A cell is connected between the points $A$ and $C$ of a circular conductor $ABCD$ of centre $O$ with angle $OC = 60^{\circ}$. If $B_{1}$ and $B_{2}$ are the magnitudes of the magnetic fields at $O$ due to the currents in $ABC$ and $ADC$ respectively, the ratio $B_{1} / B_{2}$ is 

Between $A$ and $C$, the circular conductor provides two parallel arcs: arc $ABC$ subtending $300^{\circ}$ at $O$ and arc $ADC$ subtending $60^{\circ}$ at $O$. For a wire bent into an arc, resistance is proportional to arc length, so $$R_1:R_2=300:60=5:1.$$ Since the same cell is connected across both arcs, the currents divide inversely: $$I_1:I_2=R_2:R_1=1:5.$$ Magnetic field at the centre due to an arc is proportional to $I\theta$, because $$B=\frac{\mu_0 I\theta}{4\pi r}.$$ Hence, $$\frac{B_1}{B_2}=\frac{I_1\theta_1}{I_2\theta_2}.$$ Using $\theta_1:\theta_2=300:60=5:1$ and $I_1:I_2=1:5$, $$\frac{B_1}{B_2}=\frac{1\times 5}{5\times 1}=1.$$ This matches option $C$.