Physics 9702/23 — May/June 2022

Background Concept

Density $\rho$ is defined as mass per unit volume:

$$\rho = \frac{m}{V}$$

For a sphere, the volume is

$$V = \frac{4}{3}\pi r^3$$

where $r$ is the radius (half the diameter). If mass is in $\text{g}$ and volume is in $\text{cm}^3$, then density comes out in $\text{g cm}^{-3}$.

Understanding the Question

You are given:

• diameter of a solid metal sphere: $d = (3.42 \pm 0.02)\ \text{cm}$
• mass: $m = (67 \pm 2)\ \text{g}$

Part (a) asks for the density in $\text{g cm}^{-3}$. So we need the sphere’s volume in $\text{cm}^3$ and then divide the mass by that volume.

Approach

1. Convert diameter to radius using $r = d/2$.
2. Find the volume using $V = \frac{4}{3}\pi r^3$.
3. Use $\rho = m/V$.
4. Round the final answer to a sensible number of significant figures (limited mainly by the mass, given to 2 s.f.).

Step-by-Step Reasoning

1. Radius:

$$r = \frac{d}{2} = \frac{3.42}{2} = 1.71\ \text{cm}$$

1. Volume of sphere:

$$V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(1.71)^3$$

Calculating gives $V \approx 20.9\ \text{cm}^3$.

1. Density:

$$\rho = \frac{m}{V} = \frac{67\ \text{g}}{20.9\ \text{cm}^3} \approx 3.20\ \text{g cm}^{-3}$$

1. Significant figures: the mass $67\ \text{g}$ is 2 s.f., so quoting $\rho$ as $3.2\ \text{g cm}^{-3}$ is appropriate.

Key Takeaways

• Always use radius (not diameter) in the sphere volume formula.
• Keep units consistent so the density unit matches what is asked.
• Round the final answer sensibly based on the given data.

Common Mistakes

• Using $d$ directly in $V = \frac{4}{3}\pi r^3$ (forgetting to halve it).
• Writing the unit as $\text{g cm}^3$ instead of $\text{g cm}^{-3}$.
• Giving too many significant figures (e.g. $3.19904$) compared with the data.

Things to Be Careful About

• Don’t convert to SI unless asked; here $\text{g}$ and $\text{cm}$ are already aligned with $\text{g cm}^{-3}$.
• Ensure the cube applies to the whole radius value: $(1.71)^3$, not $1.71^3$ mis-keyed as $1.7^3$ unless rounding deliberately.

Background Concept

For quantities that are multiplied and divided, the fractional (or percentage) uncertainties add.

If

$$Q = \frac{A\,B}{C}$$

then approximately

$$\frac{\Delta Q}{Q} \approx \frac{\Delta A}{A} + \frac{\Delta B}{B} + \frac{\Delta C}{C}$$

Also, if a quantity is raised to a power, the fractional uncertainty is multiplied by that power:

$$Q = A^n \Rightarrow \frac{\Delta Q}{Q} \approx n\frac{\Delta A}{A}$$

In this question, density is

$$\rho = \frac{m}{V}$$

and for a sphere $V \propto r^3$ and since $r = d/2$, it is also true that $V \propto d^3$ (the factor $1/2$ is an exact constant, so it contributes no uncertainty).

Understanding the Question

You must find the percentage uncertainty in the calculated density. The density depends on:

• the measured mass $m = 67 \pm 2\ \text{g}$
• the measured diameter $d = 3.42 \pm 0.02\ \text{cm}$ through the volume $V$.

Because volume depends on $d^3$, the diameter uncertainty gets “amplified” by a factor of 3 when transferred to the volume.

Approach

1. Compute percentage uncertainty in mass: $(\Delta m/m)\times 100\%$.
2. Compute percentage uncertainty in diameter: $(\Delta d/d)\times 100\%$.
3. Multiply the diameter percentage uncertainty by 3 to get the volume percentage uncertainty.
4. Add mass and volume percentage uncertainties to get density percentage uncertainty.

Step-by-Step Reasoning

1. Mass percentage uncertainty:

$$\%\ \Delta m = \frac{2}{67}\times 100\% \approx 2.99\% \approx 3.0\%$$

1. Diameter percentage uncertainty:

$$\%\ \Delta d = \frac{0.02}{3.42}\times 100\% \approx 0.585\%$$

1. Because $V \propto d^3$:

$$\%\ \Delta V = 3\times \%\ \Delta d \approx 3\times 0.585\% = 1.76\%$$

1. Since $\rho = m/V$ (a division), add the percentage uncertainties:

$$\%\ \Delta \rho \approx \%\ \Delta m + \%\ \Delta V \approx 2.99\% + 1.76\% = 4.75\%$$

Rounded suitably gives about $4.8\%$ (or $4.7\%$ depending on rounding during intermediate steps).

Key Takeaways

• For powers: multiply the fractional uncertainty by the power.
• For multiplication/division: add fractional (percentage) uncertainties.
• Constants (like $1/2$ in $r = d/2$) do not contribute uncertainty.

Common Mistakes

• Forgetting the factor of 3 for the cubic dependence of volume on diameter.
• Using absolute uncertainties directly (e.g. adding $2$ and $0.02$) instead of percentage/fractional uncertainties.
• Subtracting uncertainties because of the division ($m/V$); you still add the fractional uncertainties.

Things to Be Careful About

• Use diameter or radius consistently: if you use radius, you must use $\Delta r/r$; but $\Delta r/r = \Delta d/d$ because both are halved.
• Don’t over-round too early; keep a couple of extra digits until the final percentage.
• State the final uncertainty as a percentage, as asked.