An ideal fluid of density 1000kgm−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 (Pressure is greater at wider end). At wider section the velocity of fluid is x5ms−1. Then, x= --------------- (g=10m/s2)

For continuity equationav1=a2v2⇒v2=2v1from Bernoulie's theorem,P1+ρgh1+12ρv12=P2+ρgh2+12ρv22P1−P2=ρ[(v22−v122)+g(h2−h1)]4100=1000[(4v12−v122)+10×(0−1)]4110+10=3v122⇒ v1=475⇒x=47

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An ideal fluid of density $1000 k g m^{- 3}$ , 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 (Pressure is greater at wider end). At wider section the velocity of fluid is $\sqrt{\frac{x}{5}} m s^{- 1}$ . Then, $x$ = --------------- $(g = 10 m / s^{2})$

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answer is 47.

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Detailed Solution

For continuity equation

$\begin{array}{l} a v_{1} = \frac{a}{2} v_{2} \Rightarrow v_{2} = 2 v_{1} \\ f r o m B e r n o u l i e' s t h e o r e m, \\ P_{1} + ρ g h_{1} + \frac{1}{2} ρ {v_{1}}^{2} = P_{2} + ρ g h_{2} + \frac{1}{2} ρ {v_{2}}^{2} \\ P_{1} - P_{2} = ρ [(\frac{{v_{2}}^{2} - {v_{1}}^{2}}{2}) + g (h_{2} - h_{1})] \\ 4100 = 1000 [(\frac{4 {v_{1}}^{2} - {v_{1}}^{2}}{2}) + 10 \times (0 - 1)] \\ \frac{41}{10} + 10 = \frac{3 {v_{1}}^{2}}{2} \\ \Rightarrow v_{1} = \sqrt{\frac{47}{5}} \Rightarrow x = 47 \end{array}$