A force $F$ is given by $F = a t + b t^{2}$, where $t$ is time. What are the dimensions of $a$ and $b$? [PMT/NEET-2001]
a. $MLT^{-3}$ and $ML^2T^{-4}$
b. $MLT^{-3}$ and $MLT^{-4}$
c. $MLT^{-1}$ and $MLT^{0}$
d. $MLT^{-4}$ and $MLT^{1}$

Answer: (b) $a: MLT^{-3}$ and $b: MLT^{-4}$.

Solution:
Force has dimensions $ [F] = MLT^{-2} $. In the expression $F = a t + b t^{2}$, each term must carry the same dimensions as force because you can only add like-dimensioned quantities.

For the term $a t$:

$[a]\,[t] = MLT^{-2}$.

Since $[t] = T$, it follows that

$[a] = MLT^{-2}/T = MLT^{-3}$.

For the term $b t^{2}$:

$[b]\,[t]^{2} = MLT^{-2}$.

With $[t]^{2} = T^{2}$, we get

$[b] = MLT^{-2}/T^{2} = MLT^{-4}$.

Unit check for intuition: $a$ has units of newton per second (N s$^{-1}$) which is kg m s$^{-3}$, and $b$ has units of newton per second squared (N s$^{-2}$) which is kg m s$^{-4}$, matching the dimensional results.