Question

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 $g=10\, m ^{-2}$ )

Answer 1

From continuity equation
$av _{1}=\frac{ a }{2} v _{2}$
$v _{2}=2 v _{1}$
From Bernoulli's theorem,
$P _{1}+\rho g h_{1}+\frac{1}{2} \rho v _{1}^{2}= P _{2}+\rho gh _{2}+\frac{1}{2} \rho v _{2}^{2}$
$P _{1}- P _{2}=\rho\left[\left(\frac{ v _{2}^{2}- v _{1}^{2}}{2}\right)+ g \left( h _{2}- h _{1}\right)\right]$
$4100=800\left[\left(\frac{4 v _{1}^{2}- v _{1}^{2}}{2}\right)+10 \times(0-1)\right]$
$\frac{41}{8}+10=\frac{3 v _{1}^{2}}{2}$
$\frac{121}{8} \times \frac{2}{3}= v _{1}^{2}$
$v _{1}=\sqrt{\frac{ I 21}{4 \times 3} \times \frac{3}{3}}$
$v _{1}=\frac{\sqrt{363}}{6} m / s$
$X =363 .$