Q 11-01-206JEE MainJEE Main 2020 (2 Sep, Shift 2)Easy
If momentum $(P)$, area $(A)$ and time $(T)$ are taken to be the fundamental quantities then the dimensional formula for energy is:
Answer: (C) $[PA^{1/2}T^{-1}]$
Let $E = P^{x}A^{y}T^{z}$, i.e. $ML^{2}T^{-2} = (MLT^{-1})^{x}(L^{2})^{y}T^{z}$.
Mass: $x = 1$. Length: $x + 2y = 2 \Rightarrow y = \tfrac12$. Time: $-x + z = -2 \Rightarrow z = -1$.
$$E = [PA^{1/2}T^{-1}]$$
Solution by Sreeraj P, M.Sc Physics