An iron ball of radius 0.3 cm falls through a column of oil of density `0.94 gcm ^(-3)` .It is found to attain a terminal velocity of `0.5 cm s ^(-1) `. Determine the viscosity of the oil. Given that the density of iron is `7.8 gcm^(-3)`

Explanation

To determine the viscosity of the oil through which an iron ball falls, we can use the formula for terminal velocity. The terminal velocity \( v_t \) of a sphere falling through a viscous fluid is given by:

\[ v_t = \frac{2}{9} \cdot \frac{r^2 \cdot g}{\eta} \cdot (\rho_{\text{body}} - \rho_{\text{liquid}}) \]

Where:
- \( v_t \) = terminal velocity (in cm/s)
- \( r \) = radius of the sphere (in cm)
- \( g \) = acceleration due to gravity (in cm/s²)
- \( \eta \) = viscosity of the fluid (in g/cm·s)
- \( \rho_{\text{body}} \) = density of the body (in g/cm³)
- \( \rho_{\text{liquid}} \) = density of the liquid (in g/cm³)

### Step 1: Identify the given values
- Radius of the iron ball, \( r = 0.3 \, \text{cm} \)
- Density of the oil, \( \rho_{\text{liquid}} = 0.94 \, \text{g/cm}^3 \)
- Density of iron, \( \rho_{\text{body}} = 7.8 \, \text{g/cm}^3 \)
- Terminal velocity, \( v_t = 0.5 \, \text{cm/s} \)
- Acceleration due to gravity, \( g = 980 \, \text{cm/s}^2 \) (converted from \( 9.8 \, \text{m/s}^2 \))

### Step 2: Rearrange the formula to solve for viscosity \( \eta \)
Rearranging the terminal velocity formula gives:

\[ \eta = \frac{2}{9} \cdot \frac{r^2 \cdot g}{v_t} \cdot (\rho_{\text{body}} - \rho_{\text{liquid}}) \]

### Step 3: Substitute the values into the equation
Now, we can substitute the values into the rearranged formula:

\[ \eta = \frac{2}{9} \cdot \frac{(0.3)^2 \cdot 980}{0.5} \cdot (7.8 - 0.94) \]

### Step 4: Calculate the difference in densities
Calculate \( \rho_{\text{body}} - \rho_{\text{liquid}} \):

\[ 7.8 - 0.94 = 6.86 \, \text{g/cm}^3 \]

### Step 5: Calculate \( r^2 \)
Calculate \( r^2 \):

\[ (0.3)^2 = 0.09 \, \text{cm}^2 \]

### Step 6: Substitute and compute \( \eta \)
Now substitute \( r^2 \) and the density difference back into the equation:

\[ \eta = \frac{2}{9} \cdot \frac{0.09 \cdot 980}{0.5} \cdot 6.86 \]

Calculating the numerator:

\[ 0.09 \cdot 980 = 88.2 \]

Now substitute this into the equation:

\[ \eta = \frac{2}{9} \cdot \frac{88.2}{0.5} \cdot 6.86 \]

Calculating \( \frac{88.2}{0.5} \):

\[ \frac{88.2}{0.5} = 176.4 \]

Now substitute this value:

\[ \eta = \frac{2}{9} \cdot 176.4 \cdot 6.86 \]

Calculating \( 176.4 \cdot 6.86 \):

\[ 176.4 \cdot 6.86 = 1208.784 \]

Now calculate \( \frac{2}{9} \cdot 1208.784 \):

\[ \frac{2 \cdot 1208.784}{9} = \frac{2417.568}{9} \approx 268.6 \, \text{g/cm·s} \]

### Final Answer
The viscosity of the oil is approximately:

\[ \eta \approx 268.6 \, \text{g/cm·s} \]