The correct Bernoulli's equation is (symbols have their usual meaning):

$A$ jet plane flies in the air because

Water is moving with a speed of $5.0\,m/s$ through a pipe of cross-sectional area $4.0\,cm^2$. The water gradually descends $10\,m$ as the pipe increases in area to $8.0\,cm^2$. If the pressure at the upper level is $1.5 \times 10^5\,Pa$,the pressure at the lower level will be:

$A$ plane is in level flight at constant speed and each of its two wings has an area of $40 \,m^2$. If the speed of the air is $180 \,km/h$ over the lower wing surface and $252 \,km/h$ over the upper wing surface,the mass of the plane is . . . . . . $kg$. (Take air density to be $1 \,kg \,m^{-3}$ and $g=10 \,ms^{-2}$)

In this figure,an ideal liquid flows through a tube of uniform cross-section. The liquid has velocities $v_A$ and $v_B$,and pressures $P_A$ and $P_B$ at points $A$ and $B$ respectively.

$A$ closed water tank has a cross-sectional area $A$. It has a small hole at a depth of $h$ from the free surface of the water. The radius of the hole is $r$ such that $r \ll \sqrt{\frac{A}{\pi}}$. If $p_o$ is the pressure inside the tank above the water level and $p_a$ is the atmospheric pressure,the rate of flow of the water coming out of the hole is ($\rho$ is the density of water).