A spherical ball falls through viscous medium with terminal velocity `v`. If this ball is replaced by another ball of the same mass but half the radius, then the terminal velocity will be (neglect the effect of buoyancy.)

Options

A. `v`
B. `2v`
C. `4v`
D. `8v`

Explanation

To solve the problem, we need to analyze the relationship between the terminal velocity of a spherical ball falling through a viscous medium and its radius. Here’s the step-by-step solution:

### Step 1: Understand the forces acting on the ball
When a spherical ball falls through a viscous medium, it experiences two main forces:
- The gravitational force (weight) acting downward, \( F_g = mg \)
- The viscous drag force acting upward, \( F_v = 6 \pi \eta r v \)

At terminal velocity, these two forces balance each other:
\[ F_g = F_v \]
Thus, we have:
\[ mg = 6 \pi \eta r v \]

### Step 2: Express terminal velocity in terms of radius
From the equation above, we can express the terminal velocity \( v \) as:
\[ v = \frac{mg}{6 \pi \eta r} \]

### Step 3: Analyze the new ball's parameters
Now, we replace the original ball with another ball of the same mass but half the radius. Let the new radius be \( r' = \frac{r}{2} \).

### Step 4: Substitute the new radius into the terminal velocity equation
We can now substitute \( r' \) into the terminal velocity equation:
\[ v' = \frac{mg}{6 \pi \eta r'} \]
Substituting \( r' \):
\[ v' = \frac{mg}{6 \pi \eta \left(\frac{r}{2}\right)} \]
This simplifies to:
\[ v' = \frac{mg \cdot 2}{6 \pi \eta r} \]
\[ v' = 2 \cdot \frac{mg}{6 \pi \eta r} \]
Thus, we can see that:
\[ v' = 2v \]

### Step 5: Conclusion
The terminal velocity of the new ball with half the radius is twice the terminal velocity of the original ball. Therefore, the answer is:
\[ v' = 2v \]

### Final Answer
The terminal velocity of the new ball is \( 2v \).

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