In the given figure a block of mass m is tied on a wedge by an ideal string as shown in figure. String is parallel to inclined plane. All the surface involved are smooth.  Wedge is spring moved towards right with a time varying velocity $v=t^2/2 (m/s)$. At what time block will just leave the contact with the wedge (take $g=10 m/s^2$)

The block will lift once the pseudo force acting on the wedge becomes equal to the weight of the block.

As the wedge is moving with the velocity of $\frac{t^2}{t}m/s$, its acceleration will become $\frac{d}{dt}\left(\frac{t^2}{2}\right)=t m/s^2$

The component of weight along the surface is equal to $mg \cos 45°$.

The pseudo force acting opposite to it will be $mt \sin 45°$

Both the forces will be equal at the time when the block lifts off the wedge.

$$\begin{gathered} mt \sin 45°=mg \cos 45° \\ t=g \end{gathered}$$

So at time $t=10 sec$ the block will lift from the wedge.