A liquid flows steadily through a series combination of three capillary tubes of radii $r, 2r and 3r$, all of the same length $L$. If the pressure difference across the combination is $28 cm$ of mercury, the pressure difference $(\mathrm{in} \mathrm{cm} \mathrm{of} \mathrm{Hg})$ across the tube of radius $2r$ is very nearly equal to

$$\begin{gathered} Q_1=Q_2=Q_3. Therefore, \\ \frac{\pi p_1{r}^4}{8\eta L}=\frac{\pi p_2(2r{)}^4}{8\eta L}=\frac{\pi p_3(3r{)}^4}{8\eta L} \\ \Rightarrow p_1=16p_2=81p_3 \\ p_1=16p_2 \\ and p_3=\frac{{\displaystyle 16}}{{\displaystyle 81}}{p}_2 \end{gathered}$$

Given            $$\begin{gathered} Given p_1 + p_2 + p_3 = 28 cm of Hg \\ \Rightarrow 16p_2+p_2+\frac{16}{81}{p}_2=28 \\ \Rightarrow (16+1+\frac{16}{81})p_2=28 \\ \Rightarrow p_2=\frac{81\times 28}{1393}=1.6 cm of Hg \end{gathered}$$