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
JEE MAIN 2014, Diffcult
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The Dimensional formulae of
$e=\left[M^0 L^0 T^1 A^1\right]$
$\varepsilon_0=\left[M^{-1} L^{-3} T^4 A^2\right]$
$G=\left[M^{-1} L^3 T^{-2}\right]$
and $m_e=\left[M^1 L^0 T^0\right]$
Now, $\frac{e^2}{2 \pi \varepsilon_0 G m_e^2}$
$=\frac{\left[M^0 L^0 T^1 A^1\right]^2}{2 \pi\left[M^{-1} L^{-3} T^4 A^2\right]\left[M^{-1} L^3 T^{-2}\right]\left[M^1 L^0 T^0\right]^2}$
$=\frac{\left[T^2 A^2\right]}{2 \pi\left[M^{-1+12} L^{-3+3} T^{4-2} A^2\right]}$
$=\frac{\left[T^2 A^2\right]}{2 \pi\left[M^0 L^0 T^2 A^2\right]}$
$=\frac{1}{2 \pi}$
$\frac{1}{2 \pi}$ is Dimensionless thus the combination
$\frac{e^2}{2 \pi \varepsilon_0 G m_e^2}$
would $d$ have the same value in diffierent systems of units
$e=\left[M^0 L^0 T^1 A^1\right]$
$\varepsilon_0=\left[M^{-1} L^{-3} T^4 A^2\right]$
$G=\left[M^{-1} L^3 T^{-2}\right]$
and $m_e=\left[M^1 L^0 T^0\right]$
Now, $\frac{e^2}{2 \pi \varepsilon_0 G m_e^2}$
$=\frac{\left[M^0 L^0 T^1 A^1\right]^2}{2 \pi\left[M^{-1} L^{-3} T^4 A^2\right]\left[M^{-1} L^3 T^{-2}\right]\left[M^1 L^0 T^0\right]^2}$
$=\frac{\left[T^2 A^2\right]}{2 \pi\left[M^{-1+12} L^{-3+3} T^{4-2} A^2\right]}$
$=\frac{\left[T^2 A^2\right]}{2 \pi\left[M^0 L^0 T^2 A^2\right]}$
$=\frac{1}{2 \pi}$
$\frac{1}{2 \pi}$ is Dimensionless thus the combination
$\frac{e^2}{2 \pi \varepsilon_0 G m_e^2}$
would $d$ have the same value in diffierent systems of units
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