If `P` is the pressure of a gas and `rho` is its density , then find the dimension of velocity in terms of `P` and `rho` .

Options

A. `P^(1//2) rho^(-1//2)`
B. `P^(1//2) rho^(1//2)`
C. `P^(-1//2) rho^(-1//2)`
D. `P^(-1//2) rho^(-1//2)`

Explanation

To find the dimension of velocity in terms of pressure \( P \) and density \( \rho \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definitions**:
- Pressure \( P \) is defined as force per unit area.
- Density \( \rho \) is defined as mass per unit volume.

2. **Write the Dimensions**:
- The dimension of pressure \( P \):
\[ P = \frac{\text{Force}}{\text{Area}} = \frac{F}{A} \]
The dimension of force \( F \) is given by:
\[ F = \text{mass} \times \text{acceleration} = \text{kg} \cdot \text{m/s}^2 = [M][L][T^{-2}] \]
The dimension of area \( A \) is:
\[ A = \text{length}^2 = [L^2] \]
Therefore, the dimension of pressure \( P \) is:
\[ [P] = \frac{[M][L][T^{-2}]}{[L^2]} = [M][L^{-1}][T^{-2}] \]

- The dimension of density \( \rho \):
\[ \rho = \frac{\text{mass}}{\text{volume}} = \frac{[M]}{[L^3]} = [M][L^{-3}] \]

3. **Assume the Relationship**:
- We assume that the velocity \( v \) can be expressed in terms of \( P \) and \( \rho \) as:
\[ v = P^x \cdot \rho^y \]

4. **Write the Dimensions of Velocity**:
- The dimension of velocity \( v \) is:
\[ [v] = [L][T^{-1}] \]

5. **Set Up the Equation**:
- Substitute the dimensions of \( P \) and \( \rho \):
\[ [P^x] = [M^x][L^{-x}][T^{-2x}] \]
\[ [\rho^y] = [M^y][L^{-3y}] \]
- Therefore, the combined dimension of \( v \) becomes:
\[ [v] = [M^{x+y}][L^{-x-3y}][T^{-2x}] \]

6. **Equate the Dimensions**:
- Now, we equate the dimensions of velocity:
\[ [M^{x+y}][L^{-x-3y}][T^{-2x}] = [L^1][T^{-1}] \]

- This gives us the following equations:
- For mass: \( x + y = 0 \) (1)
- For length: \( -x - 3y = 1 \) (2)
- For time: \( -2x = -1 \) (3)

7. **Solve the Equations**:
- From equation (3):
\[ 2x = 1 \implies x = \frac{1}{2} \]
- Substitute \( x \) into equation (1):
\[ \frac{1}{2} + y = 0 \implies y = -\frac{1}{2} \]

8. **Final Expression**:
- Substitute \( x \) and \( y \) back into the expression for velocity:
\[ v = P^{\frac{1}{2}} \cdot \rho^{-\frac{1}{2}} \]

### Conclusion:
The dimension of velocity in terms of pressure \( P \) and density \( \rho \) is:
\[ v = P^{\frac{1}{2}} \cdot \rho^{-\frac{1}{2}} \]