A solid metallic sphere of radius r is allowed to fall freely through air. If the frictional resistance due to air is proportional to the cross-sectional area and to the square of the velocity, then the terminal velocity of the sphere is proportional to which of the following ?

Options

A. `r^(2)`
B. `r`
C. `r^(3//2)`
D. `r^(1//2)`

Answer

To find the terminal velocity of a solid metallic sphere falling through air, we need to analyze the forces acting on the sphere and how they relate to the terminal velocity. Here's a step-by-step solution:

### Step 1: Identify the forces acting on the sphere
When the sphere falls, two main forces act on it:
- The weight of the sphere (downward force)
- The frictional resistance due to air (upward force)

### Step 2: Write the expression for the frictional resistance
The frictional resistance (viscous force) \( F_b \) due to air is given to be proportional to the cross-sectional area and the square of the velocity. Mathematically, this can be expressed as:
\[ F_b = k \cdot A \cdot v^2 \]
where:
- \( k \) is a proportionality constant,
- \( A \) is the cross-sectional area of the sphere,
- \( v \) is the velocity of the sphere.

### Step 3: Calculate the cross-sectional area
The cross-sectional area \( A \) of a sphere with radius \( r \) is given by:
\[ A = \pi r^2 \]

### Step 4: Substitute the area into the frictional force equation
Substituting the expression for the area into the frictional force equation gives:
\[ F_b = k \cdot (\pi r^2) \cdot v^2 \]

### Step 5: Set up the equation for terminal velocity
At terminal velocity, the upward frictional force equals the downward weight of the sphere. The weight \( W \) of the sphere can be expressed as:
\[ W = m \cdot g \]
where \( m \) is the mass of the sphere and \( g \) is the acceleration due to gravity. The mass can be expressed in terms of density \( \rho \) and volume \( V \):
\[ m = \rho \cdot V \]
The volume \( V \) of the sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Thus, the weight becomes:
\[ W = \rho \cdot \left(\frac{4}{3} \pi r^3\right) \cdot g \]

### Step 6: Equate the forces at terminal velocity
Setting the frictional force equal to the weight gives:
\[ k \cdot (\pi r^2) \cdot v_t^2 = \rho \cdot \left(\frac{4}{3} \pi r^3\right) \cdot g \]
where \( v_t \) is the terminal velocity.

### Step 7: Simplify the equation
We can cancel \( \pi \) from both sides:
\[ k \cdot r^2 \cdot v_t^2 = \frac{4}{3} \rho r^3 g \]
Now, divide both sides by \( r^2 \):
\[ k \cdot v_t^2 = \frac{4}{3} \rho r g \]

### Step 8: Solve for terminal velocity
Rearranging the equation to solve for \( v_t^2 \):
\[ v_t^2 = \frac{4}{3} \cdot \frac{\rho g}{k} \cdot r \]
Taking the square root to find \( v_t \):
\[ v_t = \sqrt{\frac{4}{3} \cdot \frac{\rho g}{k} \cdot r} \]

### Step 9: Determine the proportionality
From the final expression for terminal velocity, we can see that:
\[ v_t \propto \sqrt{r} \]
This indicates that the terminal velocity of the sphere is proportional to the square root of its radius.

### Conclusion
The terminal velocity of the sphere is proportional to \( r^{1/2} \).