Mass — Question 10

Question 10

Equal forces in air need not mean equal masses

For an object of mass $m$ and displaced volume $V$ in air of density $\rho_a$, use the supplied model for the downward force on a balance pan: $$F=(m-\rho_a V)g.$$ An ideal equal-arm balance is level when these forces are equal. The field $g$ is the same, nonzero value on both sides; pan effects cancel. Ignore all other errors and changes in the objects.

The reference has exact mass $m_r=100.000\ \mathrm{g}$ and displaced volume $V_r=12.50\ \mathrm{cm^3}$. The sample’s central displaced volume is $V_x=125.0\ \mathrm{cm^3}$ and the central air density is $\rho_a=0.001200\ \mathrm{g\,cm^{-3}}$.

1. Derive a formula for $m_x$ at balance. Find its central value and explain the sign of the correction relative to 100.000 g.

2. The actual values satisfy the hard bounds $$V_x\in[124.5,125.5]\ \mathrm{cm^3},\qquad \rho_a\in[0.001100,0.001300]\ \mathrm{g\,cm^{-3}}.$$ Every pair is allowed. Find the full mass interval. Reference data remain exact.

3. A procedure assigns the sample exactly 100.000 g and permits an absolute mass error of at most 0.010 g. Is the assignment compatible with that requirement for any allowed values? Could adjusting the empty-balance zero remove this discrepancy for all sample volumes?

4. If air were removed without changing either object’s composition or volume, would their masses change in this model? Would these same objects still balance?

Question 10 - Solution

Equal forces in air need not mean equal masses

(a) Cancel the common field, not the volume-dependent term.

At balance, $(m_x-\rho_a V_x)g=(m_r-\rho_a V_r)g$, so $$\boxed{m_x=m_r+\rho_a(V_x-V_r)}.$$ The central correction is $0.001200(125.0-12.50)=0.1350\ \mathrm{g}$, giving $\boxed{m_{x,0}=100.1350\ \mathrm{g}}$. The larger sample displaces more air and receives more upward buoyant force. It must have greater mass to exert the same downward force as the smaller reference.

(b) Both uncertain positive factors increase the correction.

Here $V_x-V_r>0$ throughout. The extreme corrections therefore use the paired low endpoints and paired high endpoints: $$m_{x,\min}=100.000+0.001100(124.5-12.50) =100.1232\ \mathrm{g},$$ $$m_{x,\max}=100.000+0.001300(125.5-12.50) =100.1469\ \mathrm{g}.$$ Thus $\boxed{m_x\in[100.1232,100.1469]\ \mathrm{g}}$. The two deviations from the central value need not be equal; the bound comes from products of allowed values.

(c) The omitted correction exceeds the permitted error.

Assigning 100.000 g underestimates the mass by at least 0.1232 g, greater than 0.010 g for every allowed pair. The assignment fails the requirement. An empty-balance zero adjustment cannot remove a term proportional to the difference in displaced volumes for arbitrary objects. A supported buoyancy correction or appropriate comparison conditions are needed.

(d) Removing air changes the measured force, not the masses.

$$F_x-F_r=(m_x-m_r)g>0\qquad(\rho_a=0,\ g>0).$$

With $\rho_a=0$, each force is $mg$. The unchanged sample is more massive than the reference, so the sample side is heavier and they no longer balance. Their masses have not changed under the stated assumptions; only the surrounding-air force has disappeared.