If x times momentum is work , then the dimensions of x are

Options

A. `LT^(-1)`
B. `L^(-1)T`
C. `ML^(-1)T^(-1)`
D. `MLT`

Answer

To solve the problem, we need to find the dimensions of \( x \) given that \( x \) times momentum is equal to work. Let's break this down step by step.

### Step 1: Understand the relationship
We know that:
\[ x \times \text{momentum} = \text{work} \]
This can be rearranged to express \( x \):
\[ x = \frac{\text{work}}{\text{momentum}} \]

### Step 2: Write down the dimensions of work and momentum
1. **Dimensions of Work (W)**:
Work is defined as force times distance. The dimensions of force (F) are:
\[ [F] = [m \cdot a] = [m \cdot \frac{l}{t^2}] = [m][l][t^{-2}] \]
Therefore, the dimensions of work are:
\[ [W] = [F][l] = [m][l][t^{-2}][l] = [m][l^2][t^{-2}] \]
Thus, the dimensions of work are:
\[ [W] = [M][L^2][T^{-2}] \]

2. **Dimensions of Momentum (p)**:
Momentum is defined as mass times velocity. The dimensions of velocity (v) are:
\[ [v] = \frac{[l]}{[t]} = [L][T^{-1}] \]
Therefore, the dimensions of momentum are:
\[ [p] = [m][v] = [m][l][t^{-1}] = [M][L][T^{-1}] \]

### Step 3: Substitute the dimensions into the equation for \( x \)
Now we can substitute the dimensions of work and momentum into the equation for \( x \):
\[ [x] = \frac{[W]}{[p]} = \frac{[M][L^2][T^{-2}]}{[M][L][T^{-1}]} \]

### Step 4: Simplify the dimensions
When we simplify the right-hand side:
\[ [x] = \frac{[M][L^2][T^{-2}]}{[M][L][T^{-1}]} = [L^{2-1}][T^{-2 - (-1)}] = [L^1][T^{-1}] \]
Thus, we find:
\[ [x] = [L][T^{-1}] \]

### Final Answer
The dimensions of \( x \) are:
\[ [x] = [L][T^{-1}] \]
This can also be written as:
\[ [x] = L \cdot T^{-1} \]