$A$ liquid of density $750\,kg\,m^{-3}$ flows smoothly through a horizontal pipe that tapers in cross-sectional area from $A_{1} = 1.2 \times 10^{-2}\,m^{2}$ to $A_{2} = \frac{A_{1}}{2}$. The pressure difference between the wide and narrow sections of the pipe is $4500\,Pa$. The rate of flow of liquid is . . . . . . $\times 10^{-3}\,m^{3}\,s^{-1}$.

Water from a tap emerges vertically downwards with an initial velocity of $4 \,m/s$. The cross-sectional area of the tap is $A$. The flow is steady and the pressure is constant throughout the stream of water. The distance $h$ vertically below the tap,where the cross-sectional area of the stream becomes $\frac{2}{3} A$,is (take $g = 10 \,m/s^2$): (in $\,m$)

$A$ 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).

Bernoulli’s theorem is a consequence of

In the case of a fluid,Bernoulli's theorem expresses the application of the principle of conservation of :

Water is flowing with a velocity of $2 \ m/s$ in a horizontal pipe where the cross-sectional area is $2 \times 10^{-2} \ m^2$ at a pressure of $4 \times 10^4 \ Pa$. The pressure at a cross-section of area $0.01 \ m^2$ in Pascal will be: