The dimensions of physical quantity X in the equation force = `X/(sqrt("Density"))` is given by :

Options

A. `M^1 L^4 T^(-2)`
B. `M^2 L^(-2) T^(-1)`
C. `M^(3//2) L^(-1//2) T^(-2)`
D. `M^(1) L^(-2) T^(-1)`

Explanation

To find the dimensions of the physical quantity \( X \) in the equation \( \text{Force} = \frac{X}{\sqrt{\text{Density}}} \), we will follow these steps:

### Step 1: Write down the equation
We start with the equation:
\[ \text{Force} = \frac{X}{\sqrt{\text{Density}}} \]

### Step 2: Rearrange the equation to solve for \( X \)
To isolate \( X \), we can rearrange the equation:
\[ X = \text{Force} \times \sqrt{\text{Density}} \]

### Step 3: Determine the dimensions of Force
The dimensional formula for force is given by:
\[ \text{Force} = \text{mass} \times \text{acceleration} = M \cdot L \cdot T^{-2} \]
Thus, the dimensions of force are:
\[ [F] = M L T^{-2} \]

### Step 4: Determine the dimensions of Density
Density (\( \rho \)) is defined as mass per unit volume:
\[ \rho = \frac{\text{mass}}{\text{volume}} = \frac{M}{L^3} \]
Thus, the dimensions of density are:
\[ [\rho] = M L^{-3} \]

### Step 5: Find the dimensions of \( \sqrt{\text{Density}} \)
Now, we need to find the dimensions of \( \sqrt{\text{Density}} \):
\[ \sqrt{\rho} = \sqrt{M L^{-3}} = M^{1/2} L^{-3/2} \]

### Step 6: Substitute the dimensions into the equation for \( X \)
Now we substitute the dimensions of force and \( \sqrt{\text{Density}} \) into the equation for \( X \):
\[ X = (M L T^{-2}) \times (M^{1/2} L^{-3/2}) \]

### Step 7: Combine the dimensions
Now we combine the dimensions:
\[ X = M^{1 + 1/2} \cdot L^{1 - 3/2} \cdot T^{-2} \]
This simplifies to:
\[ X = M^{3/2} \cdot L^{-1/2} \cdot T^{-2} \]

### Step 8: Write the final dimensional formula for \( X \)
Thus, the dimensions of \( X \) are:
\[ [X] = M^{3/2} L^{-1/2} T^{-2} \]

### Final Answer
The dimensions of the physical quantity \( X \) are:
\[ [X] = M^{3/2} L^{-1/2} T^{-2} \]