If the velocity of light `(c)` , gravitational constant `(G)` and Planck's constant `(h)` are chosen as fundamental units, then the dimensions of mass in new system is

Options

A. `c^(1//2)G^(1//2)h^(1//2)`
B. `c^(1//2)G^(1//2)h^(-1//2)`
C. `c^(1//2)G^(-1//2)h^(1//2)`
D. `c^(-1//2)G^(1//2)h^(1//2)`

Answer

To find the dimensions of mass in a new system where the fundamental units are the velocity of light \( c \), the gravitational constant \( G \), and Planck's constant \( h \), we can follow these steps:

### Step 1: Identify the dimensions of the constants
1. **Velocity of light \( c \)**:
- Dimension: \( [c] = [L][T]^{-1} \)

2. **Gravitational constant \( G \)**:
- Dimension: \( [G] = [M]^{-1}[L]^{3}[T]^{-2} \)

3. **Planck's constant \( h \)**:
- Dimension: \( [h] = [M][L]^{2}[T]^{-1} \)

### Step 2: Express mass \( M \) in terms of \( c \), \( G \), and \( h \)
Assume the dimension of mass \( [M] \) can be expressed as:
\[ [M] = [c]^A [G]^B [h]^C \]

### Step 3: Substitute the dimensions into the equation
Substituting the dimensions of \( c \), \( G \), and \( h \) into the equation gives:
\[ [M] = ([L][T]^{-1})^A \cdot ([M]^{-1}[L]^3[T]^{-2})^B \cdot ([M][L]^2[T]^{-1})^C \]

### Step 4: Expand the equation
This expands to:
\[ [M] = [L]^A[T]^{-A}[M]^{-B}[L]^{3B}[T]^{-2B}[M]^{C}[L]^{2C}[T]^{-C} \]

### Step 5: Combine like terms
Combining the dimensions gives:
\[ [M] = [M]^{C-B} [L]^{A + 3B + 2C} [T]^{-A - 2B - C} \]

### Step 6: Set up the equations for dimensions
For the equation to hold true, the powers of \( M \), \( L \), and \( T \) must equal zero:
1. For mass (\( M \)):
\[ C - B = 1 \quad \text{(1)} \]

2. For length (\( L \)):
\[ A + 3B + 2C = 0 \quad \text{(2)} \]

3. For time (\( T \)):
\[ -A - 2B - C = 0 \quad \text{(3)} \]

### Step 7: Solve the equations
From equation (1):
\[ C = B + 1 \]

Substituting \( C \) into equations (2) and (3):
- From (2):
\[ A + 3B + 2(B + 1) = 0 \implies A + 5B + 2 = 0 \implies A = -5B - 2 \quad \text{(4)} \]

- From (3):
\[ -A - 2B - (B + 1) = 0 \implies -A - 3B - 1 = 0 \implies A = -3B - 1 \quad \text{(5)} \]

### Step 8: Equate equations (4) and (5)
Setting equations (4) and (5) equal to each other:
\[ -5B - 2 = -3B - 1 \]
Solving for \( B \):
\[ -5B + 3B = -1 + 2 \implies -2B = 1 \implies B = -\frac{1}{2} \]

### Step 9: Find \( C \) and \( A \)
Using \( B = -\frac{1}{2} \) in equation (1):
\[ C = -\frac{1}{2} + 1 = \frac{1}{2} \]

Using \( B \) in equation (4) to find \( A \):
\[ A = -5(-\frac{1}{2}) - 2 = \frac{5}{2} - 2 = \frac{1}{2} \]

### Step 10: Conclusion
Thus, the dimensions of mass in the new system are:
\[ [M] = [c]^{\frac{1}{2}} [G]^{-\frac{1}{2}} [h]^{\frac{1}{2}} = \sqrt{\frac{c h}{G}} \]

### Final Answer
The dimensions of mass in the new system are given by:
\[ [M] = \sqrt{\frac{c h}{G}} \]