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(A) According to Bernoulli's principle,for a streamline flow,the sum of pressure energy,kinetic energy,and potential energy per unit volume remains co

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

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(A) According to Bernoulli's principle,for a streamline flow,the sum of pressure energy,kinetic energy,and potential energy per unit volume remains constant.Let point $1$ be at the opening of the tube in the flowing water and point $2$ be at the top of the water column inside the vertical part of the tube.At point $1$: Pressure is $P_1$,velocity is $V$,and height is $0$ (reference level).At point $2$: Pressure is $P_0$ (atmospheric pressure),velocity is $0$ (stagnation point),and height is $h$.Applying Bernoulli's equation between point $1$ and point $2$:$P_1 + \frac{1}{2} \rho V^2 + \rho g(0) = P_0 + \frac{1}{2} \rho (0)^2 + \rho gh$Since the pressure at the opening of the tube in the flowing stream is equal to the static pressure plus the dynamic pressure,$P_1 = P_0 + \frac{1}{2} \rho V^2$.Substituting this into the equation:$(P_0 + \frac{1}{2} \rho V^2) + \frac{1}{2} \rho V^2 = P_0 + \rho gh$$P_0 + \rho V^2 = P_0 + \rho gh$$\rho V^2 = \rho gh$$h = \frac{V^2}{g}$Wait,re-evaluating the stagnation pressure: The pressure at the opening of the tube is the stagnation pressure $P_s = P_0 + \frac{1}{2} \rho V^2$. The pressure at the top of the column is $P_0$. The pressure at the bottom of the column (inside the tube) is $P_0 + \rho gh$. Equating the stagnation pressure to the pressure at the bottom of the column:$P_0 + \frac{1}{2} \rho V^2 = P_0 + \rho gh$$\frac{1}{2} \rho V^2 = \rho gh$$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 water rises in the tube will be

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