JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
A liquid of density \(750\,kgm ^{-3}\) flows smoothly through a horizontal pipe that tapers in crosssectional area from \(A _{1}=1.2 \times 10^{-2}\,m ^{2}\) to \(A_{2}=\frac{A_{1}}{2}\). The pressure difference between the wide and narrow sections of the pipe is \(4500\,Pa\). The rate of flow of liquid is________\(\times 10^{-3}\,m ^{3} s ^{-1}\)
- A \(20\)
- B \(23\)
- C \(24\)
- D \(29\)
Answer & Solution
Correct Answer
(C) \(24\)
Step-by-step Solution
Detailed explanation
\(A _{2}=\frac{ A _{1}}{2}\) \(P _{1}- P _{2}=4500\,Pa\) \(P _{1}+\frac{1}{2} \rho V _{1}^{2}+\rho gh = P _{2}+\frac{1}{2} \rho V _{2}^{2}+\rho gh\) \(P _{1}- P _{2}=\frac{1}{2} \rho\left( V _{2}^{2}- V _{1}^{2}\right)\) And \(A _{1} V _{1}= A _{2} V _{2}\)…
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