Q 11-01-043JEE MainHard
The energy of a system is given by $E = A\,e^{-\beta x^2/(kT)}$, where $x$ is a distance, $k$ is Boltzmann's constant and $T$ is temperature. The dimensions of $A\beta$ are
Answer: (B) $[M^2L^2T^{-4}]$
The exponent must be dimensionless. Since $kT$ is an energy:
$$[\beta][L^2] = [ML^2T^{-2}] \Rightarrow [\beta] = [MT^{-2}]$$
The exponential is dimensionless, so $A$ has the dimensions of energy: $[A] = [ML^2T^{-2}]$.
$$[A\beta] = [ML^2T^{-2}][MT^{-2}] = [M^2L^2T^{-4}]$$
Solution by Sreeraj P, M.Sc Physics