Units & Dimensions

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134 of 149

Question 134 of 149

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:

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Answer: (B) e22πεoGme2\frac{\mathrm{e}^2}{2 \pi \varepsilon_{\mathrm{o}} \mathrm{Gm}_{\mathrm{e}}^2}

The dimensional formulae of
e=[M0 L0 T1 A1]ε0=[M−1 L3 T4 A2]G=[M−1 L3 T−2] and me=[M1 L0 T0] Now, e22πε0Gme2=[M0 L0 T1 A1]22π[M−1 L−3 T4 A2][M−1 L3 T−2][M1 L0 T0]2\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}

=[T2 A2]2π[M−1−1+2 L−3+3 T4−2 A2]=[T2 A2]2π[M0 L0 T2 A2]=12π\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}
∵12π\because \frac{1}{2 \pi} is dimensionless thus the combination

e22πε0Gme2\frac{\mathrm{e}^2}{2 \pi \varepsilon_0 \mathrm{Gm}_{\mathrm{e}}^2} would have the same value in different systems of units.

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