I used to do this experiment with 1st year engineering student.
Let's consider a resistor that is 10cm long, not covered in insulation with resistance of 10 Ohms. Something like a piece of coal paper perhaps. We will apply a voltage of 1V across both sides of the resistor. We now measure the voltage across the resistor or parts of it using a multi meter.
First of all, the multi meter has two probes. If you put a single probe anywhere on the circuit, nothing happens. There is no such thing as an "absolute" potential: you can only measure the potential difference between two points (or between a single point and some explicit or implicit reference, like earth ground).
If we put the the probes at both ends of the resistor we get 1V. That's the total voltage drop over the resistor and equal to the source voltage.
Now we hold one probe steady at one end of the resistor and slide the other probe from the other end towards the first probe. 395
You will see that voltage will steadily decrease from 1V to 0V. Once you are in the middle it's 500mV. Once you down to 1cm, it's 100mV.
Now we can put a probe, at, 2cm and the other at 8cm. We read 600mV. It turns out that the voltage is simply proportional to the distance between the probes with 100mv/cm.
And that's all there is to it. Assigning a single potential to an entire resistor is not possible and makes no sense. You will find all the potentials between 0 and the terminal voltage in the resistor, just depending on where you look.
Ohm's law and the power laws hold true for the entire resistor as well as any arbitrary part of the resistor. The current is 100mA, so the the overall power is $$P = (100mA)^2 \cdot 10 \Omega = (1V)^2/ 10\Omega = 100mW$$ For any section of relative length $x$ we get
$$Px = (100mA)^2 \cdot 10 \Omega \cdot x = (1V \cdot x)^2/ (10\Omega \cdot x) = $$ $$ (1V)^2 \cdot x/ (10\Omega) = 100mW \cdot x $$
where $x$ is the length of the segment divided by the total length. 71