An L-shaped glass tube is just immersed in flowing water such that its opening is pointing against flowing water. If the speed of water current is v, then

To understand why the water rises in the L-shaped glass tube, we can apply Bernoulli's principle to the flowing water near the opening of the tube.

Let us consider two points in the fluid flow system:
1. Point 1: A point in the undisturbed flowing water stream, far upstream from the tube's opening, where the water flows with a velocity $v$ and has static pressure $P_0$.
2. Point 2: A point exactly at the entrance (opening) of the L-shaped tube. Since the tube is open and pointing against the flow, water enters it until it reaches a height $h$ and stops flowing inside the tube. Therefore, the velocity of the water at the opening of the tube (Point 2) becomes zero. This point is a stagnation point, and the pressure here is the stagnation pressure $P_s$.

Applying Bernoulli's equation along the streamline connecting Point 1 and Point 2 (neglecting gravitational potential differences at the horizontal immersion level):

$P_0+\frac{1}{2}\rho v^2=P_s+0$

Where:
$\rho$ is the density of the water.
$v$ is the velocity of the water current.

From the above relation, the pressure at the opening of the tube is:
$P_s=P_0+\frac{1}{2}\rho v^2$

The excess pressure at the opening relative to the static pressure of the surrounding water at that depth is:
$\Delta P=P_s-P_0=\frac{1}{2}\rho v^2$

This excess pressure supports a column of water of height $h$ inside the tube above the free surface of the flowing water. The hydrostatic pressure exerted by a water column of height $h$ is given by:
$\Delta P=h\rho g$

Equating the two expressions for the excess pressure:

$h\rho g=\frac{1}{2}\rho v^2$

Solving for $h$, the density $\rho$ cancels out from both sides:

$h=\frac{v^2}{2g}$

Thus, the water in the tube rises to a height of $v^2/2g$.