If speed of light c, accleratio due to gravity g and pressure p are taken as fundamental units, the dimension of gravitational constant (G) are

Options

A. `c^(0)g p^(-3)`
B. `c^(2)g^(3)p^(-2)`
C. `c^(0)g^(2)p^(-1)`
D. `c^(2)g^(2)p^(-2)`

Explanation

To find the dimensions of the gravitational constant \( G \) when speed of light \( c \), acceleration due to gravity \( g \), and pressure \( p \) are taken as fundamental units, we can follow these steps:

### Step 1: Identify the dimensions of the fundamental quantities

1. **Speed of light \( c \)**:
- Unit: meters per second (m/s)
- Dimension: \( [c] = L T^{-1} \)

2. **Acceleration due to gravity \( g \)**:
- Unit: meters per second squared (m/s²)
- Dimension: \( [g] = L T^{-2} \)

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

### Step 2: Write the formula for gravitational constant \( G \)

According to Newton's law of gravitation:
\[ F = \frac{G m_1 m_2}{r^2} \]
Rearranging gives:
\[ G = \frac{F r^2}{m_1 m_2} \]

### Step 3: Substitute dimensions into the formula for \( G \)

Substituting the dimensions we have:
- Dimension of force \( F \): \( M L T^{-2} \)
- Dimension of distance \( r \): \( L \)
- Dimension of mass \( m_1 \) and \( m_2 \): \( M \)

Thus, we can express the dimensions of \( G \):
\[ [G] = \frac{M L T^{-2} \cdot L^2}{M^2} = \frac{M L^3 T^{-2}}{M^2} = M^{-1} L^3 T^{-2} \]

### Step 4: Express \( G \) in terms of \( c \), \( g \), and \( p \)

We want to express \( G \) in terms of the fundamental quantities \( c \), \( g \), and \( p \):
\[ [G] = c^x g^y p^z \]

Substituting the dimensions we found:
- \( [c] = L T^{-1} \)
- \( [g] = L T^{-2} \)
- \( [p] = M L^{-1} T^{-2} \)

This gives:
\[ M^{-1} L^3 T^{-2} = (L T^{-1})^x (L T^{-2})^y (M L^{-1} T^{-2})^z \]

### Step 5: Expand and equate dimensions

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

Now, equate the powers of \( M \), \( L \), and \( T \):

1. For \( M \): \( z = -1 \)
2. For \( L \): \( x + y - z = 3 \)
3. For \( T \): \( -x - 2y - 2z = -2 \)

### Step 6: Solve the equations

Substituting \( z = -1 \) into the other equations:
1. \( x + y + 1 = 3 \) → \( x + y = 2 \)
2. \( -x - 2y + 2 = -2 \) → \( -x - 2y = -4 \) → \( x + 2y = 4 \)

Now we have a system of equations:
1. \( x + y = 2 \)
2. \( x + 2y = 4 \)

Subtract the first from the second:
\[ (x + 2y) - (x + y) = 4 - 2 \implies y = 2 \]
Substituting \( y = 2 \) into \( x + y = 2 \):
\[ x + 2 = 2 \implies x = 0 \]

### Step 7: Substitute back to find \( G \)

Now substituting \( x = 0 \), \( y = 2 \), and \( z = -1 \):
\[ G = c^0 g^2 p^{-1} = g^2 \frac{1}{p} \]

### Final Answer

Thus, the dimensions of the gravitational constant \( G \) in terms of the fundamental units \( c \), \( g \), and \( p \) are:
\[ [G] = g^2 p^{-1} \]