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 understand the relationship between the mass of the steel ball, its radius, and the terminal velocity when it falls through a viscous liquid.

### Step-by-Step Solution:

1. **Understanding Terminal Velocity**:
The terminal velocity \( v_t \) of an object falling through a viscous fluid is given by the balance of forces acting on it. The gravitational force is balanced by the drag force due to the viscosity of the liquid.

2. **Relation Between Mass and Radius**:
The mass \( m \) of the steel ball is related to its volume \( V \) and density \( \rho \) as follows:
\[ m = \rho V \]
For a sphere, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, we can express mass in terms of radius:
\[ m = \rho \left( \frac{4}{3} \pi r^3 \right) \]
This shows that mass is directly proportional to the cube of the radius:
\[ m \propto r^3 \]

3. **Finding the New Radius**:
If the mass of the new steel ball is \( 8m \), we can set up the equation:
\[ 8m = \rho \left( \frac{4}{3} \pi r'^3 \right) \]
Since \( m \propto r^3 \), we can write:
\[ 8m \propto r'^3 \]
Thus, if \( m \propto r^3 \), then:
\[ r'^3 = 8r^3 \implies r' = 2r \]

4. **Finding the Terminal Velocity**:
The terminal velocity \( v_t \) is proportional to the square of the radius:
\[ v_t \propto r^2 \]
For the new radius \( r' = 2r \):
\[ v' \propto (r')^2 = (2r)^2 = 4r^2 \]
Therefore, if the original terminal velocity is \( v \), the new terminal velocity \( v' \) will be:
\[ v' = 4v \]

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