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Question: The dimensional formula of \(\dfrac{{{e^2}}}{{\varepsilon _0^2mpm_c^2{c^3}G}}\) where e \( = \) elec...

The dimensional formula of e2ε02mpmc2c3G\dfrac{{{e^2}}}{{\varepsilon _0^2mpm_c^2{c^3}G}} where e == electronic charge, c == speed of light, mp == mass of proton; me == mass of electron, G == gravitational constant is
(A) [M][M]
(B) [I−2T−1][{I^{ - 2}}{T^{ - 1}}]
(C) [IT][IT]
(D) [I4T2M][{I^4}{T^2}M]

Explanation

Solution

We will convert the quantities given in question in terms of fundamental quantities. i.e., L,M,T.
And for that first we will write formulas for all the quantities. Then we will solve the above question step by step.

Complete step by step solution:
Here in the above question some quantities are given, let’s write their dimensions first.
e-electric charge
We know that charge, q == current ×\times time
Dimensions of current =I = I
Dimensions of time =T = T
So, e=[I1T1]e = [{I^1}{T^1}]
Mp and Me -
These are masses of proton and electron respectively.
Dimensions of mass == M
So,
Mp=[M1]Mp = [{M^1}]
Me=[M1]Me = [{M^1}]
C-speed of light -
Formula for speed == distance // time
So, dimensions of speed is,
Distance // time =L/T = L/T
So, c=[L1T−1]c = [{L^1}{T^{ - 1}}]
G, gravitational constant
As we know that G=Fr2/m2G = F{r^2}/{m^2}
r, is distance between two objects. So its dimensions is L1{L^1}
m, mass of object
So, dimensions of m=Mm = M
Force, F == mass // acceleration
Mass == M
Acceleration == speed // time
From above speed is [L1T−1][{L^1}{T^{ - 1}}]
So acceleration =[L1T−1]/[T] = [{L^1}{T^{ - 1}}]/[T]
So dimension of F is,
F=mass×acceleration=M×LT−2F = mass \times acceleration = M \times L{T^{ - 2}}
So, F=[M1L1T−2]F = [{M^1}{L^1}{T^{ - 2}}]
Or, dimension of G=[M1L1T−2]×[L2]/[M2]G = [{M^1}{L^1}{T^{ - 2}}] \times [{L^2}]/[{M^2}]
G=M−1L3T−2G = {M^{ - 1}}{L^3}{T^{ - 2}}
Eo={E^o} = epsilon note (vacuum permittivity)
Formula for electric force :
Fq=kq1q2r2{F_q} = \dfrac{{k{q_1}{q_2}}}{{{r^2}}}
or Fq=14πε0q1q2r2{F_q} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}
∈0=14πFq×q1q2r2{ \in _0} = \dfrac{1}{{4\pi {F_q}}} \times \dfrac{{{q_1}{q_2}}}{{{r^2}}}
Dimensions of Fq=M1L1T−2{F_q} = {M^1}{L^1}{T^{ - 2}}
Dimensions of charge q=I1T1q = {I^1}{T^1}
Dimensions of distance, r=Lr = L
So, ∈0[I1T1]2I[M1L1T−2][L2]{ \in _0}\dfrac{{{{[{I^1}{T^1}]}^2}I}}{{[{M^1}{L^1}{T^{ - 2}}][{L^2}]}} (14π\dfrac{1}{{4\pi }} is constant)
∈0=I2T2M1L3T−2=M−1L−3T4I2{ \in _0} = \dfrac{{{I^2}{T^2}}}{{{M^1}{L^3}{T^{ - 2}}}} = {M^{ - 1}}{L^{ - 3}}{T^4}{I^2}
We have dimensions of all the terms given in question.
So, substituting the above calculated dimensions.
Dimensional formula of e2ε02mpmc2c3G\dfrac{{{e^2}}}{{\varepsilon _0^2mpm_c^2{c^3}G}}
[I1T1]2[M−1L−3T4I2]2[M1][M1]2[LT−1]3[M−1L3T−2]\dfrac{{{{[{I^1}{T^1}]}^2}}}{{{{[{M^{ - 1}}{L^{ - 3}}{T^4}{I^2}]}^2}[{M^1}]{{[{M^1}]}^2}{{[L{T^{ - 1}}]}^3}[{M^{ - 1}}{L^3}{T^{ - 2}}]}}
=I2T2M+2+1+2−1L−6+3+3T8−3−2I4= \dfrac{{{I^2}{T^2}}}{{{M^{ + 2 + 1 + 2 - 1}}{L^{ - 6 + 3 + 3}}{T^{8 - 3 - 2}}{I^4}}}
=I2T2M0L0T3I4= \dfrac{{{I^2}{T^2}}}{{{M^0}{L^0}{T^3}{I^4}}}
=I2−4T2−3M0L0= {I^{2 - 4}}{T^{2 - 3}}{M^0}{L^0}
=I−2T−1M0L0= {I^{ - 2}}{T^{ - 1}}{M^0}{L^0}
or, dimensional formula for e2ε02mpmc2c3G\dfrac{{{e^2}}}{{\varepsilon _0^2mpm_c^2{c^3}G}} is T−1I−2{T^{ - 1}}{I^{ - 2}}

Therefore, option B i.e., I−2T−1{I^{ - 2}}{T^{ - 1}} is the correct option.

Note: 1. To write dimensional formulas for any quantity we should first write them in the terms of fundamental quantities. Then one by one we can write dimensions of all the quantities like we did in the hint section or in the solution section.
2. We should know all the basic formulas like that of speed, acceleration, force etc.
3. We can learn the dimensions of basic quantities so that we can skip the lengthy process of finding.