Question

A liquid flows through two capillary tubes connected in series. Their lengths are 4l and l and radii are 2r and r respectively. The ratio of the pressure differences across the first and 2nd tube is:

• A. 1 : 4
• B. 1 : 64
• C. 8 :1
• D. 1 :1

Answer 1

The correct option is A 1 : 4

As, the volume or flow rate of liquid from them is constant.
Let their pressure differnce be $P_1$ and $P_2$ respectively.
from the Poiseuille's equation,
$\Rightarrow \frac{\pi P_1(2D{)}^4}{8\eta 4L}=\frac{\pi P_2{D}^4}{8\eta L}$
$\Rightarrow \frac{16P_1}{4}=P_2\Rightarrow P_1:P_2=1:4$