Sections §1.6–§1.9 gave you four tools: scientific notation, dimensional analysis, significant figures, and factor-label conversion. Here are six worked problems, each combining several of those tools. The goal: when you meet a new problem, you should know where to start — pick the right tool, keep sig figs honest, and finish with a dimensional check that convinces you the answer is sensible.
None of these problems come from Halliday. The scenarios are original so you learn the method, not the answer.
Solution template
Every problem follows five steps:
- Statement + givens — what’s known, and to how many sig figs?
- Tool selection — which §1.x is the roadmap?
- Solution — carry units through every step
- Dimensional check — do the units of the answer make sense?
- Sig-fig audit — how many digits do we keep, and why?
Problem 1: Pressure at the Mariana Trench
Statement: A research submersible descends to 10.9 km in the Mariana Trench — the deepest known point in Earth’s oceans. What absolute pressure does it experience? Give the answer in both Pa and atm.
Givens:
– h = 10.9 km (depth, 3 sig figs)
– ρ_sw = 1025 kg/m^3 (seawater density, 4 sig figs)
– g = 9.81 m/s^2 (surface gravity, 3 sig figs)
– 1 atm = 101,325 Pa (definition)
– We ignore atmospheric pressure (or add it at the end)
Tools: §1.7 (dimensional check of the formula), §1.9 (km → m and Pa → atm), §1.8 (sig figs)
Solution:
Fluid column pressure formula:
$$
P = \rho\,g\,h
$$
Before plugging in numbers, do a dimensional check:
$$
[\rho\,g\,h] = \frac{\mathrm{kg}}{\mathrm{m}^3} \cdot \frac{\mathrm{m}}{\mathrm{s}^2} \cdot \mathrm{m}
= \frac{\mathrm{kg}}{\mathrm{m}\cdot\mathrm{s}^2}
= \frac{\mathrm{N}}{\mathrm{m}^2}
= \mathrm{Pa}\quad ✓
$$
Good. Convert h to SI:
$$
h = 10.9\ \mathrm{km}\ \times\ \frac{1000\,\mathrm{m}}{1\,\mathrm{km}} = 1.09 \times 10^4\ \mathrm{m}
$$
Plug in:
$$
P = (1025)\,(9.81)\,(1.09\times 10^4) = 1.10\times 10^{8}\ \mathrm{Pa}
$$
Convert to atm:
$$
P = 1.10 \times 10^{8}\ \mathrm{Pa}\ \times\ \frac{1\,\mathrm{atm}}{101{,}325\,\mathrm{Pa}} = 1.08 \times 10^3\ \mathrm{atm}
$$
Sig figs: Inputs range from 3 to 4 sig figs. By the multiplication rule (§1.8), report the answer to the smallest, i.e. 3 sig figs.
Answer:
$$
\boxed{P \approx 1.10 \times 10^{8}\ \mathrm{Pa} \approx 1.08 \times 10^3\ \mathrm{atm}}
$$
That’s about 1,080 times sea-level atmospheric pressure. For intuition: a human fingertip has roughly 1 cm^2 of surface area; at this depth, a force of about 10,800 N (the weight of a small car) presses on each fingertip. That’s why diving this deep requires purpose-built submersibles like Trieste or Deepsea Challenger.
Problem 2: Distance to Alpha Centauri
Statement: Alpha Centauri is the nearest star system to the Sun, at a distance of 4.37 ly (light-years). How many meters is that? If we receive a radio signal from it right now, how long ago was it sent?
Givens:
– d = 4.37 ly (3 sig figs)
– 1 ly = 9.461 × 10^15 m (IAU definition, 4 sig figs)
– c = 2.998 × 10^8 m/s (speed of light, 4 sig figs)
Tools: §1.6 (scientific notation for astronomical numbers), §1.9 (ly → m, s → year)
Solution:
Part 1 — distance in meters:
$$
d = 4.37\ \mathrm{ly}\ \times\ \frac{9.461 \times 10^{15}\,\mathrm{m}}{1\,\mathrm{ly}}
$$
Compute:
$$
d = 4.37 \times 9.461 \times 10^{15}\ \mathrm{m}
= 41.34 \times 10^{15}\ \mathrm{m}
= 4.13 \times 10^{16}\ \mathrm{m}
$$
Part 2 — light-travel time:
$$
t = \frac{d}{c} = \frac{4.13 \times 10^{16}\ \mathrm{m}}{2.998 \times 10^{8}\ \mathrm{m/s}}
= 1.38 \times 10^{8}\ \mathrm{s}
$$
Convert to years:
$$
t = 1.38 \times 10^{8}\ \mathrm{s}\ \times\ \frac{1\,\mathrm{year}}{3.156 \times 10^{7}\,\mathrm{s}} = 4.37\ \mathrm{year}
$$
Independent check: By definition, a light-year is the distance light travels in one year. So if a source is 4.37 ly away, light takes 4.37 year to arrive — exactly what we got. This is a simple, useful circular check.
