An ideal fluid of density 800 kgm–3, flows smoothly through a bent pipe (as shown in figure) that tapers 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 = 10 ms–2)

From continuity equationav1=a2v2v2=2v1From Bernoulli's theorem,P1+ρgh1+12ρv12=P2+ρgh2+12ρv22P1−P2=ρv22−v122+gh2−h14100=8004v12−v122+10×(0−1)418+10=3v1221218×23=v12v1=1214×3×33v1=3636m/s X = 363.

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An ideal fluid of density 800 kgm^–3, flows smoothly through a bent pipe (as shown in figure) that tapers 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 $\frac{\sqrt{x}}{6} {ms}^{- 1}$ for x = ...............(Given g = 10 ms^–2)

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answer is 363.

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Detailed Solution

From continuity equation
$\begin{array}{l} {av}_{1} = \frac{a}{2} v_{2} \\ v_{2} = 2 v_{1} \end{array}$
From Bernoulli's theorem,
$\begin{array}{l} P_{1} + ρ {gh}_{1} + \frac{1}{2} {ρv}_{1}^{2} = P_{2} + ρ {gh}_{2} + \frac{1}{2} {ρv}_{2}^{2} \\ P_{1} - P_{2} = ρ [(\frac{v_{2}^{2} - v_{1}^{2}}{2}) + g (h_{2} - h_{1})] \\ 4100 = 800 [(\frac{4 v_{1}^{2} - v_{1}^{2}}{2}) + 10 \times (0 - 1)] \\ \frac{41}{8} + 10 = \frac{3 v_{1}^{2}}{2} \\ \frac{121}{8} \times \frac{2}{3} = v_{1}^{2} \\ v_{1} = \sqrt{\frac{121}{4 \times 3} \times \frac{3}{3}} \\ v_{1} = \frac{\sqrt{363}}{6} m / s X = 363 . \end{array}$