U-tube gauge with enlarged ends

Data: tube $d=5$ mm, enlarged ends $D$ (equal), oil $S=0.97$, interface moves $x=25$ mm for a pressure difference of 1 mm of water.

Set-up. Water is in one limb and oil in the other, the interface lying in the tube. When a pressure difference $\Delta p$ is applied, the interface moves a distance $x$ in the tube. The displaced volume $\dfrac{\pi}{4}d^{2}x$ makes both free surfaces in the enlarged ends move by

$$y=x\left(\frac{d}{D}\right)^{2}$$

(the water surface falls $y$, the oil surface rises $y$). Before the pressure change the pressures balance at the initial interface level, so only the changes count. Balancing at the new interface level, with $s=0.97$:

$$\frac{\Delta p}{\gamma_w}=y(1+s)+x(1-s)$$ $$\frac{\Delta p}{\gamma_w}=x\left[(1+s)\left(\frac{d}{D}\right)^{2}+(1-s)\right]$$

Substitute $\dfrac{\Delta p}{\gamma_w}=1$ mm, $x=25$ mm:

$$1=25\left[1.97\left(\frac{d}{D}\right)^{2}+0.03\right]\ \Rightarrow\ \left(\frac{d}{D}\right)^{2}=\frac{0.04-0.03}{1.97}=0.005076$$ $$\frac{d}{D}=0.07125\ \Rightarrow\ D=\frac{5}{0.07125}=70.2\ \text{mm}$$

Answer: diameter of each enlarged end $\approx70$ mm.