Q 11-01-123JEE MainJEE Main 2024 (31 Jan, Shift 1)Medium
A force is represented by $F = ax^2 + bt^{1/2}$, where $x$ = distance and $t$ = time. The dimensions of $\dfrac{b^2}{a}$ are:
Answer: (A) $[\mathrm{ML^3T^{-3}}]$
$[a] = \dfrac{[F]}{[x^2]} = \mathrm{ML^{-1}T^{-2}}$ and $[b] = \dfrac{[F]}{[t^{1/2}]} = \mathrm{MLT^{-5/2}}$.
$$\left[\frac{b^2}{a}\right] = \frac{\mathrm{M^2L^2T^{-5}}}{\mathrm{ML^{-1}T^{-2}}} = \mathrm{ML^3T^{-3}}$$
Solution by Sreeraj P, M.Sc Physics