Question

Two capillary tubes of lengths in the ratio $2:1$ and radii in the ratio $1:2$ are connected in series. Assume the flow of the liquid through the tube is steady. Then, the ratio of pressure difference across the tubes is:

• A. $1:16$
• B. $32:1$
• C. $1:1$
• D. $1:8$

Answer 1

Answer: (B)

$32:1$

The rate of flow of liquid through both tubes are same.
$\therefore Q=\frac{\pi P_1{r}_1^4}{8\eta l_1}=\frac{\pi P_2{r}_2^4}{8\eta l_2}$

$\frac{P_1}{P_2}=\frac{r_2^4{l}_1}{r_1^4{l}_2}=32:1$