Answer
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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.
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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