Which of the following is the dimension of the coefficient of frintion ?

Options

A. `[M^(2)L^(2)T]`
B. `[M^(0)L^(0)T^(0)]`
C. `[ML^(2)T^(-2)]`
D. `[M^(2)L^(2)T^(-2)]`

Explanation

To find the dimension of the coefficient of friction (µ), we can follow these steps:

### Step 1: Understand the relationship
The coefficient of friction (µ) is defined as the ratio of the frictional force (F) to the normal force (N). Mathematically, this can be expressed as:
\[ \mu = \frac{F}{N} \]

### Step 2: Identify the dimensions of force
The dimension of force (F) can be derived from Newton's second law, which states that force is mass times acceleration:
\[ F = m \cdot a \]
Where:
- \( m \) is mass with dimension [M]
- \( a \) is acceleration with dimension [L T^{-2}]

Thus, the dimension of force (F) is:
\[ [F] = [M][L T^{-2}] = [M L T^{-2}] \]

### Step 3: Identify the dimensions of normal force
The normal force (N) is also a force, so it has the same dimension as the frictional force:
\[ [N] = [M L T^{-2}] \]

### Step 4: Calculate the dimensions of the coefficient of friction
Now, substituting the dimensions of force into the equation for µ:
\[ \mu = \frac{F}{N} = \frac{[M L T^{-2}]}{[M L T^{-2}]} \]

### Step 5: Simplify the dimensions
When we divide the dimensions of the frictional force by the dimensions of the normal force, we get:
\[ \mu = \frac{[M L T^{-2}]}{[M L T^{-2}]} = 1 \]

This means that the dimensions cancel out completely, leading to:
\[ \mu = [M^0 L^0 T^0] \]

### Conclusion
The dimension of the coefficient of friction (µ) is:
\[ \mu = [M^0 L^0 T^0] \]
This indicates that it is a dimensionless quantity.

### Final Answer
The dimension of the coefficient of friction is \( m^0, l^0, t^0 \).