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  • A liquid of density  flows smoothly through a horizontal pip

A liquid of density \mathrm{750 \; \mathrm{kgm}^{-3}} flows smoothly through a horizontal pipe that tapers in cross-sectional area from \mathrm{\mathrm{A}_{1}=1.2 \times 10^{-2} \mathrm{~m}^{2} \text { to } \mathrm{A}_{2}=\frac{\mathrm{A}_{1}}{2}}. The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is ______________\mathrm{\times 10^{-3} \mathrm{~m}^{3} \mathrm{~s}^{-1}}.

Option: 1

24


Option: 2

12


Option: 3

48


Option: 4

36


Answers (1)

best_answer

\begin{aligned} A_{1} V_{1} &=A_{2} V_{2} \\ 2 A V &=AV_{2} \\ V_{2} &=2 V \end{aligned} ___________(1)

Bernoulli's equation

\begin{aligned} &P_{1}+\frac{1}{2} \rho V_{1}^{2}=P_{2}+\frac{1}{2} \rho V_{2}^{2} \\ &\frac{3}{2} \rho v^{2}=P_{1}-P_{2} \\ &V=\sqrt{\frac{2\left(P_{1}-P_{2}\right)}{3 \rho}} \\ &=\sqrt{\frac{2 \pi\times 4500}{3 \times 750}}=2 \mathrm{~m} / \mathrm{s} \\ \end{aligned} 

So,

Q=A_{1} v=24 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{sec}

Hence 24 is the correct answer

Posted by

Shailly goel

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