If C ( the velocity of light ) g , ( the acceleration due to gravity), P ( the atmospheric pressure) are the fundamental quantities in MKS system , then the dimensions of length will be same as that of

Options

A. `C//g`
B. `C//P`
C. `PCg`
D. `C^(2)//g`

Explanation

To solve the problem, we need to find the dimensions of length in terms of the given fundamental quantities: the velocity of light (C), the acceleration due to gravity (g), and atmospheric pressure (P).

### Step-by-Step Solution:

1. **Identify the Dimensions of Each Quantity:**
- The velocity of light (C) has dimensions of velocity, which is given by:
\[ [C] = LT^{-1} \]
- The acceleration due to gravity (g) has dimensions of acceleration, which is:
\[ [g] = LT^{-2} \]
- Atmospheric pressure (P) is defined as force per unit area. The dimensions of force (mass × acceleration) are:
\[ [F] = [M][a] = MLT^{-2} \]
Therefore, the dimensions of pressure are:
\[ [P] = \frac{[F]}{[A]} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \]

2. **Formulate Relationships:**
- We will now form ratios of these quantities to find a relationship with length (L).
- First, consider the ratio \( \frac{C}{g} \):
\[ \frac{[C]}{[g]} = \frac{LT^{-1}}{LT^{-2}} = T \]
This shows that the dimensions of \( \frac{C}{g} \) yield time (T).

- Next, consider the ratio \( \frac{C}{P} \):
\[ \frac{[C]}{[P]} = \frac{LT^{-1}}{ML^{-1}T^{-2}} = \frac{L^2T}{M} \]
This indicates that the dimensions of \( \frac{C}{P} \) yield \( \frac{L^2T}{M} \).

3. **Combine Ratios to Find Length:**
- Now, we can combine the ratios to isolate length. Consider the expression \( \frac{C^2}{gP} \):
\[ \frac{[C]^2}{[g][P]} = \frac{(LT^{-1})^2}{(LT^{-2})(ML^{-1}T^{-2})} \]
This simplifies to:
\[ = \frac{L^2T^{-2}}{(LT^{-2})(ML^{-1}T^{-2})} = \frac{L^2T^{-2}}{ML^{0}T^{-4}} = \frac{L^2T^{-2}}{M} \cdot T^4 \]
Thus, we have:
\[ = \frac{L^2}{M} \cdot T^2 \]
Rearranging gives us:
\[ L = \frac{C^2}{gP} \]
This shows that the dimensions of length can be expressed in terms of the quantities C, g, and P.

4. **Conclusion:**
- The dimensions of length are the same as the dimensions derived from the combination of C, g, and P, confirming that length can be expressed in terms of these fundamental quantities.

### Final Answer:
The dimensions of length will be the same as that of \( \frac{C^2}{gP} \).

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