Question

Three capillary tubes of same length but internal radii 0.3 mm, 0.45 mm and 0.6 min are connected in series and a liquid flows steadily -through them. If the pressure difference across the third capillary is 8.1 mm of mercury, the pressure difference across the first capillary (in mm of mercury) is

• A. 16.2
• B. 32.4
• C. 129.6
• D. 2.025
• E. 4.05

Answer 1

Answer: (C)

129.6

We know that the equation,
$\frac{Q}{t}=\frac{\pi p{r}^4}{8nl}$
Here, pressure difference
$p\propto \frac{1}{r^4}$
$p\propto r^{-4}$
$\frac{p_1}{p_3}={\left(\begin{array}{c}\frac{r_3}{r_1}\end{array}\right)}^4$
$\frac{p_1}{8.1}={\left(\begin{array}{c}\frac{0.6}{0.3}\end{array}\right)}^4$
$P_1=8.1\times 16=129.6$