A conducting rod of length $l$  is hinged at point O. It is free to rotate in a vertical plane. There exists a uniform magnetic field $\overrightarrow{B}$  in horizontal direction. The rod is released from the horizontal position as shown. The potential difference between the ends of the rod when it has turned by $\text{θ}$ (as shown in figure) is proportional to

The induced electro motive force $\left(\Delta v\right)$ across ends of rotating rod about one end in magnetic field $=\frac{B}{2}{l}^2\omega$ .
$\omega$ can be calculated according to law of conservation energy.
Increase in rotational kinetic energy = decrease in gravitational potential energy 
 $\frac{1}{2}I{\omega }^2=mgh$
$\Rightarrow \frac{1}{2}\frac{m{l}^2}{3}{\omega }^2=mg.\frac{l}{2}\sin \theta$
$\Rightarrow \omega =\sqrt{\frac{3g}{l}\sin \theta }$
Now    $\Delta V=\frac{1}{2}B{l}^2\sqrt{\frac{3g}{l}\sin \text{θ}}$ ……( 1 )
$\therefore \Delta V\propto \sqrt{\sin \theta }$
Also from ( 1 ),  $\Delta V\propto l^{\frac{3}{2}}$
Therefore, the correct answer is ( D )