The relation between force `F` and density `d` is `F=(x)/(sqrt(d))`.
The dimension of `x` is

Options

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

Explanation

To find the dimensions of \( x \) in the relation \( F = \frac{x}{\sqrt{d}} \), where \( F \) is force and \( d \) is density, we can follow these steps:

### Step 1: Understand the given relation
The equation given is:
\[ F = \frac{x}{\sqrt{d}} \]
Here, \( F \) is force and \( d \) is density.

### Step 2: Express density in terms of mass and volume
Density \( d \) can be defined as:
\[ d = \frac{m}{V} \]
where \( m \) is mass and \( V \) is volume.

### Step 3: Substitute density into the equation
Now substituting for \( d \):
\[ F = \frac{x}{\sqrt{\frac{m}{V}}} \]
This can be rewritten as:
\[ F = \frac{x \sqrt{V}}{\sqrt{m}} \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ x = F \cdot \frac{\sqrt{m}}{\sqrt{V}} \]

### Step 5: Apply dimensional analysis
Now, we need to find the dimensions of \( x \). We know the dimensions of \( F \), \( m \), and \( V \):
- The dimension of force \( F \) is:
\[ [F] = [M][L][T^{-2}] = M L T^{-2} \]
- The dimension of mass \( m \) is:
\[ [m] = M \]
- The dimension of volume \( V \) is:
\[ [V] = [L^3] \]

### Step 6: Substitute dimensions into the equation for \( x \)
Substituting the dimensions into the equation for \( x \):
\[ [x] = [F] \cdot \frac{[\sqrt{m}]}{[\sqrt{V}]} \]
This becomes:
\[ [x] = (M L T^{-2}) \cdot \frac{M^{1/2}}{(L^3)^{1/2}} = (M L T^{-2}) \cdot \frac{M^{1/2}}{L^{3/2}} \]

### Step 7: Simplify the dimensions
Now simplifying:
\[ [x] = M^{1 + 1/2} L^{1 - 3/2} T^{-2} = M^{3/2} L^{-1/2} T^{-2} \]

### Conclusion
Thus, the dimensions of \( x \) are:
\[ [x] = M^{3/2} L^{-1/2} T^{-2} \]