From the following combinations of physical constants (expressed through their usual symbols) the only…

From the following combinations of physical constants (expressed through their usual symbols) the only combination, that would have the same value in different systems of units, is:
  1. $\frac{\mathrm{ch}}{2 \pi \varepsilon_{\mathrm{o}}^2}$
  2. $\frac{\mathrm{e}^2}{2 \pi \varepsilon_{\mathrm{o}} \mathrm{Gm}_{\mathrm{e}}^2}$
  3. $\frac{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}{\mathrm{c}^2} \frac{\mathrm{G}}{\mathrm{he}^2}$
  4. $\frac{2 \pi \sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}}{\mathrm{ce}^2} \frac{\mathrm{h}}{\mathrm{G}}$

Solution

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} $ $ \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} $ $\because \frac{1}{2 \pi}$ is dimensionless thus the combination $\frac{\mathrm{e}^2}{2 \pi \varepsilon_0 \mathrm{Gm}_{\mathrm{e}}^2}$ would have the same value in different systems of units

Asked in: JEE Main 2014 (12 Apr Online)

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