The terminal velocity v of a small steel ball ofradius r fal ling under gravity through a column ofa viscous liquid of coefficient of viscosity `eta` depends on mass of the ball m, acceleration due to gravity g, coefficient of viscosity `eta` and radius r. Which of the following relations is dimensionally correct?

Options

A. `v_rprop(etamg)/(r )`
B. `v_rprop(mgr)/(eta)`
C. `v_rprop(mg)/(reta)`
D. `v_rpropmg" "eta r`

Explanation

To determine which relation is dimensionally correct for the terminal velocity \( v \) 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 \( m \), acceleration due to gravity \( g \), coefficient of viscosity \( \eta \), and radius \( r \).

### Step-by-step Solution:

1. **Identify the dimensions of terminal velocity \( v \)**:
- The terminal velocity \( v \) has the dimensions of velocity, which is given by:
\[ [v] = L^1 T^{-1} \]

2. **Identify the dimensions of each variable**:
- Mass \( m \):
\[ [m] = M^1 \]
- Acceleration due to gravity \( g \):
\[ [g] = L^1 T^{-2} \]
- Coefficient of viscosity \( \eta \):
- The coefficient of viscosity can be derived from the formula for force \( F \) in Stokes' law:
\[ F = 6 \pi r \eta v \]
Rearranging gives:
\[ \eta = \frac{F}{6 \pi r v} \]
The dimensions of force \( F \) are:
\[ [F] = M^1 L^1 T^{-2} \]
Therefore, the dimensions of \( \eta \) are:
\[ [\eta] = \frac{[F]}{[r][v]} = \frac{M^1 L^1 T^{-2}}{L^1 L^1 T^{-1}} = M^1 L^{-1} T^{-1} \]
- Radius \( r \):
\[ [r] = L^1 \]

3. **Combine the dimensions for the proposed relation**:
- We need to check the dimensional correctness of the relation \( v \propto \frac{mg}{\eta r} \).
- The dimensions of \( mg \) are:
\[ [mg] = [m][g] = M^1 \cdot L^1 T^{-2} = M^1 L^1 T^{-2} \]
- The dimensions of \( \eta r \) are:
\[ [\eta][r] = (M^1 L^{-1} T^{-1})(L^1) = M^1 L^0 T^{-1} = M^1 T^{-1} \]
- Therefore, the dimensions of \( \frac{mg}{\eta r} \) are:
\[ \left[\frac{mg}{\eta r}\right] = \frac{M^1 L^1 T^{-2}}{M^1 T^{-1}} = L^1 T^{-1} \]

4. **Compare dimensions**:
- The dimensions of \( v \) (LHS) and \( \frac{mg}{\eta r} \) (RHS) are both:
\[ L^1 T^{-1} \]
- Since both sides have the same dimensions, the relation is dimensionally correct.

### Conclusion:
The relation \( v \propto \frac{mg}{\eta r} \) is dimensionally correct.