Two small spherical metal balls, having equal masses, are made from materials of densities `rho_(1) and rho_(2) (rho_(1)=8rho_(2))` and have radii 1 mm and 2mm respectively, they are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equal `eta` and whose density is 0.1 `rho_(2)`. The ratio of their terminal velocities would be

Options

A. `(79)/(36)`
B. `(79)/(72)`
C. `(19)/(36)`
D. `(39)/(72)`

Answer

To find the ratio of the terminal velocities of the two spherical metal balls falling in a viscous medium, we can follow these steps:

### Step 1: Understand the formula for terminal velocity
The terminal velocity \( V_t \) of a sphere falling through a viscous fluid is given by the formula:

\[ V_t = \frac{2}{9} \frac{r^2 g (\rho_s - \rho_l)}{\eta} \]

where:
- \( r \) = radius of the sphere
- \( g \) = acceleration due to gravity
- \( \rho_s \) = density of the sphere
- \( \rho_l \) = density of the liquid
- \( \eta \) = coefficient of viscosity of the liquid

### Step 2: Define the parameters for both spheres
Let:
- Sphere 1 (radius \( r_1 = 1 \, \text{mm} = 0.001 \, \text{m} \), density \( \rho_1 = 8\rho_2 \))
- Sphere 2 (radius \( r_2 = 2 \, \text{mm} = 0.002 \, \text{m} \), density \( \rho_2 \))
- Density of the liquid \( \rho_l = 0.1\rho_2 \)

### Step 3: Calculate the terminal velocities for both spheres
Using the formula for terminal velocity, we can write:

1. For Sphere 1:
\[ V_{t1} = \frac{2}{9} \frac{(0.001)^2 g (8\rho_2 - 0.1\rho_2)}{\eta} \]
\[ = \frac{2}{9} \frac{(0.001)^2 g (7.9\rho_2)}{\eta} \]

2. For Sphere 2:
\[ V_{t2} = \frac{2}{9} \frac{(0.002)^2 g (\rho_2 - 0.1\rho_2)}{\eta} \]
\[ = \frac{2}{9} \frac{(0.002)^2 g (0.9\rho_2)}{\eta} \]

### Step 4: Simplify the expressions
Now, we simplify both expressions:

1. For Sphere 1:
\[ V_{t1} = \frac{2}{9} \frac{(0.000001) g (7.9\rho_2)}{\eta} \]

2. For Sphere 2:
\[ V_{t2} = \frac{2}{9} \frac{(0.000004) g (0.9\rho_2)}{\eta} \]

### Step 5: Take the ratio of terminal velocities
Now, we can find the ratio \( \frac{V_{t1}}{V_{t2}} \):

\[ \frac{V_{t1}}{V_{t2}} = \frac{\frac{2}{9} \frac{(0.000001) g (7.9\rho_2)}{\eta}}{\frac{2}{9} \frac{(0.000004) g (0.9\rho_2)}{\eta}} \]

The \( \frac{2}{9} \), \( g \), and \( \eta \) cancel out:

\[ = \frac{0.000001 \cdot 7.9\rho_2}{0.000004 \cdot 0.9\rho_2} \]

The \( \rho_2 \) also cancels out:

\[ = \frac{0.000001 \cdot 7.9}{0.000004 \cdot 0.9} \]

### Step 6: Perform the calculations
Calculating the above expression:

\[ = \frac{7.9}{0.000004 \cdot 0.9} = \frac{7.9}{0.0000036} = \frac{79}{0.036} = \frac{79 \times 1000}{36} = \frac{79000}{36} \approx 2194.44 \]

### Final Result
The ratio of the terminal velocities \( \frac{V_{t1}}{V_{t2}} \) is approximately \( 2194.44 \).