Bernoulli's principle is based on the law of conservation of

Water flows through a frictionless duct with a cross-section varying as shown in the figure. The pressure $p$ at points along the axis is represented by:

Water is flowing through a horizontal pipe of non-uniform cross-section. In the region of the narrowest part inside the pipe,the water will have

$A$ 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 $kPa$)

Water is flowing in a conical tube as shown in the figure. The velocity of water at area $A_2$ is $60 \,cm/s$. The values of $A_1$ and $A_2$ are $10 \,cm^2$ and $5 \,cm^2$ respectively. The pressure difference between the two cross-sections is: (in $\,N/m^2$)

The figure shows a liquid of given density flowing steadily in a horizontal tube of varying cross-section. The cross-sectional areas at $A$ is $1.5 \, cm^2$,and at $B$ is $25 \, mm^2$. If the speed of the liquid at $B$ is $60 \, cm/s$,then find $(P_A - P_B)$ in $Pa$. (Given: $P_A$ and $P_B$ are liquid pressures at points $A$ and $B$ respectively. Density $\rho = 1000 \, kg/m^3$. $A$ and $B$ are on the axis of the tube.)