§1.10 walked through six fully-solved problems that combined the tools from §1.6–§1.9. This section flips the script — 22 problems for you to solve on your own. Each one lists the final answer and a one-line hint. For the full 5-step template, refer back to §1.10.
Organized by Halliday module: Length (M1), Time (M2), Mass/Density (M3), Original (M4). The last six problems in M4 are our own scenarios that go beyond Halliday’s set.
Difficulty:
– ⭐ Easy — one tool, one conversion
– ⭐⭐ Medium — multi-step, mixed tools
– ⭐⭐⭐ Hard — build a formula or reason across multiple layers
Copyright note: No scenario here is from Halliday. Where a problem parallels a specific Halliday problem, we cite it as a reference — not a reproduction.
Module M1 — Length
Problem 1 ⭐ (parallel to Halliday Ch1 P1)
The Moon is approximately a sphere of radius 1.74 × 10^6 m. Compute:
(a) The Moon’s circumference in km
(b) The Moon’s surface area in km^2
(c) The Moon’s volume in km^3
Hint: C=2πR, A=4πR^2, V=(4/3)πR^3. Do the m→km conversion last.
Answer:
$$
\boxed{C = 1.09 \times 10^4\ \mathrm{km},\quad A = 3.80 \times 10^7\ \mathrm{km^2},\quad V = 2.21 \times 10^{10}\ \mathrm{km^3}}
$$
Problem 2 ⭐ (parallel to Halliday P3)
The nanometer (1 nm) is a common unit in atomic physics and nanotechnology.
(a) How many nm in 1 mile?
(b) What is 1 nm in mm?
(c) How many picometers (pm) in 1 foot?
Data: 1 mile = 1.609 km, 1 ft = 30.48 cm
Hint: Chained conversion with the final unit at 10^{-9} (nano) or 10^{-12} (pico).
Answer:
$$
\boxed{
\text{(a)}\ 1.609 \times 10^{12}\ \mathrm{nm},\quad
\text{(b)}\ 10^{-6}\ \mathrm{mm},\quad
\text{(c)}\ 3.048 \times 10^{11}\ \mathrm{pm}
}
$$
Problem 3 ⭐⭐ (parallel to Halliday P5)
In Safavid-era Iran, the traditional length units were:
$$
1\ \mathrm{farsang} \approx 6.24\ \mathrm{km},\quad 1\ \mathrm{gaz} \approx 1.04\ \mathrm{m},\quad 1\ \mathrm{zar} \approx 1.04\ \mathrm{m}
$$
(a) How many km is a trip of 12.5 farsang?
(b) What is the area (in m^2) of a carpet measuring 4.5 gaz × 3.2 gaz?
(c) If a Safavid soldier marched 8 farsang per day, what was his average speed in m/s? Assume 12 hours of travel.
Hint: Part (c) needs hour → s and then factor-label.
Answer:
$$
\boxed{
\text{(a)}\ 78.0\ \mathrm{km},\quad
\text{(b)}\ 15.6\ \mathrm{m^2},\quad
\text{(c)}\ v \approx 1.16\ \mathrm{m/s}
}
$$
Problem 4 ⭐⭐ (parallel to Halliday P7)
A thunderstorm drops 85 mm of rain over a 340 km^2 area of Tehran. What is the total volume of water in m^3 and in liters? If this water were stored in Amir Kabir Dam (capacity 205 × 10^6 m^3), what percentage of the reservoir would it fill?
Hint: V = h · A — watch the mm→m and km^2 → m^2 conversions.
Answer:
$$
\boxed{
V = 2.89 \times 10^7\ \mathrm{m^3} = 2.89 \times 10^{10}\ \mathrm{L},\quad \eta \approx 14.1\%
}
$$
Problem 5 ⭐⭐ (parallel to Halliday P9)
Lake Urmia is approximately elliptical with semi-axes 70 km and 28 km. Its historical average depth is 5.5 m. Compute the volume of water at full capacity (in km^3 and m^3).
Hint: Ellipse area: A = π · a · b. Assume constant depth.
Answer:
$$
\boxed{V \approx 33.9\ \mathrm{km^3} = 3.39 \times 10^{10}\ \mathrm{m^3}}
$$
Problem 6 ⭐⭐⭐ (parallel to Halliday P8)
A basketball player of height 2.06 m is used as an informal “unit” for measuring a basketball court. If the court is 28.0 m long, what is its length in “player-units”? If we use the same measure for a football field of length 100.0 m and width 60.0 m, what is the area in “player-units squared”? (Interpret the result: why are non-standard units a problem in physics?)
Hint: Simple division. For the square, the unit is raised to power 2 — per the §1.9 rule.
Answer:
$$
\boxed{
28.0/2.06 = 13.6\ \text{player-units},\quad
\frac{100.0 \times 60.0}{2.06^2} = 1414\ \text{player-units}^2
}
$$
Lesson: Without a fixed, official definition, every person of a different height gets a different answer. That’s why SI exists.
Module M2 — Time
Problem 7 ⭐ (parallel to Halliday P17)
Three cesium atomic clocks (A, B, C) are being tested in a national laboratory. Compared to UTC reference, after exactly 1 year:
– Clock A is behind by 0.00042 s
– Clock B is ahead by 0.00089 s
– Clock C is behind by 0.00012 s
Which is the most accurate? Compute the relative error of each clock (as a percent and as ppb = parts per billion).
Hint: Relative error = |Δt| / t_total. For ppb, multiply by 10^9.
