The velocity of a body falling through a viscous medium varies with time $t$ as $v = \dfrac{a}{b}\left(1 - e^{-bt}\right)$, where $a$ and $b$ are constants. The dimensions of $a$ and $b$ are, respectively,
Answer: (A) $[\text{LT}^{-2}]$ and $[\text{T}^{-1}]$
Concept. The argument of an exponential must be dimensionless, and the bracket $\left(1 - e^{-bt}\right)$ is a pure number, so $\dfrac ab$ alone carries the dimensions of velocity.
Step 1. Dimensions of $b$.
$$[bt] = 1 \;\Rightarrow\; [b] = [\text{T}^{-1}]$$
Step 2. Dimensions of $a$.
$$\left[\frac ab\right] = [\text{LT}^{-1}] \;\Rightarrow\; [a] = [\text{LT}^{-1}][\text{T}^{-1}] = [\text{LT}^{-2}]$$
($a$ is the initial acceleration and $\dfrac ab$ the terminal speed.)
Answer. $[\text{LT}^{-2}]$ and $[\text{T}^{-1}]$ (Option A).
Solution by Sreeraj P, M.Sc Physics