Water is in streamline flow along a horizontal pipe with non-uniform cross-section. At a point in the pipe where the area of cross-section is $10 \,cm^{2}$, the velocity of water is $1 \,ms^{-1}$ and the pressure is $2000 \,Pa$. The pressure at another point where the cross-sectional area is $5 \,cm^{2}$ is (in $\,Pa$)

An ideal fluid of density $800 \; kg \cdot m^{-3}$ flows smoothly through a bent pipe (as shown in the figure) that tapers in cross-sectional area from $a$ to $\frac{a}{2}$. The pressure difference between the wide and narrow sections of the pipe is $4100 \; Pa$. At the wider section,the velocity of the fluid is $\frac{\sqrt{x}}{6} \; m \cdot s^{-1}$. Find the value of $x$. (Given $g = 10 \; m \cdot s^{-2}$)

State Bernoulli's principle in words.

The velocity of kerosene oil in a horizontal pipe is $5 \ m/s$. If $g = 10 \ m/s^2$,then the velocity head of the oil will be ....... $m$.

$A$ horizontal pipe of non-uniform cross-section allows water to flow through it with a velocity $1 \ m/s$ when the pressure is $50 \ kPa$ at a point. If the velocity of flow has to be $2 \ m/s$ at some other point, the pressure at that point should be: (in $kPa$)

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.