For the situation shown in figure, mark the correct options.

Maximum force of friction between 2 kg and 4 kg   $=0.4 X 2 X 10 = 8 N$

2 kg moves due to friction. Therefore its maximum acceleration may be  ${\text{a}}_{\text{max}}=\frac{8}{2}=4m/s^2$
Slip will start when their combined acceleration becomes $4 m/s^2$
$\begin{array}{l}\therefore \text{a }=\frac{F}{m}or 4=\frac{2t}{6}or t=12s\\ At t=3s\\ {\text{a}}_{\text{2}}=a_4=\frac{F}{m}=\frac{2t}{6}=\frac{2\times 3}{6}\\ =1m/s^2\end{array}$

Both $a_2$ and $a_4$ are towards right. Therefore pseudo forces $F_1$ (on 2 kg from 4 kg) and $F_2$ (on 4 kg from 2kg) are towards left
 $F_1 = \left(2\right) \left(1\right) =2N$
 $F_2= \left(4\right) \left(1\right) = 4N$

From here we can see that $F_1$ and $F_2$ do not make a pair of equal and opposite forces.