JEE Main 2021 — Units and Dimensions Question with Solution

The relation between wavelength and the principal quantum number is, $\frac{1}{\lambda }=R\left(\frac{1}{{\left(n_1\right)}^2}-\frac{1}{{\left(n_2\right)}^2}\right)$, for hydrogen-like atom, so the unit to Rydberg constant is, ${\mathrm{m}}^{-1}$.

Now recall the formula of energy of a photon in terms of frequency, $E=h\nu$, it means, $h=\frac{E}{\nu }=\frac{\mathrm{kg} {\mathrm{m}}^2 {\mathrm{s}}^{-2}}{{\mathrm{s}}^{-1}}=\mathrm{kg} {\mathrm{m}}^2 {\mathrm{s}}^{-1}$.

The coefficient of viscosity,
 $\eta =\frac{F}{A\times {\displaystyle \frac{dv}{dz}}}=\frac{\mathrm{kg} \mathrm{m} {\mathrm{s}}^{-2} \mathrm{m}}{{\mathrm{m}}^2\times \mathrm{m} {\mathrm{s}}^{-1}}=\mathrm{kg} {\mathrm{m}}^{-1} {\mathrm{s}}^{-1}.$ 

Energy density is defined as, $\frac{E}{V}=\frac{\mathrm{kg} {\mathrm{m}}^2 {\mathrm{s}}^{-2}}{{\mathrm{m}}^3}=\mathrm{kg} {\mathrm{m}}^{-1} {\mathrm{s}}^{-2}$.