Physics 9702/23 — October/November 2016

Background Concept

For a sphere of diameter $d$, the volume is

$$V = \frac{\pi d^3}{6}$$

Mass is related to density by

$$m = \rho V$$

Combining these gives the provided formula:

$$m = \frac{\pi d^3 \rho}{6}$$

Understanding the Question

You are given $m = 7.5\ \text{kg}$ and $\rho = 8100\ \text{kg m}^{-3}$ and asked to calculate the diameter $d$ using

$$m = \frac{\pi d^3 \rho}{6}$$

This part (i) is just the numerical value of $d$ (no uncertainty yet).

Approach

Rearrange the equation to make $d$ the subject:

1. Isolate $d^3$.
2. Take the cube root to obtain $d$.
3. Substitute the given values (already in SI units).

Step-by-Step Reasoning

Start with

$$m = \frac{\pi d^3 \rho}{6}$$

Multiply both sides by $6$ and divide by $\pi\rho$:

$$d^3 = \frac{6m}{\pi\rho}$$

Now take the cube root:

$$d = \left(\frac{6m}{\pi\rho}\right)^{1/3}$$

Substitute $m = 7.5\ \text{kg}$ and $\rho = 8100\ \text{kg m}^{-3}$:

$$d = \left(\frac{6 \times 7.5}{\pi \times 8100}\right)^{1/3}$$

Evaluating gives

$$d \approx 0.121\ \text{m}$$

The unit is metres because the density is in $\text{kg m}^{-3}$ and mass in $\text{kg}$, so the volume comes out in $\text{m}^3$ and hence $d$ in $\text{m}$.

Key Takeaways

• For relationships like $m \propto d^3$, you must take a cube root at the end.
• Keep units consistent (SI) to avoid hidden conversion errors.

Common Mistakes

• Forgetting the cube root and leaving the answer as $d^3$.
• Using radius instead of diameter (the formula already uses diameter).
• Calculator error with brackets (e.g. not dividing by $\pi\rho$ correctly).

Things to Be Careful About

• Ensure $\pi\rho$ is in the denominator together.
• Quote the final $d$ with a sensible number of significant figures (typically 2–3 here).

Background Concept

When a quantity is calculated from measured values, its uncertainty must be propagated.

Key rules (worst-case addition of fractional/percentage uncertainties, as used in A Level):

• For multiplication or division, percentage uncertainties add.
• For a power $y = x^n$, the percentage uncertainty multiplies by $|n|$.

Here,

$$d = \left(\frac{6m}{\pi\rho}\right)^{1/3}$$

The constants $6$ and $\pi$ are treated as exact, so only uncertainties in $m$ and $\rho$ contribute.

Understanding the Question

You already found $d$ in (i). Now you must:

1. Use the given uncertainties in $m$ ($\pm 4\%$) and $\rho$ ($\pm 5\%$).
2. Find the absolute uncertainty in $d$.
3. Quote $d \pm \Delta d$ to an appropriate number of significant figures.

Approach

1. Simplify the dependence: $d \propto (m/\rho)^{1/3}$.
2. Combine percentage uncertainties for $m/\rho$ by addition.
3. Apply the power rule for the cube root (power $1/3$).
4. Convert percentage uncertainty in $d$ into absolute uncertainty using $\Delta d = (\%\,\text{uncertainty}/100)\,d$.
5. Round the uncertainty (usually 1 s.f.), then round $d$ to the same decimal place.

Step-by-Step Reasoning

From

$$d = \left(\frac{6m}{\pi\rho}\right)^{1/3}$$

we note

$$d \propto \left(\frac{m}{\rho}\right)^{1/3}$$

1) Uncertainty in the ratio $m/\rho$

For division, percentage uncertainties add:

$$\%\Delta\left(\frac{m}{\rho}\right) = 4\% + 5\% = 9\%$$

2) Uncertainty in $d = (m/\rho)^{1/3}$

For a power, multiply the percentage uncertainty by the power:

$$\%\Delta d = \frac{1}{3} \times 9\% = 3\%$$

This makes sense physically: taking a cube root reduces the relative uncertainty.

3) Convert to absolute uncertainty

Using $d \approx 0.121\ \text{m}$ from part (i):

$$\Delta d = 0.03 \times 0.121 = 0.00363\ \text{m}$$

Round the uncertainty to 1 significant figure:

$$\Delta d \approx 0.004\ \text{m}$$

Then quote $d$ to the same decimal place (thousandths of a metre):

$$d = 0.121\ \text{m}$$

So

$$d = 0.121 \pm 0.004\ \text{m}$$

Key Takeaways

• For $d \propto (m/\rho)^{1/3}$, add uncertainties for $m/\rho$, then divide by 3 for the cube root.
• Convert percentage uncertainty to absolute using $\Delta d = (\%/100)d$.
• Round uncertainty first, then round the value to match its decimal place.

Common Mistakes

• Subtracting percentage uncertainties for a division (they should be added for worst-case).
• Forgetting to multiply by $1/3$ for the cube root.
• Giving a percentage uncertainty when the question asks for absolute uncertainty.
• Rounding $d$ and $\Delta d$ to inconsistent decimal places (e.g. $0.12 \pm 0.004$).

Things to Be Careful About

• Do not include uncertainty contributions from constants like $\pi$ and 6.
• Ensure the final value and its uncertainty are quoted in metres.
• Keep rounding sensible: typically $\Delta d$ to 1 s.f. and $d$ to the same decimal place as $\Delta d$.