JEE Advanced 2021 Paper 1 Physics: Paragraph 3

The pressure difference between the two sections is read from the manometer. Because the connecting tubes are filled with water, the difference in the mercury levels $h$ supports $\Delta P=(\rho_{Hg}-\rho_w)gh=(13600-1000)\times9.8\times0.05=6174$ Pa. The continuity equation gives $A_1v_1=A_2v_2$, so $v_2=(20/10)^2v_1=4v_1$. Bernoulli's equation for the horizontal pipe gives $P_1-P_2=\tfrac12\rho_w(v_2^2-v_1^2)=\tfrac12\rho_w\cdot15\,v_1^2$. Hence $v_1^2=\frac{2\times6174}{1000\times15}=0.8232$ and $v_1=0.907\ \mathrm{m\,s^{-1}}\approx0.91\ \mathrm{m\,s^{-1}}$. Ignoring the water in the connecting tubes (using $\rho_{Hg}gh$ only) would give a wrong speed of about $0.98$ m/s.

Using the pressure difference found from the manometer, $P_1-P_2=6174$ Pa, the gauge pressure in the narrow section is $P_2=1.5\times10^{4}-6174=8826$ Pa. The open vertical tube is a static column of water connected to the pipe at that point, so the water rises until the hydrostatic gauge pressure at the base equals the static pressure there: $\rho_wgh=P_2$. Hence $h=\frac{8826}{1000\times9.8}=0.9006$ m $\approx0.90$ m. The tube measures static pressure only, because its opening is perpendicular to the flow and does not stagnate the stream. A tube facing the flow would rise higher by $v_2^2/2g$, with $v_2=4v_1=3.63\ \mathrm{m\,s^{-1}}$.