If `"Force"` =`X/("density")+C` is dimensionally correct, the dimension of x are -

Options

A. `MLT^(-2)`
B. `MLT^(-3)`
C. `ML^(2)T^(-3)`
D. `M^(2)L^(-2)T^(-2)`

Explanation

To determine the dimensions of \( x \) in the equation \( \text{Force} = \frac{x}{\text{density}} + C \), we need to ensure that both terms on the right side of the equation have the same dimensions as the left side (Force). Let's break this down step by step.

### Step 1: Identify the dimensions of Force
The dimension of force is given by the formula:
\[ \text{Force} = \text{mass} \times \text{acceleration} \]
In terms of dimensions, this can be expressed as:
\[ [F] = [M][L][T^{-2}] \]
where \( [M] \) is mass, \( [L] \) is length, and \( [T] \) is time.

### Step 2: Identify the dimensions of density
Density is defined as mass per unit volume:
\[ \text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{[M]}{[L^3]} \]
Thus, the dimension of density is:
\[ [\text{density}] = [M][L^{-3}] \]

### Step 3: Set up the equation for dimensional analysis
From the equation \( \text{Force} = \frac{x}{\text{density}} + C \), we can rearrange it to find the dimensions of \( x \):
\[ \frac{x}{\text{density}} = \text{Force} \]
This implies:
\[ x = \text{Force} \times \text{density} \]

### Step 4: Substitute the dimensions
Substituting the dimensions we found:
\[ [x] = [F] \times [\text{density}] \]
\[ [x] = ([M][L][T^{-2}]) \times ([M][L^{-3}]) \]

### Step 5: Simplify the dimensions
Now, we multiply the dimensions:
\[ [x] = [M^2][L][T^{-2}][L^{-3}] = [M^2][L^{1-3}][T^{-2}] = [M^2][L^{-2}][T^{-2}] \]

### Conclusion
Thus, the dimensions of \( x \) are:
\[ [x] = [M^2][L^{-2}][T^{-2}] \]