The velocity of a body moving in a viscous medium is given by $v = \dfrac{A}{B+C}\left[1 - e^{\frac{-t}{B}}\right]$ where t is time; dimensions of A and C are

The exponent of an exponential function must be dimensionless. Therefore, the dimension of $\dfrac{t}{B}$ is $[M^0L^0T^0]$.

$[B] = [t] = [M^0L^0T^1]$

By the principle of homogeneity, quantities added or subtracted must have the same dimensions. Since $B$ and $C$ are added in the denominator, $[C] = [B] = [M^0L^0T^1]$.

The term $\left[1 - e^{\dfrac{-t}{B}}\right]$ is dimensionless. Thus, the dimensions of velocity $v$ must equal the dimensions of $\dfrac{A}{B+C}$.

$[v] = \dfrac{[A]}{[B+C]}$

$[M^0L^1T^{-1}] = \dfrac{[A]}{[M^0L^0T^1]}$

$[A] = [M^0L^1T^{-1}] \times [M^0L^0T^1] = [M^0L^1T^0]$

The dimensions of $A$ and $C$ are $[M^0L^1T^0]$ and $[M^0L^0T^1]$ respectively.

Answer: $[M^0L^1T^0],\ [M^0L^0T^1]$