The speed (v) of sound in a gas is given by `v = k P^(x) ρ^(y)`
Where K is dimensionless constant, P is pressure, and `ρ` is the density, then

Options

A. `x = ½, y = ½`
B. `x = –½, y = –½`
C. `x = ½, y = –½`
D. `x = –½, y = ½`

Explanation

To solve the problem, we need to determine the values of \( x \) and \( y \) in the equation \( v = k P^x \rho^y \) using dimensional analysis.

### Step-by-Step Solution:

1. **Identify the Dimensions**:
- The dimensions of speed \( v \) are:
\[ [v] = M^0 L^1 T^{-1} \]
- The dimensions of pressure \( P \) are:
\[ [P] = M^1 L^{-1} T^{-2} \]
- The dimensions of density \( \rho \) are:
\[ [\rho] = M^1 L^{-3} \]
- The constant \( k \) is dimensionless:
\[ [k] = M^0 L^0 T^0 \]

2. **Write the Dimensional Equation**:
- The right-hand side of the equation can be expressed in terms of dimensions:
\[ [v] = [k] [P]^x [\rho]^y = M^0 L^0 T^0 \cdot (M^1 L^{-1} T^{-2})^x \cdot (M^1 L^{-3})^y \]
- This simplifies to:
\[ [v] = M^{x+y} L^{-x-3y} T^{-2x} \]

3. **Set Up the Equation**:
- Now, we equate the dimensions from both sides:
\[ M^0 L^1 T^{-1} = M^{x+y} L^{-x-3y} T^{-2x} \]

4. **Compare the Exponents**:
- For mass (M):
\[ x + y = 0 \quad \text{(1)} \]
- For length (L):
\[ -x - 3y = 1 \quad \text{(2)} \]
- For time (T):
\[ -2x = -1 \quad \text{(3)} \]

5. **Solve for \( x \)**:
- From equation (3):
\[ -2x = -1 \implies x = \frac{1}{2} \]

6. **Substitute \( x \) into Equation (1)**:
- Substitute \( x = \frac{1}{2} \) into equation (1):
\[ \frac{1}{2} + y = 0 \implies y = -\frac{1}{2} \]

7. **Final Values**:
- Thus, the values of \( x \) and \( y \) are:
\[ x = \frac{1}{2}, \quad y = -\frac{1}{2} \]

### Conclusion:
The final answer is:
\[ x = \frac{1}{2}, \quad y = -\frac{1}{2} \]