b
Physical pendulum
$\mathrm{T}_{1}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{mgd}}}=2 \pi \sqrt{\frac{\frac{\mathrm{m} \ell^{2}}{12}+\mathrm{m}\left(\frac{(\sqrt{3})}{2}\right)^{2}}{\mathrm{mg}\left(\frac{\ell \sqrt{3}}{2}\right)}}$
$\mathrm{T}_{2}=2 \pi \sqrt{\frac{\mathrm{d}}{\mathrm{g}}}=2 \pi \sqrt{\frac{\frac{\ell \sqrt{3}}{2}}{\frac{2}{\mathrm{g}}}}$
$\left(\frac{T_{1}}{T_{2}}\right)^{2}=\frac{10}{9}$
