JEE Main 2014 — Physics Mechanics

The dimensional formulae of $$\begin{aligned} &\mathrm{e}=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^1 \mathrm{~A}^1\right] \\ &\varepsilon_0=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^4 \mathrm{~A}^2\right] \\ &\mathrm{G}=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right] \\ &\text { and } \mathrm{m}_{\mathrm{e}}=\left[\mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0\right] \\ &\text { Now, } \frac{\mathrm{e}^2}{2 \pi \varepsilon_0 \mathrm{Gm}_{\mathrm{e}}^2} \\ &=\frac{\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^1 \mathrm{~A}^1\right]^2}{2 \pi\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]\left[\mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0\right]^2} \end{aligned}$$​e=[M0 L0 T1 A1]ε0​=[M−1 L3 T4 A2]G=[M−1 L3 T−2] and me​=[M1 L0 T0] Now, 2πε0​Gme2​e2​=2π[M−1 L−3 T4 A2][M−1 L3 T−2][M1 L0 T0]2[M0 L0 T1 A1]2​​ $$\begin{aligned} &=\frac{\left[\mathrm{T}^2 \mathrm{~A}^2\right]}{2 \pi\left[\mathrm{M}^{-1-1+2} \mathrm{~L}^{-3+3} \mathrm{~T}^{4-2} \mathrm{~A}^2\right]} \\ &=\frac{\left[\mathrm{T}^2 \mathrm{~A}^2\right]}{2 \pi\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^2 \mathrm{~A}^2\right]}=\frac{1}{2 \pi} \end{aligned}$$​=2π[M−1−1+2 L−3+3 T4−2 A2][T2 A2]​=2π[M0 L0 T2 A2][T2 A2]​=2π1​​ $\because \frac{1}{2 \pi}$∵2π1​ is dimensionless thus the combination $\frac{\mathrm{e}^2}{2 \pi \varepsilon_0 \mathrm{Gm}_{\mathrm{e}}^2}$2πε0​Gme2​e2​ would have the same value in different systems of units.