Question 60 of 192

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

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Answer: 363

From equation of continuity,
$a{v}_{1}=\dfrac{a}{2}{v}_{2}\Rightarrow {v}_{2}=2{v}_{1}$
From Bernoulli's theorem,
${P}_{1}+\dfrac{1}{2}\rho {{v}_{1}}^{2}+\rho g{h}_{1}={P}_{2}+\dfrac{1}{2}\rho {{v}_{2}}^{2}+\rho g{h}_{2} \Rightarrow {P}_{1}-{P}_{2}=\rho \left[g\left({h}_{2}-{h}_{1}\right)+\dfrac{1}{2}\left({{v}_{2}}^{2}-{{v}_{1}}^{2}\right)\right] \Rightarrow 4100=800\left[10\left(-1\right)+\dfrac{1}{2}3{{v}_{1}}^{2}\right] \Rightarrow \dfrac{4100}{800}+10=\dfrac{3}{2}{{v}_{1}}^{2} \Rightarrow {v}_{1}=\sqrt{\dfrac{121}{4\times 3}}=\dfrac{\sqrt{363}}{6}m{s}^{-1}$
Therefore, $x=363$.