the speed of light c, gravitational constant G, and Planck’s constant h are taken as the fundamental units in a system. The dimension of time in this new system should be,

Options

A. `G^(1//2) h^(1//2) C^(-5//2)`
B. `G^(1//2h^(1//2) C^(1//2)`
C. `G^(1//2)h^(1//2)C^(-3//2)`
D. `G^(1//2)h^(1//2) C^(1//2)`

Explanation

To find the dimension of time in a system where the speed of light (c), gravitational constant (G), and Planck's constant (h) are taken as fundamental units, we can follow these steps:

### Step 1: Express Time in Terms of c, G, and h
We start by expressing time \( T \) in terms of the fundamental quantities \( c \), \( G \), and \( h \):
\[ T \propto c^x \cdot G^y \cdot h^z \]
where \( x \), \( y \), and \( z \) are the powers we need to determine.

### Step 2: Introduce a Dimensionless Constant
We can introduce a dimensionless constant \( K \):
\[ T = K \cdot c^x \cdot G^y \cdot h^z \]
Since \( K \) is dimensionless, we can ignore its dimensions for our analysis.

### Step 3: Determine the Dimensions of c, G, and h
Next, we need to find the dimensions of \( c \), \( G \), and \( h \):
- The speed of light \( c \) has dimensions:
\[ [c] = L T^{-1} \]
- The gravitational constant \( G \) can be derived from Newton's law of gravitation:
\[ G = \frac{F \cdot R^2}{M_1 \cdot M_2} \implies [G] = M^{-1} L^3 T^{-2} \]
- Planck's constant \( h \) is related to energy and frequency:
\[ h = \frac{E}{\nu} \implies [h] = M L^2 T^{-1} \]

### Step 4: Substitute Dimensions into the Equation
Now we substitute the dimensions into our equation:
\[ [T] = [c]^x \cdot [G]^y \cdot [h]^z \]
This gives us:
\[ [T] = (L T^{-1})^x \cdot (M^{-1} L^3 T^{-2})^y \cdot (M L^2 T^{-1})^z \]

### Step 5: Expand the Dimensions
Expanding the dimensions, we have:
\[ [T] = L^x T^{-x} \cdot M^{-y} L^{3y} T^{-2y} \cdot M^z L^{2z} T^{-z} \]
Combining the dimensions:
\[ [T] = M^{z - y} L^{x + 3y + 2z} T^{-x - 2y - z} \]

### Step 6: Set Up the Equations
Since the left side has dimensions of time, we can set up the following equations by comparing the powers of \( M \), \( L \), and \( T \):
1. For mass \( M \): \( z - y = 0 \)
2. For length \( L \): \( x + 3y + 2z = 0 \)
3. For time \( T \): \( -x - 2y - z = 1 \)

### Step 7: Solve the Equations
From equation 1, we have:
\[ z = y \]
Substituting \( z = y \) into equation 2:
\[ x + 3y + 2y = 0 \implies x + 5y = 0 \implies x = -5y \]
Now substituting \( x = -5y \) and \( z = y \) into equation 3:
\[ -(-5y) - 2y - y = 1 \implies 5y - 2y - y = 1 \implies 2y = 1 \implies y = \frac{1}{2} \]
Thus, we find:
\[ z = \frac{1}{2}, \quad x = -\frac{5}{2} \]

### Step 8: Write the Final Expression for Time
Substituting \( x \), \( y \), and \( z \) back into the equation for \( T \):
\[ T = K \cdot c^{-\frac{5}{2}} \cdot G^{\frac{1}{2}} \cdot h^{\frac{1}{2}} \]

### Conclusion
Thus, the dimension of time \( T \) in this new system is expressed in terms of \( c \), \( G \), and \( h \).