The quantities `A and B` are related by the relation `A//B = m`, where `m` is the linear mass density and `A` is the force , the dimensions of `B` will be

Options

A. Same as that of pressure
B. Same as that of work
C. That of momentum
D. Same as that of learnt heat

Explanation

To find the dimensions of quantity \( B \) in the relation \( \frac{A}{B} = m \), where \( A \) is force and \( m \) is linear mass density, we can follow these steps:

### Step-by-step Solution:

1. **Identify the quantities involved**:
- \( A \) is the force.
- \( B \) is the quantity we need to find the dimensions for.
- \( m \) is the linear mass density.

2. **Write down the dimensions of force \( A \)**:
- The dimension of force \( A \) is given by Newton's second law, \( F = ma \) (force = mass × acceleration).
- Thus, the dimension of force is:
\[ [A] = [M][L][T^{-2}] \]
- Where \( [M] \) is the dimension of mass, \( [L] \) is the dimension of length, and \( [T] \) is the dimension of time.

3. **Understand the dimension of linear mass density \( m \)**:
- Linear mass density \( m \) is defined as mass per unit length.
- Therefore, the dimension of linear mass density is:
\[ [m] = \frac{[M]}{[L]} = [M][L^{-1}] \]

4. **Rearranging the equation**:
- From the relation \( \frac{A}{B} = m \), we can rearrange it to find \( B \):
\[ B = \frac{A}{m} \]

5. **Substituting the dimensions**:
- Substitute the dimensions of \( A \) and \( m \) into the equation for \( B \):
\[ [B] = \frac{[A]}{[m]} = \frac{[M][L][T^{-2}]}{[M][L^{-1}]} \]

6. **Simplifying the dimensions**:
- When we simplify the expression:
\[ [B] = \frac{[M][L][T^{-2}]}{[M][L^{-1}]} = [L^{2}][T^{-2}] \]
- The \( [M] \) cancels out, and we are left with:
\[ [B] = [L^{2}][T^{-2}] \]

### Final Result:
The dimensions of \( B \) are:
\[ [B] = [L^{2}][T^{-2}] \]