A liquid flows through a horizontal tube. The | vedclass

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(B) According to the equation of continuity,the volume flow rate is constant: $A_1v_1 = A_2v_2$. Thus,option $(A)$ is correct.Applying Bernoulli's pri

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${v_2} - {v_1} = \sqrt {2gh} $

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(B) According to the equation of continuity,the volume flow rate is constant: $A_1v_1 = A_2v_2$. Thus,option $(A)$ is correct.Applying Bernoulli's principle for a horizontal tube $(h_1 = h_2)$: $P_1 + \frac{1}{2}ho v_1^2 = P_2 + \frac{1}{2}ho v_2^2$$P_1 - P_2 = \frac{1}{2}ho(v_2^2 - v_1^2)$Since the pressure difference is given by the height difference $h$ in the manometers,$P_1 - P_2 = ho gh$. Therefore,$ho gh = \frac{1}{2}ho(v_2^2 - v_1^2)$,which simplifies to $v_2^2 - v_1^2 = 2gh$. Thus,option $(C)$ is correct.Option $(B)$ states $v_2 - v_1 = \sqrt{2gh}$,which is mathematically incorrect because $\sqrt{v_2^2 - v_1^2} 
eq v_2 - v_1$.Option $(D)$ is incorrect because,in a real fluid flow,energy is dissipated due to viscosity,and even in an ideal fluid,the total energy per unit mass (Bernoulli's constant) is conserved,but the distribution between pressure energy and kinetic energy changes. However,in the context of this specific question,option $(B)$ is the most explicitly incorrect mathematical statement.

$A$ liquid flows through a horizontal tube. The velocities of the liquid in the two sections which have areas of cross-section $A_1$ and $A_2$ are $v_1$ and $v_2$ respectively. The difference in the levels of the liquid in the two vertical tubes is $h$. The incorrect statement is

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