Sig figs: The smallest input has 3 sig figs, so the answer is 3 sig figs.
Answer:
$$
\boxed{d = 4.13 \times 10^{16}\ \mathrm{m},\qquad t = 4.37\ \mathrm{year} \approx 138\ \mathrm{million\ seconds}}
$$
The signal we’re receiving now was sent when, say, a teenager who is 20 today was still 15 or 16. Any two-way conversation would have a round-trip lag of more than 8.7 years.
Problem 3: Fever or not?
Statement: A US clinician reports a patient’s temperature as 100.4°F. To compare with normal body temperature (37.0°C), we need both on the same scale. Compute the difference from normal in both °C and °F.
Givens:
– T_F = 100.4 °F (measured temperature, 4 sig figs)
– T_C_normal = 37.0 °C (normal body temperature, 3 sig figs)
– Conversion formula: T_C = (5/9)(T_F − 32)
Tools: §1.9 (nonlinear temperature conversion; absolute vs difference)
Solution:
Step 1 — absolute conversion:
The Fahrenheit scale has both an offset and a scale factor. For absolute values, factor-label alone doesn’t work; use the full formula:
$$
T_C = \tfrac{5}{9}(T_F – 32) = \tfrac{5}{9}(100.4 – 32) = \tfrac{5}{9}(68.4) = 38.0\ ^{\circ}\mathrm{C}
$$
Step 2 — difference from normal in °C:
$$
\Delta T_C = 38.0\ ^{\circ}\mathrm{C} – 37.0\ ^{\circ}\mathrm{C} = 1.0\ ^{\circ}\mathrm{C}
$$
Step 3 — difference in °F:
Normal body temperature is 98.6 °F (Anglo-American medicine’s convention for 37.0 °C):
$$
\Delta T_F = 100.4 – 98.6 = 1.8\ ^{\circ}\mathrm{F}
$$
Ratio check (§1.9 audit): For a temperature difference, the offset cancels and only the scale factor remains:
$$
\frac{\Delta T_F}{\Delta T_C} = \frac{1.8}{1.0} = 1.80 = \tfrac{9}{5}\quad ✓
$$
This distinction matters: absolute values convert with the full formula (including the offset), but differences convert with the ratio 9/5 alone (no offset). This is the rule we drilled in §1.9.
Sig figs: 100.4 had 4 sig figs; 37.0 had 3. For subtraction, what matters is the position of the last digit (§1.8) — both are precise to the tenth of a degree, so 1.0 °C is meaningful.
Answer:
$$
\boxed{T = 38.0\ ^{\circ}\mathrm{C},\qquad \Delta T = 1.0\ ^{\circ}\mathrm{C} = 1.8\ ^{\circ}\mathrm{F}}
$$
Clinical note: The standard fever threshold is usually 38.0 °C (100.4 °F). This patient is exactly on the border. Consumer digital thermometers have a precision of about ±0.1 °C, so reporting more digits would be meaningless — the §1.8 rule again: don’t quote more precision than the instrument delivers.
Problem 4: Fuel efficiency — km/L ↔ mpg
Statement: An Iranian car advertises a fuel economy of 15 km/L. A US friend wants to know the equivalent in mpg (miles per US gallon). And the reverse: what’s 52 mpg in km/L?
Givens:
– 1 mile = 1.609 km (4 sig figs)
– 1 US gallon = 3.785 L (4 sig figs)
– Inputs: 15 km/L (2 sig figs), 52 mpg (2 sig figs)
Tools: §1.9 (chained conversion with units in both numerator and denominator)
Solution:
Conversion 1: 15 km/L → mpg
Move km to mile (so mile ends up on top) and L to gallon (so gallon ends up on the bottom):
$$
15\ \frac{\mathrm{km}}{\mathrm{L}}
\ \times\ \frac{1\,\mathrm{mile}}{1.609\,\mathrm{km}}
\ \times\ \frac{3.785\,\mathrm{L}}{1\,\mathrm{gal}}
= \frac{15 \times 3.785}{1.609}\ \frac{\mathrm{mile}}{\mathrm{gal}}
$$
$$
= \frac{56.775}{1.609} = 35.29\ \mathrm{mpg}
$$
Conversion 2: 52 mpg → km/L
Reverse — take mile to km (on top), and gallon to L (on bottom):
$$
52\ \frac{\mathrm{mile}}{\mathrm{gal}}
\ \times\ \frac{1.609\,\mathrm{km}}{1\,\mathrm{mile}}
\ \times\ \frac{1\,\mathrm{gal}}{3.785\,\mathrm{L}}
= \frac{52 \times 1.609}{3.785}\ \frac{\mathrm{km}}{\mathrm{L}}
$$
$$
= \frac{83.668}{3.785} = 22.11\ \mathrm{km/L}
$$
Dimensional check: In Conversion 1, km cancels top-and-bottom, L cancels top-and-bottom. What’s left: mile/gal ✓. Conversion 2 mirrors this ✓.
