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_Tetarmg`
D. `v_Tprop(mgr)/(eta)`

Explanation

To determine which relation is dimensionally correct for the terminal velocity \( v_T \) of a small steel ball falling through a viscous liquid, we need to analyze the dimensions of the quantities involved. The terminal velocity depends on the mass of the ball \( m \), the coefficient of viscosity \( \eta \), the radius \( r \), and the acceleration due to gravity \( g \).

### Step-by-Step Solution:

1. **Identify the dimensions of each variable:**
- Terminal velocity \( v_T \): The dimension is \( [v_T] = [L][T^{-1}] \) (length per time).
- Mass \( m \): The dimension is \( [m] = [M] \).
- Coefficient of viscosity \( \eta \): Defined as \( \eta = \frac{F}{A \cdot \frac{dv}{dx}} \).
- Force \( F \): The dimension is \( [F] = [M][L][T^{-2}] \).
- Area \( A \): The dimension is \( [A] = [L^2] \).
- Velocity gradient \( \frac{dv}{dx} \): The dimension is \( [\frac{dv}{dx}] = [L][T^{-1}] \cdot [L^{-1}] = [T^{-1}] \).
- Therefore, the dimension of viscosity \( \eta \) is:
\[ [\eta] = \frac{[M][L][T^{-2}]}{[L^2][T^{-1}]} = [M][L^{-1}][T^{-1}] \]

2. **Identify the dimension of radius \( r \):**
- The dimension of radius \( r \) is \( [r] = [L] \).

3. **Identify the dimension of acceleration due to gravity \( g \):**
- The dimension of \( g \) is \( [g] = [L][T^{-2}] \).

4. **Construct the right-hand side of the proposed relation:**
- The proposed relation is \( v_T \propto \frac{mg}{\eta r} \).
- The dimensions of the right-hand side are:
\[ [\frac{mg}{\eta r}] = \frac{[M][L][T^{-2}]}{[M][L^{-1}][T^{-1}][L]} = \frac{[M][L][T^{-2}]}{[M][L^0][T^{-1}]} = [L][T^{-1}] \]

5. **Compare the dimensions:**
- Left-hand side (LHS) dimension of \( v_T \): \( [L][T^{-1}] \).
- Right-hand side (RHS) dimension of \( \frac{mg}{\eta r} \): \( [L][T^{-1}] \).

6. **Conclusion:**
- Since the dimensions of both sides are equal, the relation \( v_T \propto \frac{mg}{\eta r} \) is dimensionally correct.

### Final Answer:
The relation \( v_T \propto \frac{mg}{\eta r} \) is dimensionally correct.