Question

Three capillaries of length l, l/2 and l/3 are connected in series. Their radii are r,r/2 and r/3 respectively. Then, it streamline flow is to be maintained and the pressure across the first capillary is p

Answer 1

Hello dear,

● Answer-
P2 = 8P
P3 = 27P

● Explanation-
# Formula-
dQ/dt = πP^4 / 8ηl

# Solution-
To maintain streamline flow-
dQ1/dt = dQ2/dt = dQ3/dt
πP1r1^4 / 8ηl1 = πP2r2^4 / 8ηl3 = πP3r3^4 / 8ηl3

Putting values-
πP(r)^4 / 8η(l) = πP2(r/2)^4 / 8η(l/2) = πP3(r/3)^4 / 8η(l/3)

Solving this-
P2 = 8 P
P3 = 27 P

Hope this is helpful...