Sig figs: Inputs are 2 sig figs, so the answer is too:
$$
\boxed{15\ \mathrm{km/L} \approx 35\ \mathrm{mpg},\qquad 52\ \mathrm{mpg} \approx 22\ \mathrm{km/L}}
$$
Driving trivia: km/L (bigger = better) and L/100km (smaller = better, the European convention) are reciprocals of each other. So 15 km/L = 100/15 = 6.67 L/100km. If you’re plotting fuel consumption, the unit choice matters — L/100km is linear in trip duration/distance, but km/L is not.
Problem 5: The mass of Earth’s atmosphere
Statement: Using surface pressure, Earth’s radius, and gravity, estimate the total mass of Earth’s atmosphere. This is a classic problem that requires some cleverness — no textbook formula reads “mass of the atmosphere = …”. Dimensional analysis has to build the formula for you.
Givens:
– P_0 = 101{,}325 Pa (sea-level pressure, exact by definition)
– R_E = 6.37 × 10^6 m (mean Earth radius, 3 sig figs)
– g = 9.81 m/s^2 (3 sig figs)
Tools: §1.7 (build the formula from first principles), §1.9 (implicit conversions), §1.6 (scientific notation for large numbers), §1.8 (sig figs)
Solution:
Step 1 — build the formula (dimensional analysis + physics):
The atmosphere is a gas layer whose weight rests on Earth’s surface. Surface pressure, by definition, is force per unit area:
$$
P_0 = \frac{F_{\mathrm{atm}}}{A_{\mathrm{Earth}}}
$$
The weight of the atmosphere (approximating g as roughly constant):
$$
F_{\mathrm{atm}} = M_{\mathrm{atm}}\,g
$$
Earth’s surface area (a sphere of radius R_E):
$$
A_{\mathrm{Earth}} = 4\pi R_E^{2}
$$
Combining:
$$
P_0 = \frac{M_{\mathrm{atm}}\,g}{4\pi R_E^{2}}
\quad\Rightarrow\quad
M_{\mathrm{atm}} = \frac{4\pi R_E^{2}\,P_0}{g}
$$
Dimensional check:
$$
\left[\frac{R^2 P}{g}\right] = \frac{\mathrm{m}^2\cdot\mathrm{Pa}}{\mathrm{m/s^2}}
= \frac{\mathrm{m}^2\cdot(\mathrm{kg/(m\cdot s^2)})}{\mathrm{m/s^2}}
= \frac{\mathrm{kg}\cdot\mathrm{m}}{\mathrm{s}^2}\cdot\frac{\mathrm{s}^2}{\mathrm{m}} = \mathrm{kg}\quad ✓
$$
Step 2 — numerical:
Earth’s surface area:
$$
A_E = 4\pi\,(6.37\times 10^{6})^{2} = 4\pi\,(40.58 \times 10^{12})\,\mathrm{m}^2 = 5.10 \times 10^{14}\ \mathrm{m}^2
$$
Total atmospheric force:
$$
F_{\mathrm{atm}} = P_0\,A_E = (1.013\times 10^{5})\,(5.10\times 10^{14})\ \mathrm{N} = 5.17 \times 10^{19}\ \mathrm{N}
$$
Mass:
$$
M_{\mathrm{atm}} = \frac{F_{\mathrm{atm}}}{g} = \frac{5.17 \times 10^{19}}{9.81}\ \mathrm{kg} = 5.27 \times 10^{18}\ \mathrm{kg}
$$
Compare to accepted value (NASA): M_atm ≈ 5.15 × 10^18 kg. Our estimate is within about 2% — about as accurate as this simple formula can get. The residual is because g actually varies with altitude, air is compressible, and the atmosphere isn’t a perfect shell (it has an equatorial bulge).
Sig figs: Smallest input is 3 sig figs, so the answer is 3.
