Q 11-01-042JEE MainMedium
The force on a particle is given by $F = a\sqrt{x} + bt^2$, where $x$ is position and $t$ is time. The dimensions of $\dfrac{a}{b}$ are
Answer: (B) $[L^{-1/2}T^{2}]$
Each term has the dimensions of force.
$[a][L^{1/2}] = [MLT^{-2}] \Rightarrow [a] = [ML^{1/2}T^{-2}]$
$[b][T^2] = [MLT^{-2}] \Rightarrow [b] = [MLT^{-4}]$
$$\left[\frac{a}{b}\right] = \frac{[ML^{1/2}T^{-2}]}{[MLT^{-4}]} = [L^{-1/2}T^{2}]$$
Solution by Sreeraj P, M.Sc Physics