Physics - Unit & Dimension Question with Solution | TestHub

$$\begin{array}{rl}& \text{ Hint : }[\mathrm{h}]=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-1}\right]\\ & [\mathrm{c}]=\left[{\mathrm{LT}}^{-1}\right]\\ & [\mathrm{m}]=\left[{\mathrm{M}}^1{\mathrm{L}}^0{\mathrm{T}}^0\right]\\ & \text{ Dimension of length }\\ & [\mathrm{L}]={\mathrm{h}}^{\mathrm{x}}{\mathrm{c}}^{\mathrm{y}}{\mathrm{m}}^z\\ & \left[{\mathrm{L}}^1\right]={\left[{\mathrm{ML}}^2{\mathrm{T}}^{-1}\right]}^{\mathrm{x}}{\left[{\mathrm{LT}}^{-1}\right]}^{\mathrm{y}}{\left[{\mathrm{M}}^1{\mathrm{L}}^0{\mathrm{T}}^0\right]}^{\mathrm{z}}\end{array}$$ Dimensional analysis on mass $$0=x+z\Rightarrow x=-z\Rightarrow z=-1$$ Dimensional analysis on length $$1=2x+y\Rightarrow 2x-x=1\Rightarrow x=1$$ Dimensional analysis on time $$0=-x-y\Rightarrow -x=+y\Rightarrow y=-1$$ Therefore final formula is $\frac{h}{\mathrm{mc}}$