Q. A liquid flows through a pipe of non-uniform cross-section. If $ {{\text{A}}_{\text{1}}} $ and $ {{\text{A}}_{2}} $ arc the cross-sectional areas of the pipe at two points, the ratio of velocities of the liquid at these points will be :

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A

$ {{A}_{1}}{{A}_{2}} $

B

$ \frac{{{A}_{1}}}{{{A}_{2}}} $

C

$ \frac{{{A}_{2}}}{{{A}_{1}}} $

D

$ \frac{1}{{{A}_{1}}{{A}_{2}}} $

Solution:

Volume of liquid flowing at first point $ =A{{ & }_{1}}{{v}_{1}} $ Similarly, volume of liquid flowing at second point $ =A{{ & }_{2}}{{v}_{2}} $ From equation of continuity, $ {{A}_{1}}{{v}_{1}}={{A}_{2}}{{v}_{2}} $ or $ \frac{{{v}_{1}}}{{{v}_{2}}}=\frac{{{A}_{2}}}{{{A}_{1}}} $

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