As I said, $V_{new}=A(\Delta x + \Delta s)$ and $V_{old}=A\Delta x$. So $\Delta V= A \Delta s$.

No...I'm answering. The new volume is definitely $(A\Delta x-As+A(s+\Delta s))$. So change is $A\Delta s$. Instead of doing it myself, I was giving you a hint.

Alright then, let us assume that the object whose temperature we're measuring is of a very large volume but very less density, which results in its mass being comparable to mercury. Now can we say that $T_0\approx T_i$?

Sure, that. I should also say that it does not make $any$ sense. Ever held a thermometer? How much does it weigh? What fraction of that weight is due to the glass?

That does not make much sense, because heat transfer is related to the masses of the bodies, and not the volume.

As written in the last line of my answer, you will have to apply a force greater than weight, because both weight and air drag are directed downwards.

It represents a vector along the tangential direction.

In the question you linked, air drag is not to be considered either. This is evident from the fact that coefficient of viscosity of air, $\eta$, has not been given to us. Here you have to consider the weight of the object as the only force acting on it, other than the external force supplied by you.