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 quantity \( \eta \) in the equation \( F = 6 \pi \eta r v \), we can follow these steps:

### Step 1: Understand the given equation
The equation given is:
\[ F = 6 \pi \eta r v \]
where:
- \( F \) is the force,
- \( r \) is the radius of the sphere,
- \( v \) is the velocity of the sphere,
- \( \eta \) is the quantity whose dimensions we want to find.

### Step 2: Rearrange the equation to solve for \( \eta \)
We can rearrange the equation to isolate \( \eta \):
\[ \eta = \frac{F}{6 \pi r v} \]
Since \( 6 \pi \) is a constant, it does not affect the dimensions.

### Step 3: Write down the dimensions of the quantities involved
1. **Force \( F \)**: The dimension of force is given by Newton's second law, which is:
\[ [F] = [M L T^{-2}] \]
2. **Radius \( r \)**: The dimension of radius is:
\[ [r] = [L] \]
3. **Velocity \( v \)**: The dimension of velocity is:
\[ [v] = [L T^{-1}] \]

### Step 4: Substitute the dimensions into the equation for \( \eta \)
Now we can substitute the dimensions into the expression for \( \eta \):
\[ [\eta] = \frac{[F]}{[r][v]} = \frac{[M L T^{-2}]}{[L][L T^{-1}]} \]

### Step 5: Simplify the expression
Now, simplify the right-hand side:
\[ [\eta] = \frac{[M L T^{-2}]}{[L^2 T^{-1}]} = [M] \cdot [L^{-1}] \cdot [T] = [M L^{-1} T^{-1}] \]

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