Time (T), velocity (C) and angular momentum (h) are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be:

Options

A. `[M] - [T^(1)C^2 h]`
B. `[M] = [T^(1)Ch^(-1)]`
C. `[M] = [T^(-1)C^(-2)h]`
D. `[M] = [TC^(-2)h]`

Answer

To find the dimensions of mass in terms of time (T), velocity (C), and angular momentum (h), we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Dimensions of Fundamental Quantities**:
- Time (T) has the dimension: \([T] = T^1\)
- Velocity (C) has the dimension: \([C] = [L][T]^{-1} = L^1 T^{-1}\)
- Angular momentum (h) has the dimension: \([h] = [M][L]^2[T]^{-1} = M^1 L^2 T^{-1}\)

2. **Set Up the Equation**:
We want to express mass (M) in terms of T, C, and h. We can write:
\[ M^1 = T^x \cdot C^y \cdot h^z \]
Substituting the dimensions of C and h:
\[ M^1 = T^x \cdot (L^1 T^{-1})^y \cdot (M^1 L^2 T^{-1})^z \]

3. **Expand the Right Side**:
Expanding the right side gives:
\[ M^1 = T^x \cdot L^y T^{-y} \cdot M^z L^{2z} T^{-z} \]
Combining the terms:
\[ M^1 = M^z \cdot L^{y + 2z} \cdot T^{x - y - z} \]

4. **Equate the Powers**:
Now, we equate the powers of M, L, and T from both sides:
- For M: \(1 = z\)
- For L: \(0 = y + 2z\)
- For T: \(0 = x - y - z\)

5. **Solve the Equations**:
From \(1 = z\), we have:
\[ z = 1 \]
Substituting \(z = 1\) into \(0 = y + 2z\):
\[ 0 = y + 2 \cdot 1 \implies y = -2 \]
Now substituting \(y = -2\) and \(z = 1\) into \(0 = x - y - z\):
\[ 0 = x - (-2) - 1 \implies x + 2 - 1 = 0 \implies x = -1 \]

6. **Final Expression**:
Now we have:
- \(x = -1\)
- \(y = -2\)
- \(z = 1\)

Therefore, we can express mass (M) as:
\[ M^1 = T^{-1} \cdot C^{-2} \cdot h^{1} \]
Thus, the dimensions of mass in terms of T, C, and h are:
\[ [M] = T^{-1} C^{-2} h^{1} \]

### Final Answer:
The dimensions of mass in terms of time (T), velocity (C), and angular momentum (h) are:
\[ [M] = T^{-1} C^{-2} h^{1} \]