A small steel ball of radius r is allowed to fall under gravity through a column of a viscous liquid of coefficient of viscosity `eta`. After some time the velocity of the ball attains a constant value known as terminal velocity `upsilon_T`. The terminal velocity depends on (i) the mass of the ball m (ii) `eta`, (iii) r and (iv) acceleration due to gravity g . Which of the following relations is dimensionally correct?

Options

A. `v_Tprop(mg)/(etar)`
B. `v_Tprop(etar)/(mg)`
C. `v_Tpropetarmg`
D. `v_Tprop(mgr)/(eta)`

Explanation

To determine the correct dimensional relation for the terminal velocity \( v_T \) of a small steel ball falling through a viscous liquid, we need to analyze how \( v_T \) depends on the parameters given: mass \( m \), coefficient of viscosity \( \eta \), radius \( r \), and acceleration due to gravity \( g \).

### Step-by-Step Solution:

1. **Identify the Parameters and Their Dimensions**:
- Mass \( m \): Dimension \( [M] \)
- Coefficient of viscosity \( \eta \): Dimension \( [M L^{-1} T^{-1}] \)
- Radius \( r \): Dimension \( [L] \)
- Acceleration due to gravity \( g \): Dimension \( [L T^{-2}] \)

2. **Formulate the Relationship**:
We are looking for a relationship of the form:
\[ v_T \propto \frac{m^a \cdot \eta^b \cdot r^c \cdot g^d}{1} \]
where \( a, b, c, d \) are the powers to which each parameter is raised.

3. **Write the Dimensions of Each Parameter**:
The dimensions of the right-hand side can be expressed as:
\[ [v_T] = [L T^{-1}] = [M^a][M L^{-1} T^{-1}]^b [L]^c [L T^{-2}]^d \]

4. **Combine the Dimensions**:
Expanding the dimensions gives:
\[ [M^a][M^b L^{-b} T^{-b}][L^c][L^d T^{-2d}] = [M^{a+b} L^{c-b+d} T^{-b-2d}] \]

5. **Set Up the Equation**:
For the dimensions to be consistent, we need:
- For mass: \( a + b = 0 \)
- For length: \( c - b + d = 1 \)
- For time: \( -b - 2d = -1 \)

6. **Solve the Equations**:
From \( a + b = 0 \), we have \( a = -b \).

From \( -b - 2d = -1 \), we can express \( b \) in terms of \( d \):
\[ b + 2d = 1 \implies b = 1 - 2d \]

Substituting \( b \) into \( c - b + d = 1 \):
\[ c - (1 - 2d) + d = 1 \implies c + d - 1 = 1 \implies c + d = 2 \implies c = 2 - d \]

7. **Choose Values for \( d \)**:
Let's choose \( d = 0 \):
- Then \( b = 1 \), \( a = -1 \), \( c = 2 \).

This gives us:
\[ v_T \propto \frac{m^0 \cdot \eta^1 \cdot r^2 \cdot g^0}{1} \implies v_T \propto \frac{r^2}{\eta} \]

8. **Final Relation**:
However, we need to include the mass and gravitational force:
\[ v_T \propto \frac{mg}{\eta r} \]

Thus, we conclude:
\[ v_T \propto \frac{mg}{\eta r} \]

### Conclusion:
The correct dimensional relation for the terminal velocity \( v_T \) is:
\[ v_T \propto \frac{mg}{\eta r} \]