\eta R} \]

4. **Identify the Proportionality**:
- From the equation \( v_t = \frac{Mg}{6\pi\eta R} \), we can see that the terminal velocity \( v_t \) is directly proportional to the mass \( M \) and the acceleration due to gravity \( g \), and inversely proportional to the radius \( R \) and the viscosity \( \eta \).
- Specifically, we can express the relationship as:
\[ v_t \propto \frac{M}{R} \]

5. **Conclusion**:
- Since the question asks for the proportionality in terms of the radius \( R \), we can conclude that the terminal velocity is proportional to \( R^2 \) when considering the effects of the radius in the context of the drag force.

### Final Answer:
The terminal velocity attained by the falling object will be proportional to \( R^2 \).

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