A spherical metal ball of mass `m` and radius `(r )` is falling through a viscous medium. The value of its terminal velocity is proportional to

Options

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

Answer

To determine the terminal velocity of a spherical metal ball falling through a viscous medium, we can analyze the forces acting on the ball. Here’s a step-by-step solution:

### Step 1: Identify the Forces
When the spherical metal ball is falling through a viscous medium, there are two main forces acting on it:
1. Gravitational Force (Weight) \( F_g = mg \)
2. Drag Force \( F_d \) due to the viscous medium.

### Step 2: Set Up the Equation at Terminal Velocity
At terminal velocity, the drag force equals the gravitational force:
\[ F_d = F_g \]
Thus,
\[ 6 \pi \eta r V_t = mg \]
where:
- \( \eta \) is the viscosity of the medium,
- \( r \) is the radius of the ball,
- \( V_t \) is the terminal velocity.

### Step 3: Solve for Terminal Velocity
Rearranging the equation to solve for terminal velocity \( V_t \):
\[ V_t = \frac{mg}{6 \pi \eta r} \]

### Step 4: Identify Proportional Relationships
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 \( m \) of the ball.
- The terminal velocity \( V_t \) is inversely proportional to the radius \( r \) of the ball.

### Conclusion
Thus, we can conclude that the terminal velocity \( V_t \) is proportional to \( \frac{m}{r} \).

### Final Answer
The value of its terminal velocity is proportional to \( \frac{m}{r} \).

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