Question

The figure shows a pipe of uniform cross-section inclined in a vertical A plane. A U-tube manometer is connected between the points A and B. If the liquid of density $P_0$ flows with velocity $v_0$ in the pipe, Then the the reading $h$ of the manometer is

• A. $h=0$
• B. $h=\frac{v_0^2}{2g}$
• C. $h=\frac{{\rho }_0}{\rho }\left(\frac{v_0^2}{2g}\right)$
• D. $h=\frac{{\rho }_{0}H}{\rho -p_0}$

Answer 1

The correct option is A $h=0$

From continuity equation,
$v_A=v_B=v_0$
$\therefore$ $P_A+\rho gh=P_B+0$
$\therefore$ $P_B-P_A=\rho gh$ (i)
Now let us make pressure equation from manometer.
$P_A+\rho g(h+H)-{\rho }_{Hg}gh=P_3$
Putting $P_B-P_A=\rho gh$ we get h=0