Match List - I with List - II, where $h$, $G$ and $c$ are taken as fundamental units.
$$\begin{array}{|l|l|l|l|}\hline & \text{List - I} & & \text{List - II} \\ \hline \text{A.} & \text{Meter (L)} & \text{I.} & \sqrt{\dfrac{hc}{G}} \\ \hline \text{B.} & \text{Second (S)} & \text{II.} & \sqrt{\dfrac{Gh}{c^5}} \\ \hline \text{C.} & \text{Kilogram (M)} & \text{III.} & \sqrt{\dfrac{K^2L^2c^3}{Gh}} \\ \hline \text{D.} & \text{Kelvin (K)} & \text{IV.} & \sqrt{\dfrac{Gh}{c^3}} \\ \hline \end{array}$$
Choose the correct answer from the options given below :
Answer: (B) A-IV, B-II, C-I, D-III
Concept. With $h$, $G$ and $c$ as base units, length, time and mass are the Planck units. Find each one by making combinations of $h$, $G$, $c$ in which the unwanted dimensions cancel.
Step 1. Mass. As in the previous question, $\left[\dfrac{hc}{G}\right]=M^2$, so
$$M=\sqrt{\frac{hc}{G}}\quad\to\ \text{I}$$
Step 2. Length. Try $\dfrac{Gh}{c^3}$:
$$\left[\frac{Gh}{c^3}\right]=\frac{M^{-1}L^3T^{-2}\cdot ML^2T^{-1}}{L^3T^{-3}}=\frac{L^5T^{-3}}{L^3T^{-3}}=L^2$$
So length $=\sqrt{\dfrac{Gh}{c^3}}\ \to$ IV.
Step 3. Time. Time $=$ length$/$speed $=\dfrac1c\sqrt{\dfrac{Gh}{c^3}}=\sqrt{\dfrac{Gh}{c^5}}\ \to$ II.
Step 4. Temperature. The only expression left, III, is the one that brings in the temperature; it is matched with kelvin by elimination.
Answer. A-IV, B-II, C-I, D-III (Option B). Remember: length $\sqrt{Gh/c^3}$, time $\sqrt{Gh/c^5}$, mass $\sqrt{hc/G}$.
Solution by Sreeraj P, M.Sc Physics
