Measuring Pressure — Question 2

Question 2

One meniscus displacement is not the height difference

A U-tube has vertical arms of areas $A_L=1.0\,\mathrm{cm^2}$ and $A_R=4.0\,\mathrm{cm^2}$. Its liquid density is $1000\,\mathrm{kg/m^3}$. Starting with level surfaces, the left gas pressure is increased while the right gas pressure is unchanged. The left surface falls $8.0\,\mathrm{mm}$; use $g=10\,\mathrm{m/s^2}$.

1. Find the right-surface rise and the final pressure difference.

2. What pressure difference would be inferred if the left displacement alone were used as the column difference? Quantify the fractional underestimate.

Question 2 - Solution

One meniscus displacement is not the height difference

(a) Derivation and calculation

Liquid-volume conservation gives$$A_Ld_L=A_Rd_R\Rightarrow\boxed{d_R=(1/4)(8.0)=2.0\,\mathrm{mm}}.$$$$h=d_L+d_R=10.0\,\mathrm{mm}=.0100\,\mathrm m.$$$$\boxed{p_L-p_R=\rho gh=(1000)(10)(.0100)=100\,\mathrm{Pa}}.$$

(b) Result and interpretation

$$\Delta p_{\mathrm{wrong}}=\rho gd_L=80\,\mathrm{Pa}.$$$$\boxed{\frac{100-80}{100}=.200=20.0\%\ \text{underestimate}}.$$$$d_R/d_L=A_L/A_R.$$Neglecting reservoir motion is justified only when this area ratio is sufficiently small.

Physical check

The displacement of either single meniscus is smaller than their total separation.