Measuring Things — Question 9

Question 9

The volume that the water actually measures

A fully submerged, nonabsorbing solid object is measured by water displacement. The initial reading is $V_i=40.0\pm0.5$ mL and the final reading is $V_f=57.0\pm0.5$ mL. Each $\pm$ is a hard reading-error bound.

External air bubbles cling to the submerged object. A separate check bounds their total volume as $V_b=3.0\pm1.0$ mL at the same immersion conditions. Assume no spill, evaporation, trapped water change, or other displacement contribution. The supplied physical model is $$V_f-V_i=V_{\mathrm{object}}+V_b.$$

1. Find the central estimate of the object’s volume. How would neglecting the bubbles bias the result?

2. Compute the smallest and largest object volumes allowed by all three bounds, without statistical cancellation assumptions.

3. After carefully removing the external bubbles, predict the central final water reading if the initial volume is unchanged. Explain why repeating the original bubble-contaminated measurement many times is not an adequate substitute for this control.

Question 9 - Solution

The volume that the water actually measures

(a) Correct the quantity being measured.

The central water rise is $57.0-40.0=17.0$ mL, but it includes the bubbles: $$\boxed{V_{\mathrm{object}}=57.0-40.0-3.0=14.0\ \mathrm{mL}}.$$ Neglecting the bubbles would give $17.0$ mL, an overestimate of $3.0$ mL for the central inputs.

(b) Take the worst permitted combinations.

$$\delta V_o\le\delta V_f+\delta V_i+\delta V_b=0.5+0.5+1.0=2.0\ \mathrm{mL}.$$

The intervals are $V_f\in[56.5,57.5]$, $V_i\in[39.5,40.5]$, and $V_b\in[2.0,4.0]$ mL. Hence $$V_{\min}=56.5-40.5-4.0=12.0\ \mathrm{mL},$$ $$V_{\max}=57.5-39.5-2.0=16.0\ \mathrm{mL}.$$ Therefore $\boxed{V_{\mathrm{object}}\in[12.0,16.0]\ \mathrm{mL}}$, equivalently a central estimate of $14.0$ mL with a worst-case bound of $2.0$ mL. The three absolute bounds add; no independence or probability model was supplied.

(c) Remove the nuisance contribution.

With no external bubbles, the predicted central final reading is $40.0+14.0=\boxed{54.0\ \mathrm{mL}}$. This is a central prediction, not an uncertainty-free new measurement. Repeated readings with persistent bubbles can be highly consistent yet still measure object-plus-bubble volume. A control that removes bubbles addresses this mechanism; repetition alone does not.