For the equation F `prop A^(a)V^(b)D^(c )`, where F is the force, A is the area, V is the velocity D is the density, the dimensional formula gives the following values

Options

A. a = 1, b = 2, c = 1
B. a = 2, b = 1, c = 1
C. a = 1, b = 1, c = 2
D. a = 0, b = 1, c = 1

Answer

To solve the problem \( F \propto A^a V^b D^c \) where \( F \) is the force, \( A \) is the area, \( V \) is the velocity, and \( D \) is the density, we will perform dimensional analysis to find the values of \( a \), \( b \), and \( c \).

### Step 1: Write the dimensional formulas
- The dimensional formula for force \( F \) is:
\[ [F] = M L T^{-2} \]
- The dimensional formula for area \( A \) is:
\[ [A] = L^2 \]
- The dimensional formula for velocity \( V \) is:
\[ [V] = L T^{-1} \]
- The dimensional formula for density \( D \) is:
\[ [D] = M L^{-3} \]

### Step 2: Substitute the dimensional formulas into the equation
We can substitute these formulas into the equation \( F \propto A^a V^b D^c \):
\[ [M L T^{-2}] \propto [L^2]^a [L T^{-1}]^b [M L^{-3}]^c \]

### Step 3: Expand the right-hand side
Expanding the right-hand side gives:
\[ [M L T^{-2}] \propto M^c L^{2a} L^b L^{-3c} T^{-b} \]
This simplifies to:
\[ [M L T^{-2}] \propto M^c L^{2a + b - 3c} T^{-b} \]

### Step 4: Equate the dimensions
Now we equate the coefficients of \( M \), \( L \), and \( T \) from both sides:
1. For mass \( M \):
\[ 1 = c \quad \text{(1)} \]
2. For length \( L \):
\[ 1 = 2a + b - 3c \quad \text{(2)} \]
3. For time \( T \):
\[ -2 = -b \quad \text{(3)} \]

### Step 5: Solve the equations
From equation (1):
\[ c = 1 \]

Substituting \( c = 1 \) into equation (2):
\[ 1 = 2a + b - 3(1) \]
This simplifies to:
\[ 1 = 2a + b - 3 \implies 2a + b = 4 \quad \text{(4)} \]

From equation (3):
\[ b = 2 \]

Substituting \( b = 2 \) into equation (4):
\[ 2a + 2 = 4 \implies 2a = 2 \implies a = 1 \]

### Conclusion
The values are:
\[ a = 1, \quad b = 2, \quad c = 1 \]
Thus, the final answer is \( (1, 2, 1) \).