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

Options

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

Explanation

To find the dimensions of mass when the velocity of light `(c)`, gravitational constant `(G)`, and Planck's constant `(h)` are chosen as fundamental units, we will follow these steps:

### Step 1: Write down the dimensions of the constants
- The dimension of the speed of light `(c)` is given by:
\[ [c] = L T^{-1} \]
- The dimension of the gravitational constant `(G)` is:
\[ [G] = M^{-1} L^3 T^{-2} \]
- The dimension of Planck's constant `(h)` is:
\[ [h] = M L^2 T^{-1} \]

### Step 2: Express the dimension of mass in terms of `c`, `G`, and `h`
Assume the dimension of mass `[M]` can be expressed as:
\[ [M] = c^a \cdot G^b \cdot h^c \]
This means:
\[ [M] = (L T^{-1})^a \cdot (M^{-1} L^3 T^{-2})^b \cdot (M L^2 T^{-1})^c \]

### Step 3: Expand the dimensions
Now, we will expand the right-hand side:
\[ [M] = L^a T^{-a} \cdot M^{-b} L^{3b} T^{-2b} \cdot M^c L^{2c} T^{-c} \]
Combining the dimensions, we get:
\[ [M] = M^{c-b} L^{a + 3b + 2c} T^{-a - 2b - c} \]

### Step 4: Set up equations for dimensions
Since we want the dimension of mass to be `[M]`, we can equate the powers of `M`, `L`, and `T` to zero:
1. For mass (M):
\[ c - b = 1 \quad (1) \]
2. For length (L):
\[ a + 3b + 2c = 0 \quad (2) \]
3. For time (T):
\[ -a - 2b - c = 0 \quad (3) \]

### Step 5: Solve the equations
From equation (1):
\[ c = b + 1 \]
Substituting \(c\) in equations (2) and (3):
- Substitute \(c\) in equation (2):
\[ a + 3b + 2(b + 1) = 0 \implies a + 5b + 2 = 0 \quad (4) \]
- Substitute \(c\) in equation (3):
\[ -a - 2b - (b + 1) = 0 \implies -a - 3b - 1 = 0 \implies a + 3b = -1 \quad (5) \]

Now we have two equations (4) and (5):
From equation (5):
\[ a = -1 - 3b \]
Substituting \(a\) in equation (4):
\[ (-1 - 3b) + 5b + 2 = 0 \implies 2b + 1 = 0 \implies b = -\frac{1}{2} \]
Now substituting \(b\) back to find \(a\) and \(c\):
\[ a = -1 - 3(-\frac{1}{2}) = -1 + \frac{3}{2} = \frac{1}{2} \]
\[ c = -\frac{1}{2} + 1 = \frac{1}{2} \]

### Step 6: Write the final result
Now we have:
- \(a = \frac{1}{2}\)
- \(b = -\frac{1}{2}\)
- \(c = \frac{1}{2}\)

Thus, the dimension of mass in the new system is:
\[ [M] = c^{\frac{1}{2}} \cdot G^{-\frac{1}{2}} \cdot h^{\frac{1}{2}} \]