If acceleration due to gravity g, the speed of light c and pressure p are taken as the fundamental quantities then find the dimensions of length.

Explanation

To find the dimensions of length (L) in terms of acceleration due to gravity (g), speed of light (c), and pressure (p), we will follow these steps:

### Step 1: Write the dimensional formulas
We need to start by writing the dimensional formulas for each of the quantities involved:

1. **Length (L)**: The dimension of length is represented as \( [L] \).
2. **Acceleration due to gravity (g)**: The dimension of acceleration is \( [L T^{-2}] \).
3. **Speed of light (c)**: The dimension of speed is \( [L T^{-1}] \).
4. **Pressure (p)**: Pressure is defined as force per unit area. The dimension of pressure is \( [M L^{-1} T^{-2}] \).

### Step 2: Set up the relationship
We express length (L) in terms of the fundamental quantities \( g \), \( c \), and \( p \):
\[ L \propto g^x c^y p^z \]
Where \( x \), \( y \), and \( z \) are the powers we need to determine.

### Step 3: Write the dimensional equation
Substituting the dimensional formulas into the equation gives:
\[ [L] = [g]^x [c]^y [p]^z \]
This translates to:
\[ [L] = [L T^{-2}]^x [L T^{-1}]^y [M L^{-1} T^{-2}]^z \]

### Step 4: Expand the right-hand side
Expanding the right-hand side:
\[ [L] = [L^x T^{-2x}] [L^y T^{-y}] [M^z L^{-z} T^{-2z}] \]
Combining the dimensions:
\[ [L] = [L^{x+y-z} M^z T^{-2x-y-2z}] \]

### Step 5: Compare the dimensions
Now we compare the powers of \( L \), \( M \), and \( T \) on both sides of the equation:

1. For \( L \):
\[ x + y - z = 1 \quad \text{(1)} \]
2. For \( M \):
\[ z = 0 \quad \text{(2)} \]
3. For \( T \):
\[ -2x - y - 2z = 0 \quad \text{(3)} \]

### Step 6: Solve the equations
From equation (2), we have \( z = 0 \). Substituting \( z = 0 \) into equations (1) and (3):

1. From (1):
\[ x + y = 1 \quad \text{(4)} \]
2. From (3):
\[ -2x - y = 0 \quad \text{(5)} \]

Now we can solve equations (4) and (5):
From (5):
\[ y = -2x \]
Substituting into (4):
\[ x - 2x = 1 \implies -x = 1 \implies x = -1 \]
Substituting \( x = -1 \) back into (4):
\[ -1 + y = 1 \implies y = 2 \]

### Step 7: Conclusion
We have found:
- \( x = -1 \)
- \( y = 2 \)
- \( z = 0 \)

Thus, we can express length \( L \) as:
\[ L \propto g^{-1} c^{2} p^{0} \]
This simplifies to:
\[ L = \frac{c^2}{g} \]

### Final Answer:
The dimensions of length in terms of \( g \), \( c \), and \( p \) are given by:
\[ L \propto \frac{c^2}{g} \]