A) 3 m/s
B) 6 m/s
C) 12 m/s
D) 8 m/s
Correct Answer: C
Solution :
According to principle of continuity, when an incompressible and non-viscous liquid flows in a streamlined motion through a tube of non-uniform cross-section, then the product of the area of cross-section and the velocity of flow is same at every point in the tube. Thus, Av = constant or \[{{A}_{1}}{{v}_{1}}={{A}_{2}}{{v}_{2}}\] or \[\pi {{r}_{1}}^{2}{{v}_{1}}=\pi {{r}_{2}}^{2}{{v}_{2}}\] Given, \[{{r}_{1}}=\frac{4}{2}cm=0.02\,m\], \[{{r}_{2}}=\frac{2}{2}\,cm=0.02\,m\], \[{{v}_{1}}=3\,m/s\] \[\therefore \] \[\pi {{(0.02)}^{2}}\times 3=\pi {{(0.01)}^{2}}{{v}_{2}}\] or \[{{v}_{2}}={{\left( \frac{0.02}{0.01} \right)}^{2}}\times 3\]
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