Answer:
[tex] 0.12 mm^{2}[/tex]
Explanation:
[tex]v_{1}[/tex] = flow speed in the region of the partial blockage = 5100 mm/h
[tex]v_{2}[/tex] = flow speed in the clear region of nearby vessel = 1200 mm/h
[tex]A_{2}[/tex] = Cross-sectional area for the clear region = 0.51 mm²
[tex]A_{1}[/tex] = Cross-sectional area for the region near the blockage
According to equation of continuity, Volume flow rate must remain same at all points. hence
[tex]A_{1} v_{1} = A_{2} v_{2} \\A_{1} (5100) = (0.51) (1200)\\ A_{1} (5100) = 612\\\\ A_{1} = \frac{612}{5100} \\\\ A_{1} = 0.12 mm^{2}[/tex]