A steel ball of radius `2 xx 10^(-3) `m is released in an oil of viscosity `0.232 Ns m^(-2)` and density `840 kg m^(-3)`. Calculate the terminal velocity of ball. Take density of steel as `7800 kg m^(-3) `

Explanation

To calculate the terminal velocity of a steel ball released in oil, we can use the formula for terminal velocity \( V_t \):

\[ V_t = \frac{2}{9} \cdot \frac{R^2 (\rho - \sigma) g}{\eta} \]

Where:
- \( R \) = radius of the ball
- \( \rho \) = density of the steel ball
- \( \sigma \) = density of the oil
- \( g \) = acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \))
- \( \eta \) = viscosity of the oil

### Step 1: Identify the given values
- Radius of the steel ball, \( R = 2 \times 10^{-3} \, \text{m} \)
- Density of steel, \( \rho = 7800 \, \text{kg/m}^3 \)
- Density of oil, \( \sigma = 840 \, \text{kg/m}^3 \)
- Viscosity of oil, \( \eta = 0.232 \, \text{Ns/m}^2 \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Substitute the values into the formula
Now, substituting the values into the terminal velocity formula:

\[ V_t = \frac{2}{9} \cdot \frac{(2 \times 10^{-3})^2 (7800 - 840) \cdot 9.8}{0.232} \]

### Step 3: Calculate \( R^2 \)
Calculate \( R^2 \):

\[ R^2 = (2 \times 10^{-3})^2 = 4 \times 10^{-6} \, \text{m}^2 \]

### Step 4: Calculate \( \rho - \sigma \)
Calculate the difference in densities:

\[ \rho - \sigma = 7800 - 840 = 6960 \, \text{kg/m}^3 \]

### Step 5: Substitute and simplify
Now substitute \( R^2 \) and \( \rho - \sigma \) back into the equation:

\[ V_t = \frac{2}{9} \cdot \frac{(4 \times 10^{-6}) \cdot (6960) \cdot (9.8)}{0.232} \]

### Step 6: Calculate the numerator
Calculate the numerator:

\[ 4 \times 10^{-6} \cdot 6960 \cdot 9.8 = 4 \times 10^{-6} \cdot 68388.8 \approx 0.27355 \, \text{(approximately)} \]

### Step 7: Calculate the entire expression
Now calculate \( V_t \):

\[ V_t = \frac{2}{9} \cdot \frac{0.27355}{0.232} \]

Calculating the fraction:

\[ \frac{0.27355}{0.232} \approx 1.176 \]

Now multiply by \( \frac{2}{9} \):

\[ V_t \approx \frac{2 \cdot 1.176}{9} \approx \frac{2.352}{9} \approx 0.2613 \, \text{m/s} \]

### Final Answer
Thus, the terminal velocity \( V_t \) of the steel ball is approximately:

\[ V_t \approx 0.26 \, \text{m/s} \]

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