Let P represent radiation pressure, c represent speed of light and l represents radiation energy striking a unit area per second, then `P^(x)I^(y)c^(z)` will be dimensionless for

Options

A. x = 0, y = z
B. x = y = z
C. x = z = -y
D. x = y = -z

Answer

To solve the problem, we need to find the values of \(x\), \(y\), and \(z\) such that the expression \(P^x I^y c^z\) is dimensionless. Let's break it down step by step.

### Step 1: Determine the dimensions of each variable

1. **Radiation Pressure (P)**:
- Radiation pressure is defined as force per unit area.
- Force (F) has dimensions of mass times acceleration:
\[ [F] = [M][L][T^{-2}] \]
- Area (A) has dimensions of length squared:
\[ [A] = [L^2] \]
- Therefore, the dimensions of radiation pressure (P) are:
\[ [P] = \frac{[F]}{[A]} = \frac{[M][L][T^{-2}]}{[L^2]} = [M][L^{-1}][T^{-2}] \]

2. **Speed of Light (c)**:
- The speed of light has dimensions of length per time:
\[ [c] = \frac{[L]}{[T]} = [L][T^{-1}] \]

3. **Radiation Intensity (I)**:
- Intensity is defined as energy per unit area per unit time.
- Energy has dimensions of mass times length squared per time squared:
\[ [E] = [M][L^2][T^{-2}] \]
- Therefore, the dimensions of intensity (I) are:
\[ [I] = \frac{[E]}{[A][T]} = \frac{[M][L^2][T^{-2}]}{[L^2][T]} = [M][T^{-3}] \]

### Step 2: Write the expression with dimensions

Now we can express \(P^x I^y c^z\) in terms of its dimensions:
\[ [P^x I^y c^z] = [P]^x [I]^y [c]^z = ([M][L^{-1}][T^{-2}])^x ([M][T^{-3}])^y ([L][T^{-1}])^z \]

### Step 3: Combine the dimensions

Combining the dimensions gives:
\[ = [M^{x+y}][L^{-x+z}][T^{-2x-3y-z}] \]

### Step 4: Set the dimensions to be dimensionless

For the expression to be dimensionless, the powers of \(M\), \(L\), and \(T\) must all equal zero:
1. For mass (M):
\[ x + y = 0 \quad \text{(1)} \]
2. For length (L):
\[ -x + z = 0 \quad \text{(2)} \]
3. For time (T):
\[ -2x - 3y - z = 0 \quad \text{(3)} \]

### Step 5: Solve the equations

From equation (1):
\[ y = -x \]

Substituting \(y = -x\) into equation (2):
\[ -x + z = 0 \implies z = x \]

Now substitute \(y = -x\) and \(z = x\) into equation (3):
\[ -2x - 3(-x) - x = 0 \implies -2x + 3x - x = 0 \implies 0 = 0 \]

This means that the equations are satisfied for any value of \(x\). Thus, we have:
\[ x = z \quad \text{and} \quad y = -x \]

### Final Result

The relationships between \(x\), \(y\), and \(z\) are:
\[ x = z \quad \text{and} \quad y = -x \]