Q 11-01-168JEE MainJEE Main 2022 (27 Jul, Shift 2)Medium
An expression of energy density is given by $u = \dfrac{\alpha}{\beta}\sin\left(\dfrac{\alpha x}{kt}\right)$, where $\alpha$, $\beta$ are constants, $x$ is displacement, $k$ is Boltzmann constant and $t$ is the temperature. The dimensions of $\beta$ will be
Answer: (D) $[\text{M}^0\text{L}^2\text{T}^0]$
The argument of sine is dimensionless, so $[\alpha] = \dfrac{[kt]}{[x]}$. Since $kt$ is an energy,
$$[\alpha] = \frac{[\text{ML}^2\text{T}^{-2}]}{[\text{L}]} = [\text{MLT}^{-2}]$$
Energy density $[u] = [\text{ML}^{-1}\text{T}^{-2}]$, and $[u] = \dfrac{[\alpha]}{[\beta]}$:
$$[\beta] = \frac{[\text{MLT}^{-2}]}{[\text{ML}^{-1}\text{T}^{-2}]} = [\text{M}^0\text{L}^2\text{T}^0]$$
Solution by Sreeraj P, M.Sc Physics