A sperical body of mass `m` and radius `r` is allowed to fall in a medium of viscosity `eta`. The time in which the velocity of the body increases from zero to `0.63 ` times the terminal velocity `(v)` is called constant `(tau)`. Dimensionally , `tau` can be represented by

Options

A. `(mr^(2))/( 6 pi eta)`
B. `sqrt((6 pi m r eta)/( g^(2)))`
C. ` (m)/( 6 pi eta r v)`
D. None of these

Explanation

To solve the problem of determining the dimensional representation of the time constant \( \tau \) for a spherical body falling through a viscous medium, we will analyze the given options step by step.

### Step 1: Understanding the Problem
We need to find the dimensional representation of the time constant \( \tau \) in terms of the parameters given: mass \( m \), radius \( r \), viscosity \( \eta \), and terminal velocity \( v \).

### Step 2: Identifying Dimensions
- Mass \( m \): Dimension is \( [M] \)
- Radius \( r \): Dimension is \( [L] \)
- Viscosity \( \eta \): Dimension is \( [M][L]^{-1}[T]^{-2} \)
- Velocity \( v \): Dimension is \( [L][T]^{-1} \)

### Step 3: Analyzing Each Option

#### Option 1: \( \frac{m r^2}{6 \pi \eta} \)
1. **Calculate dimensions:**
- \( m \): \( [M] \)
- \( r^2 \): \( [L^2] \)
- \( \eta \): \( [M][L]^{-1}[T]^{-2} \)

Now, substituting into the expression:
\[ \text{Dimension} = \frac{[M][L^2]}{[M][L]^{-1}[T]^{-2}} = \frac{[M][L^2]}{[M][L^{-1}][T^{-2}]} \]
Simplifying gives:
\[ = [L^3][T^2] = [M^0][L^3][T^2] \]
This does not represent time.

#### Option 2: \( \sqrt{\frac{6 \pi m r \eta}{g^2}} \)
1. **Calculate dimensions:**
- \( g \) (acceleration due to gravity): \( [L][T]^{-2} \)

Now substituting into the expression:
\[ \text{Dimension} = \sqrt{\frac{[M][L][M][L]^{-1}[T]^{-2}}{[L^2][T^{-4}]}} = \sqrt{\frac{[M^2][L^0][T^{-2}]}{[L^2][T^{-4}]}} \]
Simplifying gives:
\[ = \sqrt{[M^2][L^{-2}][T^2]} = [M][L^{-1}][T] \]
This also does not represent time.

#### Option 3: \( \frac{m}{6 \pi \eta r v} \)
1. **Calculate dimensions:**
- Substitute dimensions:
\[ \text{Dimension} = \frac{[M]}{[M][L]^{-1}[T]^{-2} \cdot [L] \cdot [L][T]^{-1}} = \frac{[M]}{[M][L^0][T^{-3}]} \]
Simplifying gives:
\[ = [T^3] \]
This does not represent time.

### Conclusion
After analyzing all options, none of them represent the dimension of time \( [T] \). Therefore, the conclusion is that none of the options correctly represent the dimensional formula for \( \tau \).