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  • The figure shows a liquid of a given density flowing steadily in a horizontal tube of varying cross–section. Cross-sectional areas at A are 1.5 cm2, and B is 25 mm2

The figure shows a liquid of a given density flowing steadily in a horizontal tube of varying cross–section. Cross-sectional areas at A are 1.5 cm2, and B is 25 mm2, if the speed of liquid at B is 60 cm/s then (PA – PB) is: (Given PA and PB are liquid pressures at A and B points)
Density \rho=1000 \mathrm{~kg} \mathrm{~m}^{-3}
A and B are on the axis of the tube

Option: 1

175 Pa


Option: 2

36 Pa


Option: 3

27 Pa

 


Option: 4

135 Pa


Answers (1)

best_answer

By equation of continuity,

\begin{aligned} & \mathrm{A}_1 \mathrm{~V}_1=\mathrm{A}_2 \mathrm{~V}_2 \\ & \left(1.5 \times 10^{-4}\right) \mathrm{V}_{\mathrm{A}}=\left(25 \times 10^{-6}\right) 60 \mathrm{~cm} / \mathrm{s} \\ & \mathrm{V}_{\mathrm{A}}=10 \mathrm{~cm} / \mathrm{s} \end{aligned}

By Bernoulli's theorem, 

\begin{aligned} & P_A+\frac{1}{2} \rho \mathrm{V}_{\mathrm{A}}^2=P_B+\frac{1}{2} \rho \mathrm{V}_{\mathrm{B}}^2 \\ \\& P_A-P_B=\frac{\rho}{2}\left(\mathrm{~V}_{\mathrm{B}}^2-\mathrm{V}_{\mathrm{A}}^2\right) \end{aligned}

\begin{aligned} & \mathrm{P}_{\mathrm{A}}-\mathrm{P}_{\mathrm{B}}=\frac{1000}{2}\left(60^2-10^2\right) \times 10^{-4} \\ \\& \mathrm{P}_{\mathrm{A}}-\mathrm{P}_{\mathrm{B}}=175 \mathrm{~Pa} \end{aligned}

Posted by

himanshu.meshram

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