Physics - Circular Motion Question with Solution | TestHub

$\tan \theta =\frac{v_0^2}{rg}$

For A (friction ↓ slope): $\frac{v_A^2}{rg}>\tan \theta ⟹v_A>v_0$

For B (friction ↑ slope): $\frac{v_B^2}{rg}<\tan \theta ⟹v_B<v_0$

For C (no friction): $\frac{v_C^2}{rg}=\tan \theta ⟹v_C=v_0$

$v_A>v_C>v_B$.

Ans : A, B