Q 11-01-210JEE MainJEE Main 2020 (4 Sep, Shift 2)Medium
A quantity $x$ is given by $\left(\dfrac{IFv^{2}}{WL^{4}}\right)$ in terms of moment of inertia $I$, force $F$, velocity $v$, work $W$ and length $L$. The dimensional formula for $x$ is same as that of:
Answer: (C) energy density
$$[x] = \frac{ML^{2}\cdot MLT^{-2}\cdot L^{2}T^{-2}}{ML^{2}T^{-2}\cdot L^{4}} = ML^{-1}T^{-2}$$
Energy density (energy per unit volume) has dimensions $\dfrac{ML^{2}T^{-2}}{L^{3}} = ML^{-1}T^{-2}$.
Solution by Sreeraj P, M.Sc Physics