For a fluid that is flowing steadily in a horizontal tube as shown in the figure, the level in the vertical tubes is best represented by

There are two identical small holes on the opposite sides of a tank containing a liquid. The tank is open at the top. The difference in height of the two holes is $h$ as shown in the figure. As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to:

Water is filled up to a height $\mathrm{h}$ in a beaker of radius $\mathrm{R}$ as shown in the figure. The density of water is $\rho$, the surface tension of water is $T$ and the atmospheric pressure is $P_0$. Consider a vertical section $ABCD$ of the water column through a diameter of the beaker. The force on water on one side of this section by water on the other side of this section has magnitude

A tube in a vertical plane is shown in the figure. It is filled with a liquid of density $\mathrm{\rho }$ and its end $B$ is closed Then the force exerted by the fluid on the tube at end $B$ will be: [Neglect atmospheric pressure and assume the radius of the tube to be negligible in comparison to $l$]

A $\mathrm{U}-$tube of base length $l$ filled with the same volume of two liquids of densities $\rho$ and $2\rho$ is moving with an acceleration $a$ on the horizontal plane as shown in the figure. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height $h$ is given by: