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 $\theta$  (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 $=\frac{B}{2}{1}^2\omega$
2. $\omega$  can be calculated according to law of conservation energy. Increase in rotational kinetic energy = decrease in gravitational potential energy
3.  $\frac{1}{2}I{\omega }^2=mgh$
$\begin{array}{l}\Rightarrow \frac{1}{2}\frac{m{l}^2}{3}{\omega }^2=mg.\frac{1}{2}\sin \theta \\ \Rightarrow \omega =\sqrt{\frac{3g}{1}\sin \theta }\\ Now\Delta V=\frac{1}{2}B{l}^2\sqrt{\frac{3g}{1}\sin \theta }\mathrm{......}\left(1\right)\\ \therefore \Delta V\infty \sqrt{\sin \theta }\\ Also\text{ from }\left(1\right),\Delta V\infty 1^{\frac{3}{2}}\\ Therefore,\text{ the correct answer is }\left(2\right)\end{array}$