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 \(x = \dots\) \(\left(\right.\) Given \(g =10 \; m ^{-2}\) )

\(a v _{1}=\frac{ a }{2} v _{2}\)

\(v _{2}=2 v _{1}\)

From Bernoulli's theorem,

\(P _{1}+\rho g h_{1}+\frac{1}{2} \rho v _{1}^{2}= P _{2}+\rho g h_{2}+\frac{1}{2} \rho v _{2}^{2}\)

\(P _{1}- P _{2}=\rho\left[\left(\frac{ v _{2}^{2}- v _{1}^{2}}{2}\right)+ g \left( h _{2}- h _{1}\right)\right]\)

\(4100=800\left[\left(\frac{4 v _{1}^{2}- v _{1}^{2}}{2}\right)+10 \times(0-1)\right]\)

\(\frac{41}{8}+10=\frac{3 v _{1}^{2}}{2}\)

\(\frac{121}{8} \times \frac{2}{3}= v _{1}^{2}\)

\(v _{1}=\sqrt{\frac{ I 21}{4 \times 3} \times \frac{3}{3}}\)

\(v _{1}=\frac{\sqrt{363}}{6} \; m / s\)

\(X =363\)