# Is the tension force (instead of normal force) the apparent weight here?

An object's weight, henceforth called "actual weight", is the downward force exerted upon it by the earth's gravity. By contrast, an object's apparent weight is the upward force (the normal force, or reaction force), typically transmitted through the ground, that opposes gravity and prevents a supported object from falling.

In my picture, the elevator is accelerating upwards with an acceleration $$a$$. An object is attached to the ceiling of the elevator with a massless string. The actual weight of the object is $$mg$$. No normal force exists here, so can we say that the apparent weight of the object is the tension force $$T$$?

. . . . . so can we say that the apparent weight of the object is the tension force $$T$$? - Yes
Using $$F=ma$$ with "up" as the positive direction $$t-mg = ma \Rightarrow T = Mg+Ma$$ and that is the apparent weight of the object.
If $$a=0$$ then $$T=mg$$ which is the weight of the object.
If $$a=-g$$, ie the system is in free fall, $$T=0$$ and so the onject appears to be weightless ie the reading on a spring balance would be zero.