Explain how to answer the following :

Dear student

Using Bernoulli's theorem
$$\begin{gathered} P_1+\frac{1}{2}\rho {v_1}^2+\rho g{h}_1=P_2+\frac{1}{2}\rho {v_2}^2+\rho g{h}_2 \\ but P_1=P_2 and \left(h_2-h_1\right)=l\sin \theta \\ by Continuity eq. A_1{v}_1=A_2{v}_2 \\ \Rightarrow v_1=\frac{A_2}{A_1}\sqrt{\frac{9l}{2}}=2\sqrt{\frac{9l}{2}}=\sqrt{18l} \\ thus \\ \frac{1}{2}\rho \left({v_1}^2-{v_2}^2\right)=\rho g\left(h_2-h_1\right) \\ \frac{1}{2}\left(18l-\frac{9l}{2}\right)=9.8\times l\sin \theta \\ \sin \theta =\frac{27}{4\times 9.8}\approx \frac{3}{4} \\ \theta ={\sin }^{-1}\left(\frac{3}{4}\right) \end{gathered}$$
Thus, Correct Option (B)

Regards