At Point $P, m g \sin \theta=\frac{m v^2}{l}\quad\text{..... (1)}$

By conservation of mechanical energy at point $P \in Q$

$$\frac{1}{2} m v_0^2=\frac{1}{2} m v^2+m g(I+I \sin \theta)$$

$$\begin{aligned} \& \frac{v_0^2}{2}=\frac{v^2}{2}+g l(1+\sin \theta) \\ \& \text { Put gl }=\frac{v^2}{\sin \theta} \text { using (1) } \\ \& \frac{v_0^2}{2}=\frac{v^2}{2}+\frac{v^2}{\sin \theta}(1+\sin \theta) \\ \& \frac{v_0^2}{2}=\frac{v^2}{2}+\frac{v^2}{\sin \theta}+v^2 \\ \& \frac{v_0^2}{2}=\frac{3}{2} v^2+\frac{2 v^2}{2 \sin \theta} \\ \& v_0^2=v^2\left[3+\frac{2}{\sin \theta}\right] \\ \& \frac{v}{v_0}=\left(\frac{\sin \theta}{3 \sin \theta+2}\right)^{\frac{1}{2}} \end{aligned}$$