For conical pendulum we know that
\(T=2 \pi \sqrt{\frac{l \cos \theta}{g}}\)
where \(\theta\) is the angle from vertical but in question \(\theta\) is given from horizontal hence
\(T=2 \pi \sqrt{\frac{l \sin \theta}{g}}\)
\(T^2 \propto \sin \theta\)
$(g = 10\, m/s^2 \; and\; \tan 12^o = 0.2125)$