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 \(\frac{\mathrm{a}}{2}\). The pressure difference between the wide and narrow sections of pipe is 4100 Pa. At wider section, the velocity of fluid is \(\frac{\sqrt{\mathrm{x}}}{6} \mathrm{~ms}^{-1}\) for \(\mathrm{x}=\) \_\_\_\_ (Given \(\mathrm{g}=10 \mathrm{~m}^{-2}\))

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Fluid Mechanics

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{x}{6} ms^{- 1}

for

x =

____ (Given

g = 10 m^{- 2}

)

Answer: 363

From the continuity equation:

a v_{1} = \frac{a}{2} v_{2}

, so

v_{2} = 2 v_{1}

.

Applying Bernoulli's theorem between the wide and narrow sections:

P_{1} + ρ g h_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + ρ g h_{2} + \frac{1}{2} ρ v_{2}^{2}

P_{1} - P_{2} = ρ [\frac{v _{2}^{2} - v _{1}^{2}}{2} + g (h_{2} - h_{1})]

Given

P_{1} - P_{2} = 4100

Pa,

ρ = 800

kg/m³,

h_{2} - h_{1} = - 1

m (from figure):

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

\frac{4100}{800} = \frac{3 v _{1}^{2}}{2} - 10

5.125 + 10 = \frac{3 v _{1}^{2}}{2}

15.125 = \frac{3 v _{1}^{2}}{2}

v_{1}^{2} = \frac{121}{8 \times 3} \times 3 = \frac{121}{12}

v_{1} = \frac{11}{2 3} = \frac{121}{2 3} = \frac{363}{6}

Therefore,

x = 363

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Integer answer