Oil-water interface in enlarged-end U-tube

Data: tube $d=7$ mm, enlarged ends $D=44$ mm, oil $S=0.83$ (right limb), water in the left limb and base, interface moves $x=1$ cm $=0.01$ m.

   p1 (water side)       p2 (oil side)
   [D]                   [D]
    |  water      oil     |
    |                     |
    |__________ interface moves x

Raise the pressure $p_1$ on the water side. The interface moves up the right limb by $x$. The volume displaced in the narrow tube moves the free surfaces in the enlarged ends by

$$y=x\left(\frac{d}{D}\right)^{2}=0.01\left(\frac{7}{44}\right)^{2}=2.53\times10^{-4}\ \text{m}$$

(the water surface falls $y$, the oil surface rises $y$). Initially the pressures balance at the interface level, so the change in the pressure difference is balanced by the change in column heights:

$$p_1-p_2=\gamma_w y+\gamma_o y+(\gamma_w-\gamma_o)x$$ $$p_1-p_2=\gamma_w\,x\left[(1+s)\left(\frac{d}{D}\right)^{2}+(1-s)\right]$$

With $s=0.83$:

$$p_1-p_2=9810(0.01)\left[1.83(0.02531)+0.17\right]=98.1\times0.2163$$ $$p_1-p_2=21.2\ \text{Pa}$$

Answer: a pressure difference of about 21.2 Pa (2.16 mm of water) moves the interface by 1 cm. The small value shows how sensitive the enlarged-end manometer is.