The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross-section. Cross sectional areas at $A$ is $1.5\,cm ^2$, and $B$ is $25\,mm ^2$, if the speed of liquid at $B$ is $60\,cm / s$ then $\left( P _{ A }- P _{ B }\right)$ is :(Given $P _{ A }$ and $P _{ B }$ are liquid pressures at $A$ and $B$ points.Density $\rho=1000\,kg\,m ^{-3}$
$A$ and $B$ are on the axis of tube $............\,Pa$
$175$
$27$
$135$
$36$
Prove Bernoulli’s Principle.
According to Bernoulli's equation $\frac{P}{{\rho g}} + h + \frac{1}{2}\,\frac{{{v^2}}}{g} = {\rm{constant}}$ The terms $A, B$ and $ C$ are generally called respectively:
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 increase in area to $8.0\,cm^2$ . If the pressure at the upper level is $1.5 \times 10^5\,Pa$ , the pressure at lower level will be
A wide vessel with a small hole at the bottom is filled with two liquids. The density and heightof one liquid are $\rho_1$ and $h_1$ and that of the otherare $\rho_2$ and $h_2 \ (\rho_1 >\rho_2)$. The velocity of liquid coming out of the hole is :
A body of density $\rho$ is dropped from rest from a height $h$ into a lake of density $\sigma$, where $\sigma > \rho$. Neglecting all dissipative forces, the maximum depth to which the body sinks before returning to float on surface ..........