A liquid enters at point A_{1} with a speed o | vedclass

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(A) From the equation of continuity,$A_{1}V_{1} = A_{2}V_{2}$. Since the cross-sectional area is uniform $(A_{1} = A_{2})$,the speed of the liquid rem

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(A) From the equation of continuity,$A_{1}V_{1} = A_{2}V_{2}$. Since the cross-sectional area is uniform $(A_{1} = A_{2})$,the speed of the liquid remains constant,so $V_{1} = V_{2} = 3.5 \ m/s$.Applying Bernoulli's theorem between point $A_{1}$ (at height $0$) and the maximum height $h$ reached above point $A_{2}$ (where the final velocity is $0$):$P_{atm} + \frac{1}{2} ho V_{1}^{2} + ho g(0) = P_{atm} + \frac{1}{2} ho(0)^{2} + ho gh$Simplifying the equation:$\frac{1}{2} ho V_{1}^{2} = ho gh$$h = \frac{V_{1}^{2}}{2g}$Substituting the given values ($V_{1} = 3.5 \ m/s$ and $g = 9.8 \ m/s^{2}$):$h = \frac{(3.5)^{2}}{2 	imes 9.8} = \frac{12.25}{19.6} = 0.625 \ m$Converting to centimeters:$h = 0.625 	imes 100 = 62.5 \ cm$.Note: Using $g = 10 \ m/s^{2}$ gives $h = \frac{12.25}{20} = 0.6125 \ m = 61.25 \ cm$. Thus,option $A$ is correct.

$A$ liquid enters at point $A_{1}$ with a speed of $3.5 \ m/s$ and leaves at point $A_{2}$. Find the height attained by the liquid above point $A_{2}$ (in $cm$). (in $.25$)

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