The viscous force 'F' acting on a small sphere of rtadius 'r' moving with velocity v through the liquid is gib=ven by F= `6pinrv`. Calculate the dimensions of n , the cofficent of viscosity.

Answer

To find the dimensions of the coefficient of viscosity \( n \) in the equation for viscous force \( F = 6 \pi n r v \), we will follow these steps:

### Step 1: Understand the equation
The equation for the viscous force is given by:
\[ F = 6 \pi n r v \]
where:
- \( F \) is the viscous force,
- \( n \) is the coefficient of viscosity,
- \( r \) is the radius of the sphere,
- \( v \) is the velocity of the sphere.

### Step 2: Identify the dimensions of each variable
1. **Force \( F \)**: The dimension of force is given by:
\[ [F] = [M][L][T^{-2}] \]
where \( M \) is mass, \( L \) is length, and \( T \) is time.

2. **Radius \( r \)**: The radius is a measure of length, so:
\[ [r] = [L] \]

3. **Velocity \( v \)**: Velocity is distance per unit time, so:
\[ [v] = \frac{[L]}{[T]} = [L][T^{-1}] \]

### Step 3: Substitute dimensions into the equation
Rearranging the equation \( F = 6 \pi n r v \) to solve for \( n \):
\[ n = \frac{F}{6 \pi r v} \]
Since \( 6 \pi \) is a constant, we can ignore its dimensions. Thus, we can write:
\[ [n] = \frac{[F]}{[r][v]} \]

### Step 4: Substitute the dimensions into the equation
Now substituting the dimensions we identified:
\[ [n] = \frac{[M][L][T^{-2}]}{[L][L][T^{-1}]} \]

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

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