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 finding the dimensional representation of the time constant \(\tau\) for a spherical body falling in a viscous medium, we will analyze the given options step by step.

### Step-by-Step Solution:

1. **Understanding Terminal Velocity**:
The terminal velocity \(v\) of a spherical body falling in a viscous medium can be expressed using the formula:
\[ v = \frac{mg}{6\pi \eta r} \]
where:
- \(m\) = mass of the body
- \(g\) = acceleration due to gravity
- \(\eta\) = viscosity of the medium
- \(r\) = radius of the spherical body

2. **Dimensional Analysis of Terminal Velocity**:
The dimensions of each variable are:
- Mass \(m\): \([M]\)
- Acceleration due to gravity \(g\): \([L T^{-2}]\)
- Viscosity \(\eta\): \([M L^{-1} T^{-1}]\)
- Radius \(r\): \([L]\)

Plugging these into the terminal velocity formula:
\[ v = \frac{[M][L T^{-2}]}{[M L^{-1} T^{-1}][L]} = \frac{M L T^{-2}}{M L^{-1} T^{-1} L} = \frac{M L T^{-2}}{M L^{0} T^{-1}} = L T^{-1} \]
Thus, the dimension of terminal velocity \(v\) is \([L T^{-1}]\).

3. **Finding the Time Constant \(\tau\)**:
The time constant \(\tau\) is the time taken for the velocity to reach \(0.63\) times the terminal velocity. The time constant can be expressed in terms of the dimensions of the system.

4. **Analyzing the Given Options**:
We will check each option to see if it has the dimension of time \([T]\).

**Option A**: \(\frac{m r^2}{6 \pi \eta}\)
- Dimensions:
\[ = \frac{[M][L^2]}{[M L^{-1} T^{-1}]} = \frac{M L^2}{M L^{-1} T^{-1}} = L^3 T \]
- Not dimensionally time.

**Option B**: \(\sqrt{\frac{6 \pi m r}{\eta g^2}}\)
- Dimensions:
\[ = \sqrt{\frac{[M][L]}{[M L^{-1} T^{-1}][L^2 T^{-4}]}} = \sqrt{\frac{M L}{M L^{-1} T^{-1} L^2 T^{-4}}} = \sqrt{L^3 T^3} = L^{3/2} T^{3/2} \]
- Not dimensionally time.

**Option C**: \(\frac{m}{6 \pi \eta g^2}\)
- Dimensions:
\[ = \frac{[M]}{[M L^{-1} T^{-1}][L^2 T^{-4}]} = \frac{M}{M L^{-1} T^{-1} L^2 T^{-4}} = \frac{1}{L T^2} \]
- Not dimensionally time.

**Option D**: \(\frac{1}{g}\)
- Dimensions:
\[ = \frac{1}{[L T^{-2}]} = [T] \]
- This has the dimension of time.

5. **Conclusion**:
The only option that represents the time constant \(\tau\) dimensionally is:
\[ \tau \sim \frac{1}{g} \]

### Final Answer:
The dimensional representation of the time constant \(\tau\) is given by option D.