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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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$\frac{V^2}{2g}$

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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 along a streamline.Let point $1$ be at the opening of the horizontal part of the tube where the water velocity is $V$,and point $2$ be at the top of the vertical column where the water comes to rest $(V_2 = 0)$.Applying Bernoulli's equation between point $1$ and point $2$:$P_1 + \frac{1}{2} \rho V^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho V_2^2 + \rho g h_2$Since both points are at the same horizontal level initially (before the rise),we can consider $h_1 = h_2$ relative to the reference level,or simply note that the pressure at the opening is the static pressure of the fluid plus the dynamic pressure.At the opening (point $1$): $P_1 = P_{atm} + \rho g H + \frac{1}{2} \rho V^2$At the top of the vertical column (point $2$): $P_2 = P_{atm} + \rho g (H - h)$Equating the pressure at the stagnation point (where velocity becomes zero) to the hydrostatic pressure:$\frac{1}{2} \rho V^2 = \rho g h$Solving for $h$:$h = \frac{V^2}{2g}$

An $L$-shaped glass tube is just immersed in flowing water as shown. If the speed of the water current is $V$,then the height $h$ up to which the water rises will be

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