Two capillary tubes of same length each of 50   cm but of different radii 4   mm and 2   mm…

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,

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