Answer:
$$
\boxed{M_{\mathrm{atm}} \approx 5.27 \times 10^{18}\ \mathrm{kg} \approx 5{,}270\ \mathrm{Pg}}
$$
For intuition: the atmosphere is roughly one part in a million of Earth’s total mass (M_E ≈ 5.97 × 10^24 kg). Yet even just the nitrogen fraction, if liquefied, would be about twice the mass of all Earth’s oceans.
Problem 6: Fermi estimate — water in a rain cloud
Statement: How much water is in a typical rain cloud? This is a classic Fermi estimate (§1.6) — with rough inputs and only one significant figure, we can still get the order of magnitude right.
Givens (approximate):
– Cloud dimensions: 1 km × 1 km × 1 km (assume cubic for simplicity)
– LWC (Liquid Water Content) in an ordinary cumulus cloud: 0.5 g/m^3 (1 sig fig — actual range 0.1–2 g/m^3)
Tools: §1.6 (Fermi estimation, scientific notation), §1.8 (sig figs on weak inputs)
Solution:
Cloud volume:
$$
V = (1\,\mathrm{km})^{3}
= (10^3\,\mathrm{m})^{3}
= 10^{9}\ \mathrm{m}^3
$$
§1.9 aside (powers on compound units): the exponent 3 applies to the entire km, so (10^3)^3 = 10^9. If we’d only cubed 1000 and forgot to cube 10^3 → 10^9, the answer would be a million times off.
Water mass:
$$
M_{\mathrm{water}} = V \cdot \mathrm{LWC}
= (10^{9}\,\mathrm{m}^3) \cdot (0.5\,\mathrm{g/m}^3)
= 5 \times 10^{8}\ \mathrm{g}
= 5 \times 10^{5}\ \mathrm{kg}
$$
That’s:
500,000 kg = 500 tonnes- Roughly the weight of 400 typical cars — suspended in the air as tiny droplets (
~20 μmacross)!
Sig figs: With 1-sig-fig LWC, the answer is meaningful only to order of magnitude:
$$
\boxed{M_{\mathrm{water}} \sim 5 \times 10^{5}\ \mathrm{kg} \sim 500\ \mathrm{tonne}}
$$
Take-aways:
- Fermi estimation isn’t precise — but it gets the order of magnitude right with the simplest tools. If someone tells you “a cloud has 5 kg of water” or “50 million tonnes”, you can rule both out without doing a real calculation.
- Sig figs are a property of the inputs, not our preference. With 1-sig-fig input, we report 1 sig fig. We can’t invent
5.27 × 10^5 kgfrom thin air. - Sanity check after estimation: the formula says
5 × 10^5 kg = 500 tonnes— but because droplets are tiny, when the cloud rains, the total mass reaches the ground gradually, not all at once. If it did fall at once, every rain cloud would be a disaster.
Ties to earlier sections
Each problem combined several tools:
| Problem | §1.6 (scientific notation) | §1.7 (dimensional analysis) | §1.8 (sig figs) | §1.9 (unit conversion) |
|---|---|---|---|---|
| 1 — Mariana Trench | ✓ | ✓ | ✓ | |
| 2 — Alpha Centauri | ✓ | ✓ | ||
| 3 — Fever | ✓ | ✓ (nonlinear) | ||
| 4 — Fuel efficiency | ✓ | ✓ (chained) | ||
| 5 — Atmosphere mass | ✓ | ✓ (build formula) | ✓ | ✓ |
| 6 — Rain cloud | ✓ (Fermi) | ✓ |
If any tool tripped you up, revisit its reference section. The 5-step pattern (statement, tool, solution, dimensional check, sig-fig audit) works everywhere — from Chapter 1 exercises to quantum mechanics problems.
What you should be able to do
After this section, you should be able to:
- Look at an unfamiliar problem and identify the right tool (dimensional analysis? simple conversion? Fermi estimate?)
- Carry units through your algebra and do a quick dimensional check after each step
- Report the final answer to the precision the weakest input allows — not more
- Apply the absolute vs difference distinction for nonlinear conversions (temperature)
- Start a “no-formula” problem by building the formula with dimensional analysis before plugging in numbers
Section §1.11 is the practice set — original problems for you to solve, with final answers only for you to check your work.
📚 See also: Halliday Vol 1, Ch 1 — the Worked Examples and Problems sections at the chapter’s end (problem structure is common, but scenarios here are independent).
📖 Open reference: OpenStax University Physics Vol 1 — Chapter 1: Units and Measurement (solved examples with diverse scenarios).
📖 Feynman Lectures — Vol I, Ch 5: Time and Distance (Fermi estimates from the master of the art).
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