Physics - Circular Motion Question with Solution | TestHub

In vertical circular motin tension at top is mg - mv²/r  and  tension at bottom is mg + mv²/r

$T_{\mathrm{t}\mathrm{o}\mathrm{p}}=\frac{m{v}^2}{r}-m\mathrm{g}=2m\mathrm{g}$

$\frac{T_{\mathrm{t}\mathrm{o}\mathrm{p}}}{T_{\mathrm{b}\mathrm{o}\mathrm{t}\mathrm{t}\mathrm{o}\mathrm{m}}}=\frac{2m\mathrm{g}}{2m\mathrm{g}+6m\mathrm{g}}=\frac{1}{4}$