Is a load needed to measure the voltage drop of a long cable?
To expand on the comments, a long wire obeys Ohms law, just like any other resistance, so V = IR.

With no load, no current flows.  I = 0, so V = 0 and there's no voltage drop.

If you add a load, then a current flows, and the voltage drop becomes non-zero and proportional to the current.  It's possible to calculate the drop if you know the ohms-per-metre of the cable and the expected current.  Don't forget the cable goes both ways, so the drop is double that of one length of cable.

On AC, you might see a small voltage drop in a long enough cable because there will be some capacitive leakage between the cores, and hence a small current.