Match List - I with List - II.
$$\begin{array}{|l|l|l|l|}\hline & \text{List - I} & & \text{List - II} \\ \hline \text{A.} & \text{Boltzmann constant} & \text{I.} & [\text{M}^{-1}\text{L}^3\text{T}^{-2}] \\ \hline \text{B.} & \text{Stefan's constant} & \text{II.} & [\text{ML}^2\text{T}^{-1}] \\ \hline \text{C.} & \text{Planck's constant} & \text{III.} & [\text{ML}^2\text{T}^{-2}\text{K}^{-1}] \\ \hline \text{D.} & \text{Gravitational constant} & \text{IV.} & [\text{ML}^0\text{T}^{-3}\text{K}^{-4}] \\ \hline \end{array}$$
Choose the correct answer from the options given below :
Answer: (C) A-III, B-IV, C-II, D-I
Concept. A constant gets its dimensions from the law in which it appears: rearrange the law so the constant stands alone, then put in the dimensions of everything else. Energy is $[ML^2T^{-2}]$ and force is $[MLT^{-2}]$.
Step 1. Boltzmann constant. The mean energy of a particle is of order $kT$, so $k=E/T$:
$$[k]=\frac{ML^2T^{-2}}{K}=[ML^2T^{-2}K^{-1}]\quad\to\ \text{III}$$
Step 2. Stefan's constant. Stefan's law gives power per unit area: $P/A=\sigma T^4$. Power per area is $\dfrac{ML^2T^{-3}}{L^2}=MT^{-3}$, so
$$[\sigma]=\frac{MT^{-3}}{K^4}=[ML^0T^{-3}K^{-4}]\quad\to\ \text{IV}$$
Step 3. Planck's constant. $E=h\nu$ with $[\nu]=T^{-1}$:
$$[h]=\frac{ML^2T^{-2}}{T^{-1}}=[ML^2T^{-1}]\quad\to\ \text{II}$$
Step 4. Gravitational constant. $F=\dfrac{Gm^2}{r^2}$ gives $G=\dfrac{Fr^2}{m^2}$:
$$[G]=\frac{MLT^{-2}\cdot L^2}{M^2}=[M^{-1}L^3T^{-2}]\quad\to\ \text{I}$$
Answer. A-III, B-IV, C-II, D-I (Option C). Note that $h$ has the same dimensions as angular momentum.
Solution by Sreeraj P, M.Sc Physics
