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________$\times 10^{-3}\,m ^{3} s ^{-1}$
JEE MAIN 2022, Diffcult
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$A _{2}=\frac{ A _{1}}{2}$
$P _{1}- P _{2}=4500\,Pa$
$P _{1}+\frac{1}{2} \rho V _{1}^{2}+\rho gh = P _{2}+\frac{1}{2} \rho V _{2}^{2}+\rho gh$
$P _{1}- P _{2}=\frac{1}{2} \rho\left( V _{2}^{2}- V _{1}^{2}\right)$
And $A _{1} V _{1}= A _{2} V _{2}$
$\Rightarrow \quad V _{2}=2 V _{1}$
$4500=\frac{1}{2} \times 750 \times 3 V _{1}^{2}$
$V _{1}=2\,m / s$
Volume flow rate $= A _{1} V _{1}=24 \times 10^{-3}\,m ^{3} s ^{-1}$
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