Two capillary tubes of same length each of $50 \mathrm{cm}$ but of different radii $4 \mathrm{mm}$ and $2 \mathrm{mm}$ are connected in series. When water flows, the pressure difference between the ends of the arrangement is $P$. Then the pressure difference between the ends of the first tube is

The figure below represents the capillary tubes of same
lengths but different radii.

From the Poiseuille's equation, the expression for the water
resistance,

$R=\frac{8\eta l}{\pi r^4}$

$R\propto \frac{1}{r^4}$

The ratio of water resistance in tube I and II,

$\frac{R_l}{R_{II}}=\frac{{\left(r_{II}\right)}^4}{{\left(r_l\right)}^4}=\frac{16\times {10}^{-12}}{16\times 16\times {10}^{-12}}$

$16{\mathrm{R}}_I=R_{II}$

Let $R_I=R,R_{II}=16\mathrm{R}$

Thus, the pressure difference between ends of first tube is,

$\frac{p}{R+16\mathrm{R}}\times R=\frac{p}{17}$