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**Hint:**In this question, we will simply equate the dimensions of C, g, P with the dimensions of length and compare them.

Dimensions of velocity of light is given by-

Dim $C = [{M^0}{L^1}{T^{ - 1}}]$

Dimensions of acceleration of gravity is given by-

Dim $g = [{M^0}{L^1}{T^{ - 2}}]$

Dimensions of atmospheric pressure is given by-

Dim $P = [{M^1}{L^{ - 1}}{T^{ - 2}}]$

**Complete step-by-step answer:**

Given: Velocity of light=C

Acceleration due to gravity - g

Pressure -P

Let 'L' be the length.

Let's assume,

$L = {C^x}{g^y}{P^z}$ …eq i)

Now putting all the values in above equation,

$[{M^0}L{T^0}]$=${[{M^0}{L^1}{T^{ - 1}}]^x}{[{M^0}{L^1}{T^{ - 2}}]^y}{[{M^1}{L^{ - 1}}{T^{ - 2}}]^z}$

⇒$[{M^0}{L^1}{T^0}] = [{M^z}{L^{x + y - z}}{T^{ - x - 2y - 2z}}]$

Applying the principle of homogeneity,

Comparing the powers of M,

$z = 0$

Compare the powers of L,

$x + y - z = 1$ ( As z=0)

⇒$x + y = 1$

⇒$x = 1 - y$ …eq ii)

Compare the powers of T,

$ - x - 2y - 2z = 0$ ( As z=0)

⇒$ - x - 2y = 0$ …eq iii)

Put the value of eq ii) in above equation, we get

⇒ $ - (1 - y) - 2y = 0$

⇒$y = - 1$

Putting the value of y in eq ii), we get

⇒$x = 2$

Now, put the values of x, y and z in eq i)

Therefore, the dimensions of length will be:

$L = {C^2}{g^{ - 1}}{P^0}$

⇒$L = \dfrac{{{C^2}}}{g}$

**Hence, option D is the correct answer.**

**Note:**Principle of Homogeneity: It states that dimensions of each of the terms of a dimensional equation on both sides should be the same. This principle is helpful because it helps us convert the units from one form to another.

Applications of Dimensional Analysis:

1. To check the consistency of a dimensional equation.

2. To derive the relation between physical quantities in physical phenomena.

3. To change units from one system to another.

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