The force F on a sphere of radius r moving in a medium with velocity v is given by `F = 6 pi eta rv`. The dimensions of `eta` are

Options

A. `[ML^(-3)]`
B. `[MLT^(-2)]`
C. `[MT^(-1)]`
D. `[ML^(-1)T^(-1)]`

Explanation

To find the dimensions of the viscosity coefficient \( \eta \) from the equation \( F = 6 \pi \eta rv \), we will follow these steps:

### Step 1: Rearrange the equation
We start with the given equation:
\[ F = 6 \pi \eta rv \]
We can rearrange this to solve for \( \eta \):
\[ \eta = \frac{F}{6 \pi rv} \]

### Step 2: Identify the dimensions of each variable
Next, we need to determine the dimensions of the variables involved:
- The dimension of force \( F \) is given by:
\[ [F] = M L T^{-2} \]
- The dimension of radius \( r \) is:
\[ [r] = L \]
- The dimension of velocity \( v \) is:
\[ [v] = L T^{-1} \]

### Step 3: Substitute the dimensions into the equation
Now we substitute the dimensions into the equation for \( \eta \):
\[ [\eta] = \frac{[F]}{[r][v]} = \frac{M L T^{-2}}{L \cdot (L T^{-1})} \]

### Step 4: Simplify the expression
Now we simplify the expression:
\[ [\eta] = \frac{M L T^{-2}}{L^2 T^{-1}} = \frac{M L T^{-2}}{L^2} \cdot T = M L^{-1} T^{-1} \]

### Step 5: Final result
Thus, the dimensions of \( \eta \) are:
\[ [\eta] = M L^{-1} T^{-1} \]

### Conclusion
The dimensions of the viscosity coefficient \( \eta \) are \( M L^{-1} T^{-1} \).

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