Consider a massless string
Why massless ? What does this change ?
being pulled from both ends with equal forces, increasing gradually and equally on both sides. At the point where the string can no longer withstand the tension, the string snaps (ultimate failure).
Having mass would not change that.
My question is where (meaning point on the string from an end) will the string snap, and why? Since the string is massless, I think the imperfections in the string is out of the question.
No, having mass does not mean you have no imperfections. Stating the string is homogeneous is all that is required.
The string will snap in the middle, regardless of mass, once the forces are equal and the string has no random imperfections.
Now considering a real string (having mass), but perfect in the same situation.
You are saying it is still a homogeneous string - no random imperfections.
Where will it snap, and why? I think answer to the first one will apply here as well.
Same again. Middle.
As for the third type (no answer expected, but if you have something to add, you are very welcome), that is string having mass and is imperfect
Forget the mass. Now saying it's "imperfect" doesn't mean much. What kind of imperfections ? Are the imperfections localized to one region ? Are they related to tensile strength ? You need to be precise when explaining a problem.
Anyway we'll assume it's some tensile weakness at some point along the string.
I know it will snap at the weakest point along its length.
I assume that at the weakest point, the tension is somehow greater than at any other point on the string, causing it to give in first.
The tension is no (initially) greater anywhere.
Typically you'd see the string stretch out, then some fibers (the weakest) break, but the remaining ones would continue to hold until the tension became too great for them. Obviously as the number of fiber's still intact decreases, the tension in the remaining fibers is increased. The string will break when the tension exceeds what the last remaining fibers can hold.
The precise point of final failure won't be the same as the initial point of failure. The first failure will be the weakest fiber, but that obviously won't be the last failure point.