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pl answer Q 4 (more than one correct answer type)

Frictional force will provide Centripetal force

$\frac{m{v}^{2}}{r}=\mu mg\phantom{\rule{0ex}{0ex}}v=\sqrt{\mu rg}=\sqrt{\frac{1}{\sqrt{3}}\times \sqrt{3}\times 10}=\sqrt{10}m/s$

$\theta ={\mathrm{tan}}^{-1}\left(\frac{{v}^{2}}{rg}\right)=t{\mathrm{an}}^{-1}\left(\frac{10}{10\sqrt{3}}\right)=30$

The line of action of net contact force is passing thorugh centre of gravity of cycylist system for equillibrium

Thus all options are correct

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