The figure shows a circular loop made of a wire of radius R. The resistivity of the material varies as a function of  $\theta$ such that  $\rho ={\rho }_0{\sin }^2\theta$. The positions of the Jockey such that the magnetic field at the centre (O) due to the current in the loop is zero, will be

Consider two sectors one of  $\alpha$ and other of $(2\pi -\alpha )$

$\mathrm{B}\text{ at }\mathrm{O}\text{ will be zero if }{\mathrm{I}}_1\alpha ={\mathrm{I}}_2(2\pi -\alpha )$

$\Rightarrow {\mathrm{I}}_1=\frac{\mathrm{E}}{{\mathrm{R}}_1},\Rightarrow {\mathrm{I}}_2=\frac{\mathrm{E}}{{\mathrm{R}}_2}$

$$\begin{gathered} \text{ So }\frac{\alpha }{R_1}=\left(\frac{2\pi -\alpha }{R_2}\right) \\ \Rightarrow {\mathrm{R}}_1={\int }_0^{\alpha }\frac{{\mathrm{\rho }}_0{\sin }^2\mathrm{\theta }}{\mathrm{A}}\text{ Rd }\mathrm{\theta } \end{gathered}$$

$=\frac{{\mathrm{\rho }}_0\mathrm{R}}{A}\left[\alpha -\frac{\sin 2\alpha }{2}\right]\left(\because \int {\sin }^2\theta d\theta =\frac{1}{2}\left[x-\sin \theta \cos \theta \right]+C\right)$

$$\begin{gathered} S\mathrm{imilarly} \text{ we can get } \\ {\mathrm{R}}_2={\int }_{\mathrm{\alpha }}^{2\mathrm{\pi }}\frac{{\mathrm{\rho }}_0{\sin }^2\mathrm{\theta }}{\mathrm{A}}\text{ Rd }\mathrm{\theta } =\frac{{\mathrm{\rho }}_0\mathrm{R}}{\mathrm{A}}\left[\left(2\mathrm{\pi }-\mathrm{\alpha }\right)+\frac{\sin 2\mathrm{\alpha }}{2}\right] \end{gathered}$$

$\therefore {\mathrm{\alpha R}}_2=\left(2\mathrm{\pi }-\mathrm{\alpha }\right){\mathrm{R}}_1$

$\Rightarrow \mathrm{\alpha }\frac{{\mathrm{\rho }}_0\mathrm{R}}{\mathrm{A}}\left[\left(2\mathrm{\pi }-\mathrm{\alpha }\right)+\frac{\sin 2\mathrm{\alpha }}{2}\right]$

$$\begin{gathered} =\left(2\mathrm{\pi }-\mathrm{\alpha }\right)\frac{{\mathrm{\rho }}_0\mathrm{R}}{\mathrm{A}}\left[\mathrm{\alpha }-\frac{\sin 2\mathrm{\alpha }}{2}\right] \\ \Rightarrow \sin 2\mathrm{\alpha }=0 \end{gathered}$$

$$\begin{gathered} \mathrm{On}\mathrm{solving}\mathrm{we}get \\ \mathrm{\alpha }=\frac{\mathrm{\pi }}{2},\mathrm{\pi },\frac{3\mathrm{\pi }}{2} \end{gathered}$$