An ideal fluid of density 800kgm−3,flows smoothly through a bent pipe (as shown in figure) that tappers in cross-sectional area from a to a2.The pressure difference between the wide and narrow sections of pipe is 4100 Pa. At wider section, the velocity of fluid is x6ms−1 for x=_____.(Given g=10ms−2)

From continuity equation av1=a2v2, v2=2v1From Bernoulli's theorem,P1+ρgh1+12ρv12=P1+ρgh2+12ρv22P1−P2=ρv22−v122+gh2−h14100=8004v12−v122+10×0−1418+10=3v122, 1218×23=v12v1=I214×3×33, v1=3636m/sX=363

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An ideal fluid of density $800 k g m^{- 3}$ ,flows smoothly through a bent pipe (as shown in figure) that tappers in cross-sectional area from $a$ to $\frac{a}{2}$ .The pressure difference between the wide and narrow sections of pipe is 4100 Pa. At wider section, the velocity of fluid is $\frac{\sqrt{x}}{6} m s^{- 1}$ for $x =_____$ .(Given $g = 10 m s^{- 2}$ )

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answer is 363.

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Detailed Solution

From continuity equation $a v_{1} = \frac{a}{2} v_{2}, v_{2} = 2 v_{1}$

From Bernoulli's theorem,

$P_{1} + ρ g h_{1} + \frac{1}{2} ρ v_{1}^{2} = P 1 + ρ g h_{2} + \frac{1}{2} ρ v_{2}^{2}$

$P_{1} - P_{2} = ρ [(\frac{v_{2}^{2} - v_{1}^{2}}{2}) + g (h_{2} - h_{1})]$

$4100 = 800 [(\frac{4 v_{1}^{2} - v_{1}^{2}}{2}) + 10 \times (0 - 1)]$

$\frac{41}{8} + 10 = \frac{3 v_{1}^{2}}{2}, \frac{121}{8} \times \frac{2}{3} = v_{1}^{2}$

$v_{1} = \sqrt{\frac{I 21}{4 \times 3} \times \frac{3}{3}}, v_{1} = \frac{\sqrt{363}}{6} m / s$

$X=363$

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