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 $g =10 \; m ^{-2}$ )
JEE MAIN 2022, Diffcult
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From continuity equation
$a v _{1}=\frac{ a }{2} v _{2}$
$v _{2}=2 v _{1}$
From Bernoulli's theorem,
$P _{1}+\rho g h_{1}+\frac{1}{2} \rho v _{1}^{2}= P _{2}+\rho g h_{2}+\frac{1}{2} \rho v _{2}^{2}$
$P _{1}- P _{2}=\rho\left[\left(\frac{ v _{2}^{2}- v _{1}^{2}}{2}\right)+ g \left( h _{2}- h _{1}\right)\right]$
$4100=800\left[\left(\frac{4 v _{1}^{2}- v _{1}^{2}}{2}\right)+10 \times(0-1)\right]$
$\frac{41}{8}+10=\frac{3 v _{1}^{2}}{2}$
$\frac{121}{8} \times \frac{2}{3}= v _{1}^{2}$
$v _{1}=\sqrt{\frac{ I 21}{4 \times 3} \times \frac{3}{3}}$
$v _{1}=\frac{\sqrt{363}}{6} \; m / s$
$X =363$
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