A steel ball of mass m falls in a viscous liquid with terminal velocity v, then the steel ball of mass `8m` will fall in the same liquid with terminal velocity

Options

A. `v`
B. `4v`
C. `8v`
D. `16sqrt(2)v`

Explanation

To solve the problem, we need to find the terminal velocity of a steel ball with mass \(8m\) falling in a viscous liquid, given that a steel ball of mass \(m\) falls with terminal velocity \(v\).

### Step-by-Step Solution:

1. **Understanding Terminal Velocity**:
Terminal velocity (\(v_t\)) is the constant speed reached by an object when the force of gravity is balanced by the drag force (viscous force) acting on it.

2. **Forces Acting on the Ball**:
When a ball falls through a viscous fluid, the forces acting on it are:
- Weight (\(W = mg\)) acting downwards.
- Upthrust (buoyant force) acting upwards.
- Viscous drag force (\(F_v\)) acting upwards.

At terminal velocity, these forces balance:
\[ mg = F_v + \text{Upthrust} \]

3. **Expression for Terminal Velocity**:
The terminal velocity for a sphere falling through a viscous fluid can be expressed as:
\[ v_t = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta} \]
where:
- \(r\) = radius of the ball,
- \(\rho\) = density of the ball,
- \(\sigma\) = density of the fluid,
- \(g\) = acceleration due to gravity,
- \(\eta\) = coefficient of viscosity.

4. **Relation Between Mass and Radius**:
The mass of the ball is related to its volume and density:
\[ m = \rho_{ball} \cdot V = \rho_{ball} \cdot \frac{4}{3} \pi r^3 \]
Thus, mass is proportional to the cube of the radius:
\[ m \propto r^3 \quad \Rightarrow \quad m_1 \propto r_1^3 \quad \text{and} \quad m_2 \propto r_2^3 \]

5. **Finding the New Radius**:
If the mass increases from \(m\) to \(8m\), we can set up the following relationship:
\[ 8m \propto r'^3 \quad \Rightarrow \quad 8 \cdot r^3 \propto r'^3 \]
Taking the cube root:
\[ r' = 2r \]

6. **Calculating the New Terminal Velocity**:
Now, substituting \(r' = 2r\) into the terminal velocity formula:
\[ v_{t'} = \frac{2}{9} \frac{(2r)^2 (\rho - \sigma) g}{\eta} = \frac{2}{9} \frac{4r^2 (\rho - \sigma) g}{\eta} = 4 \left(\frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta}\right) = 4v_t \]
Since \(v_t\) for mass \(m\) is \(v\), we have:
\[ v_{t'} = 4v \]

### Conclusion:
The terminal velocity of the steel ball with mass \(8m\) falling in the same viscous liquid will be \(4v\).