Given equation

$v = At^2 + \frac{Bt}{C+t}$

Since $v$ is velocity, its dimension is

$[v] = [LT^{-1}]$

Find dimension of $A$

The term $At^2$ must have the same dimension as velocity:

$[A][t^2] = [LT^{-1}]$

Since $[t]=[T]$,

$[A][T^2] = [LT^{-1}]$

$[A] = [LT^{-3}]$

Find dimension of $C$

In the denominator, $(C+t)$ is a sum, so $C$ and $t$ must have the same dimension:

$[C] = [T]$

Find dimension of $B$

The second term must also have dimension of velocity:

$\left[\frac{Bt}{C+t}\right] = [LT^{-1}]$

Since $[C+t]=[T]$, we get

$\left[\frac{Bt}{T}\right] = [B]$

because $t$ also has dimension $T$, so

$\frac{[B][T]}{[T]} = [B]$

Therefore,

$[B] = [LT^{-1}]$

Find dimension of $ABC$

$[ABC] = [A][B][C]$

$= [LT^{-3}]\,[LT^{-1}]\,[T]$

$= [L^2T^{-3}]$

Including mass exponent explicitly,

$[ABC] = [M^0L^2T^{-3}]$

Match with options

This corresponds to Option B.