Consider a water tank as shown in the figure. | vedclass

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(A) Given:Mass of load $m = 24\, kg$Cross-sectional area of tank $A = 0.4\, m^{2}$Cross-sectional area of opening $a = 1\, cm^{2} = 10^{-4}\, m^{2}$He

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$3$

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(A) Given:Mass of load $m = 24\, kg$Cross-sectional area of tank $A = 0.4\, m^{2}$Cross-sectional area of opening $a = 1\, cm^{2} = 10^{-4}\, m^{2}$Height of water $H = 40\, cm = 0.4\, m$Acceleration due to gravity $g = 10\, ms^{-2}$Density of water $ho = 1000\, kg/m^{3}$Applying Bernoulli's equation at the top surface and at the opening $B$:$P_{top} + ho gH + \frac{1}{2}ho v_{1}^{2} = P_{atm} + \frac{1}{2}ho v^{2}$Where $P_{top} = P_{atm} + \frac{mg}{A}$Substituting the values:$(P_{atm} + \frac{mg}{A}) + ho gH = P_{atm} + \frac{1}{2}ho v^{2}$ (Assuming $v_{1} \approx 0$ as $A \gg a$)$\frac{mg}{A} + ho gH = \frac{1}{2}ho v^{2}$$v = \sqrt{2gH + \frac{2mg}{Aho}}$$v = \sqrt{2 	imes 10 	imes 0.4 + \frac{2 	imes 24 	imes 10}{0.4 	imes 1000}}$$v = \sqrt{8 + \frac{480}{400}} = \sqrt{8 + 1.2} = \sqrt{9.2}$$v \approx 3.033\, m/s$Rounding to the nearest integer,$v = 3\, m/s$.

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