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Q.
Dimensions of ohm are same as (where $h$ is Planck's constant and $e$ is charge)
UP CPMTUP CPMT 2015
Solution:
$\left[\frac{h}{e^{2}}\right]=\frac{\left[ML^{2}T^{-1}\right]}{\left[AT\right]^{2}}=\left[ML^{2}T^{-3}A^{-2}\right]$
[Resistance] $=\frac{\text{[Potential difference]}}{\text{[Current]}}$
$=\frac{\left[ML^{2}T^{-3}A^{-1}\right]}{\left[A\right]}=\left[ML^{2}T^{-3}A^{-2}\right]$
The $SI$ unit of resistance is ohm
Therefore, the dimensions of ohm are same as that of $\frac{h}{e^{2}}$