A straight metallic rod of total length 1.5 m is pivoted at a point...

Let the pivot point be $O$. The rod is divided into two segments of lengths $r_1 = 1.0 \text{ m}$ and $r_2 = 0.5 \text{ m}$ on opposite sides of the pivot.

The motional EMF induced across a rotating rod of length $r$ is given by $\varepsilon = \dfrac{1}{2} B \omega r^2$.

For the longer segment:
$\varepsilon_1 = \dfrac{1}{2} \times 4.0 \times 10 \times (1.0)^2 = 20 \text{ V}$

For the shorter segment:
$\varepsilon_2 = \dfrac{1}{2} \times 4.0 \times 10 \times (0.5)^2 = 5 \text{ V}$

Since the two segments are on opposite sides of the pivot and rotate in the same sense, the velocity vectors of points on the two segments are in opposite directions. By Fleming's right-hand rule (or $\vec{v} \times \vec{B}$), the induced EMF in both segments pushes positive charges away from the pivot (or both towards the pivot). Thus, the two ends acquire the same polarity relative to the pivot.

The potential difference between the two extreme ends is the difference of their potentials relative to the pivot:
$\Delta V = |\varepsilon_1 - \varepsilon_2| = |20 - 5| = 15 \text{ V}$.

Answer: $15 \text{ V}$