The buoyant force `F` acting on a body depends on the density of medium `rho`, volume of body immerese `V` and acceleration due to gravity `g`. Establish the relation using method of dimensions.

Options

A. `Vrhog`
B. `(rhog)/V`
C. `rhogV^(2)`
D. `sqrt(rhogV)`

Answer

To establish the relationship between the buoyant force \( F \), the density of the medium \( \rho \), the volume of the body immersed \( V \), and the acceleration due to gravity \( g \) using the method of dimensions, we can follow these steps:

### Step 1: Identify the Variables and Their Dimensions
We know that the buoyant force \( F \) depends on:
- Density of the medium \( \rho \)
- Volume of the body immersed \( V \)
- Acceleration due to gravity \( g \)

The dimensions of these quantities are:
- \( F \) (force) has dimensions of \( [F] = MLT^{-2} \)
- \( \rho \) (density) has dimensions of \( [\rho] = ML^{-3} \)
- \( V \) (volume) has dimensions of \( [V] = L^3 \)
- \( g \) (acceleration due to gravity) has dimensions of \( [g] = LT^{-2} \)

### Step 2: Formulate the Relationship
Assume that the buoyant force \( F \) can be expressed as:
\[ F = K \cdot \rho^a \cdot V^b \cdot g^c \]
where \( K \) is a dimensionless constant, and \( a \), \( b \), and \( c \) are the powers to be determined.

### Step 3: Write the Dimensions of Each Side
The dimensions of the right-hand side can be expressed as:
\[ [\rho^a] = (ML^{-3})^a = M^a L^{-3a} \]
\[ [V^b] = (L^3)^b = L^{3b} \]
\[ [g^c] = (LT^{-2})^c = L^c T^{-2c} \]

Combining these, we have:
\[ [\rho^a \cdot V^b \cdot g^c] = M^a L^{-3a + 3b + c} T^{-2c} \]

### Step 4: Set Up the Equation for Dimensions
Since the dimensions on both sides must be equal, we can equate the dimensions:
\[ MLT^{-2} = M^a L^{-3a + 3b + c} T^{-2c} \]

### Step 5: Create a System of Equations
From the equality of dimensions, we can create the following equations:
1. For mass: \( a = 1 \)
2. For length: \( -3a + 3b + c = 1 \)
3. For time: \( -2c = -2 \) or \( c = 1 \)

### Step 6: Solve the Equations
From equation 1, we have \( a = 1 \).
From equation 3, we have \( c = 1 \).

Substituting \( a = 1 \) and \( c = 1 \) into equation 2:
\[ -3(1) + 3b + 1 = 1 \]
This simplifies to:
\[ -3 + 3b + 1 = 1 \implies 3b - 2 = 0 \implies 3b = 2 \implies b = 1 \]

### Step 7: Write the Final Relation
Now substituting \( a = 1 \), \( b = 1 \), and \( c = 1 \) back into the original equation gives:
\[ F = K \cdot \rho^1 \cdot V^1 \cdot g^1 \]
Thus, we can express the buoyant force as:
\[ F = K \cdot \rho \cdot V \cdot g \]

### Conclusion
The relation established using dimensional analysis is:
\[ F = K \cdot \rho \cdot V \cdot g \]
where \( K \) is a dimensionless constant.