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# If C (the velocity of light) g, (the acceleration due to gravity), and P(the atmospheric pressure) are the fundamental quantities in MKS system, then the dimensions of length will be same as that ofA. \$C/g\$ B. \$C/P\$C. \$PCg\$D. \${C^2}/g\$

Last updated date: 02nd Aug 2024
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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}}]\$

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.