An ideal fluid of density 800 ; kg cdot m^{-3 | vedclass

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(A) From the equation of continuity,$A_1 v_1 = A_2 v_2$. Given $A_1 = a$ and $A_2 = \frac{a}{2}$,we have $a v_1 = \frac{a}{2} v_2$,which implies $v_2

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$363$

Vedclass · Answer

(A) From the equation of continuity,$A_1 v_1 = A_2 v_2$. Given $A_1 = a$ and $A_2 = \frac{a}{2}$,we have $a v_1 = \frac{a}{2} v_2$,which implies $v_2 = 2 v_1$. Applying Bernoulli's theorem between the two sections: $P_1 + ho g h_1 + \frac{1}{2} ho v_1^2 = P_2 + ho g h_2 + \frac{1}{2} ho v_2^2$. Taking the lower section as reference level $(h_2 = 0)$,then $h_1 = 1 \; m$. $P_1 - P_2 = ho g (h_2 - h_1) + \frac{1}{2} ho (v_2^2 - v_1^2)$. Substituting the given values: $4100 = 800 	imes [10 	imes (0 - 1) + \frac{1}{2} ( (2v_1)^2 - v_1^2 )]$. $4100 = 800 	imes [-10 + \frac{3 v_1^2}{2}]$. $\frac{4100}{800} = -10 + \frac{3 v_1^2}{2}$. $5.125 = -10 + 1.5 v_1^2$. $15.125 = 1.5 v_1^2$. $v_1^2 = \frac{15.125}{1.5} = \frac{121}{12}$. $v_1 = \sqrt{\frac{121}{12}} = \frac{11}{\sqrt{12}} = \frac{11 \sqrt{3}}{6} = \frac{\sqrt{121 	imes 3}}{6} = \frac{\sqrt{363}}{6}$. Comparing this with $\frac{\sqrt{x}}{6}$,we get $x = 363$.

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