
Therefore, in this case, the pressure remains constant initially and then decreases as the area of cross section decreases along the neck of the tube and then remains constant along the mouth of the tube.
Hence, graph in option $A$ is correct.
$(A)$ $\frac{\left|\overrightarrow{ V }_{ P }\right|}{\left|\overrightarrow{ V }_{ Q }\right|}=\frac{\eta_1}{\eta_2}$ $(B)$ $\frac{\left|\overrightarrow{ V }_{ P }\right|}{\left|\overrightarrow{ V }_{ Q }\right|}=\frac{\eta_2}{\eta_1}$
$(C)$ $\overrightarrow{ V }_{ P } \cdot \overrightarrow{ V }_{ Q } > 0$ $(D)$ $\overrightarrow{ V }_{ P } \cdot \overrightarrow{ V }_{ Q } < 0$
|Take atmospheric pressure $=1.0 \times 10^5 \mathrm{~N} / \mathrm{m}^2$, density of water $=1000 \mathrm{~kg} / \mathrm{m}^3$ and $g=10 \mathrm{~m} / \mathrm{s}^2$. Neglect any effect of surface tension.]