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 $\rho ={\rho }_{0}si{n}^2\theta$ . 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 α and other of (2π -α)
B at O will be zero if ${\mathrm{I}}_1\mathrm{\alpha }={\mathrm{I}}_2(2\mathrm{\pi }-\mathrm{\alpha })$
${\mathrm{I}}_1=\frac{\mathrm{E}}{{\mathrm{R}}_1},{\mathrm{I}}_2=\frac{\mathrm{E}}{{\mathrm{R}}_2}$
$\text{ So }\frac{\mathrm{\alpha }}{{\mathrm{R}}_1}=\left(\frac{2\mathrm{\pi }-\mathrm{\alpha }}{{\mathrm{R}}_2}\right)$
${\mathrm{R}}_1={\int }_0^{\alpha }\frac{{\mathrm{\rho }}_0{\sin }^2\mathrm{\theta }}{\mathrm{A}}\mathrm{Rd\theta }$ 

$R_1=\frac{{\rho }_0\mathrm{R}}{A}\left[\alpha -\frac{\sin 2\alpha }{2}\right]$

Similarly we can get R₂

${\mathrm{R}}_2={\int }_{\alpha }^{2\pi }\frac{{\rho }_0{\sin }^2\theta }{\mathrm{A}}\mathrm{Rd}\theta =\frac{{\rho }_0\mathrm{R}}{\mathrm{A}}\left[(2\pi -\alpha )+\frac{\sin 2\alpha }{2}\right]$

Now using, 

$\therefore \alpha R_2=(2\pi -\alpha )R_1$

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

$\Rightarrow \sin 2\alpha =0$

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