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Q. A non-viscous liquid of constant density $1000\, kg\, m ^{-3}$ flows in a streamline motion along a tube of variable cross-section. The tube is kept inclined in the vertical plane as shown in figure. The areas of cross-section of the tube at P and Q at heights of 2 m and 5 m are respectively $4 \times 10^{-3} \,m ^{2}$ and $8 \times 10^{-3} \,m ^{2}$. The velocity of liquid at point is $1 \,m \,s ^{-1}$. The velocity of liquid at $Q$ is
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Mechanical Properties of Fluids

Solution:

$A_{P} v_{P}=A_{Q} v_{Q}$
$\Rightarrow v_{Q}=\frac{A_{P} v_{P}}{A_{O}}$
$=\frac{\left(4 \times 10^{-3} m ^{2}\right)\left(1 m s ^{-1}\right)}{\left(8 \times 10^{-3} m ^{2}\right)}$
$=0.5 \,m \,s ^{-1}$