Question

A physical quantity of the dimensions of mass that can be formed out of c, G and $\dfrac{{{e^2}}}{{8\pi {\varepsilon _o}}}$ is (c is the velocity of light, G is universal Gravitational constant and e is charge, is electrical permittivity):
A. $c{\left[ {G\dfrac{{{e^2}}}{{8\pi {\varepsilon _o}}}} \right]^{\dfrac{1}{2}}}$
B. ${c^0}{\left[ {G\dfrac{{{e^2}}}{{8\pi {\varepsilon _o}}}} \right]^2}$
C. ${c^0}{\left[ {\dfrac{{{e^2}}}{{G8\pi {\varepsilon _o}}}} \right]^{\dfrac{1}{2}}}$
D. ${c^2}{\left[ {G\dfrac{{{e^2}}}{{8\pi {\varepsilon _o}}}} \right]^{\dfrac{1}{2}}}$

Answer 1

This question belongs to the category of dimensional analysis. We want to construct a dimension of mass i.e. kilogram from the given parameters. Just by random hit and trial method we can do the problem or we can go option by option and just put away the dimensions in the given expression and find the dimension of mass.

Complete step by step answer: It is a Dimensional analysis question and is quite famous among the examiners of both boards and JEE level. In this type of questions, we put together dimensions of all the given parameters and try to adjust the dimensions to get the required dimension.
Here since the expressions are already given, we don’t have to build up our own through conjectures. So firstly, the dimensions of given parameters are:
$$
c = \dfrac{L}{T} \\
G = \dfrac{{{L^3}}}{{M{T^2}}} \\
\varepsilon = \dfrac{{{T^4}{A^2}}}{{{L^3}M}} \\
e = AT \\

Where, M= mass L= length T=time A=electric current Now, we need the dimension of mass M, so either we bring together all the parameters then assign them powers and equate it with the dimension of mass by solving 4 linear equations Or we can go option by option. The latter one is much easier here or else we have to solve 4 linear equations to get 4 unknowns, so, By option A: $$\dfrac{L}{T}{\left[ {\dfrac{{{L^3}}}{{M{T^2}}}\dfrac{{{A^2}{T^2}.{L^3}M}}{{{T^4}{A^2}}}} \right]^{\dfrac{1}{2}}} = \dfrac{{{L^4}}}{{{T^3}}} \ne M$$ By option B: $${\left[ {\dfrac{{{L^3}}}{{M{T^2}}}\dfrac{{{A^2}{T^2}.{L^3}M}}{{{T^4}{A^2}}}} \right]^{\dfrac{1}{2}}} = \dfrac{{{L^3}}}{{{T^2}}} \ne M$$ By option C: $${\left[ {\dfrac{{M{T^2}}}{{{L^3}}}\dfrac{{{A^2}{T^2}.{L^3}M}}{{{T^4}{A^2}}}} \right]^{\dfrac{1}{2}}} = M = M$$ By option D: $$\dfrac{{{L^2}}}{{{T^2}}}{\left[ {\dfrac{{M{T^2}}}{{{L^3}}}\dfrac{{{A^2}{T^2}.{L^3}M}}{{{T^4}{A^2}}}} \right]^{\dfrac{1}{2}}} = \dfrac{{{L^2}M}}{{{T^2}}} \ne M$$ Therefore, the correct answer is option C. **Note:** The other method for when the question is subjective and options are not given is as follows: Let $$ M = {c^a}{G^b}{\varepsilon ^c}{e^d} \\\ M = {L^{a + 3b - 3c}}{M^{ - b - c}}{T^{ - a - 2b + 4c + d}}{A^{2c + d}} \\\

Now solving the 4 linear equations by comparing both the sides we will get the values of a, b, c and d which we can substitute back and get the desired result.