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(A) According to Bernoulli's principle,for a streamline flow of an incompressible,non-viscous fluid,the sum of pressure energy,kinetic energy,and pote

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The water in the tube rises to height $\frac{v^2}{2g}$

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(A) According to Bernoulli's principle,for a streamline flow of an incompressible,non-viscous fluid,the sum of pressure energy,kinetic energy,and potential energy per unit volume remains constant.Let the pressure at the opening of the tube be $P_0$ and the velocity of water be $v$. Inside the tube,the water comes to rest,so the velocity at the top of the vertical column is $0$.Applying Bernoulli's equation between the opening of the tube (point $1$) and the top of the water column in the tube (point $2$):$P_1 + \frac{1}{2}\rho v^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$Here,$P_1 = P_2 = P_{atm}$ (atmospheric pressure),$v_1 = v$,$v_2 = 0$,and $h_2 - h_1 = h$ (the height of the water column above the water level).Substituting these values:$P_{atm} + \frac{1}{2}\rho v^2 + 0 = P_{atm} + 0 + \rho g h$$\frac{1}{2}\rho v^2 = \rho g h$$h = \frac{v^2}{2g}$Thus,the water in the tube rises to a height of $\frac{v^2}{2g}$.

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

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