Solve this:

​Q39. A ball is released from the top of  vertical circular pipe. Find angle $\mathrm{\theta }$ with vertical where the ball will lose contact with inner side wall of pipe and start moving with outer side wall. (Thickness of pipe is small as compared to radius of circle)

                
   $$\begin{gathered} \left(1\right) {\cos }^{-1}\frac{2}{3} \left(2\right) {\cos }^{-1}\frac{1}{3} \\ \left(3\right) {\sin }^{-1}\left(\frac{2}{3}\right) \left(4\right) {\sin }^{-1}\left(\frac{1}{3}\right) \end{gathered}$$