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An ideal fluid of density $800 \mathrm{kgm}^{-3}$ , 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 $4100 \mathrm{~Pa}$ . At wider section, the velocity of fluid is $\frac{\sqrt{x}}{6} \mathrm{~ms}-1$ for $x$ = ________________. $(Given \: \: g=10 \mathrm{~ms}^{-2} )$

Option: 1
363

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

By equation of continuity ,
$\mathrm{av_{1}= \left ( \frac{a}{2} \right )v_{2}}$
$\mathrm{\therefore v_{2}= 2v_{1}\: \rightarrow (1)}$

By Bernoulli's theorem ,
$\mathrm{p_{1}+\frac{1}{2}\, \delta v_{1}^{2}+\delta\, gh= p_{2}+\frac{1}{2}\, \delta v_{2}^{2} +0}$
$\mathrm{\left ( p_{1}-p_{2} \right )= 4100\, pa}$

$4100+800 \times 1\times 10=400(3)\left(\frac{\mathrm{x}}{36}\right)$

$\mathrm{12100= \left ( 400 \right )\left ( \frac{x}{12} \right )}$
$\mathrm{x= \frac{121\times 12}{4}}$
$\mathrm{x= 363}$

Posted by

Suraj Bhandari

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