Answer: Clock C is the most accurate.
$$
\boxed{
\begin{aligned}
\text{A:}\ &1.33 \times 10^{-11} = 0.0133\ \mathrm{ppb} \
\text{B:}\ &2.82 \times 10^{-11} = 0.0282\ \mathrm{ppb} \
\text{C:}\ &3.80 \times 10^{-12} = 0.00380\ \mathrm{ppb}
\end{aligned}
}
$$
Problem 8 ⭐⭐ (parallel to Halliday P12)
Bamboo can grow up to 91 cm/day in peak season (Guinness record). Assuming uniform growth:
(a) What is the growth rate in μm/s?
(b) How many mm does it grow in 2 minutes?
(c) If the stem cross-section is 4.0 cm^2 and the wood density is 0.60 g/cm^3, how many grams of new mass are produced per hour?
Hint: (a) chained conversion cm→μm and day→s. (c) Δm = ρ · A · Δh.
Answer:
$$
\boxed{
\text{(a)}\ 10.5\ \mathrm{μm/s},\quad
\text{(b)}\ 1.26\ \mathrm{mm},\quad
\text{(c)}\ \Delta m = 0.091\ \mathrm{g/hour}
}
$$
Problem 9 ⭐⭐ (parallel to Halliday P16)
The neutron star at pulsar PSR J1748-2446ad rotates at 716 Hz — the fastest known pulsar. Compute its rotation period in ms and μs. How many rotations does it complete in 1 hour?
Hint: T = 1/f. For rotation count: N = f · Δt.
Answer:
$$
\boxed{
T = 1.40\ \mathrm{ms} = 1.40 \times 10^3\ \mathrm{μs},\quad
N_{\mathrm{hour}} = 2.58 \times 10^6\ \text{rotations}
}
$$
Problem 10 ⭐⭐⭐ (parallel to Halliday P18)
At the time of the dinosaurs (~200 Ma ago), Earth’s rotation period was about 22.5 hour. Today it is 24.0 hour (more precisely: 86,164 s). What is the average slowing per day (in ms)? If the slowdown continued uniformly, how many years would it take for a day to reach 25 hour?
Hint:
– Slowdown: ΔT_day = (24.0 − 22.5) hour = 1.5 hour over 200 Ma
– Rate: 1.5 hour / 200 Ma
– To reach 25 hour: 1 more hour must be added.
Answer:
$$
\boxed{
\text{Rate:}\ \frac{1.5\ \mathrm{h}}{2 \times 10^8\ \mathrm{yr}} \approx 27\ \mathrm{ms/century},\quad
\text{Time to 25 h:}\ \approx 1.3 \times 10^8\ \mathrm{year}
}
$$
Note: Earth’s rotation doesn’t actually slow uniformly (tidal coupling isn’t linear) — this is only a first-order linearization.
Problem 11 ⭐ (parallel to Halliday P14)
A fun unit: 1 microcentury = 10^{-6} · century = 52.6 min. If an exam lasts 120 min, how many microcenturies is that? Conversely, how many hours is 10 microcenturies?
Hint: Direct multiplication/division.
Answer:
$$
\boxed{
120\ \mathrm{min} = 2.28\ \mathrm{μcentury},\quad
10\ \mathrm{μcentury} = 8.77\ \mathrm{hour}
}
$$
Module M3 — Mass and density
Problem 12 ⭐ (parallel to Halliday P22)
Platinum has a density of 21.45 g/cm^3 — among the densest metals. A platinum bar measures 10.0 cm × 5.0 cm × 2.0 cm:
(a) What is its volume in cm^3 and m^3?
(b) What is its mass in kg?
(c) At $31/g, what is the bar worth in dollars?
Hint: Rectangular volume, then m = ρV, then multiply by price.
Answer:
$$
\boxed{
V = 100\ \mathrm{cm^3} = 10^{-4}\ \mathrm{m^3},\quad
m = 2.145\ \mathrm{kg},\quad
\$ = 66{,}500
}
$$
Problem 13 ⭐ (parallel to Halliday P23)
Compute the mass of 1 m^3 of water (water density = 1.000 × 10^3 kg/m^3 exactly). If a pool measures 25 m × 12 m × 2.5 m and is completely filled:
(a) How many m^3 of water?
(b) What is the total water mass in tonne (metric tons)?
(c) To heat this water from 20°C to 28°C (specific heat: c = 4186 J/(kg·K)), how many MJ of energy are needed?
Hint: Q = mc·ΔT. Dimensional check: [kg × J/(kg·K) × K] = J ✓
Answer:
$$
\boxed{
V = 750\ \mathrm{m^3},\quad
m = 750\ \mathrm{tonne},\quad
Q = 2.51 \times 10^{10}\ \mathrm{J} = 25.1\ \mathrm{GJ}
}
$$
Problem 14 ⭐⭐ (parallel to Halliday P24)
A sand sample has approximately spherical grains of mean radius 80 μm. The grains are silica (ρ = 2.65 g/cm^3).
(a) Volume of one grain in m^3?
(b) Mass of one grain in μg?
(c) If a 250 mL cup is completely filled with this sand (assume packing density 1.6 g/cm^3 — accounts for voids between grains), approximately how many grains does it contain?
Hint: V_grain = (4/3)π r^3. For (c), use the packing density, not the pure density.
Answer:
$$
\boxed{
V_{\mathrm{grain}} = 2.14 \times 10^{-12}\ \mathrm{m^3},\quad
m_{\mathrm{grain}} = 5.68\ \mathrm{μg},\quad
N \approx 7.0 \times 10^{7}\ \text{grains}
}
$$
Problem 15 ⭐⭐ (parallel to Halliday P26)
An ordinary cumulus cloud contains about 200 water droplets per cm^3, each of mean radius 12 μm.
(a) Mass of one droplet in μg.
(b) Water mass per 1 m^3 of cloud (in g).
(c) If the cloud is a cube of side 500 m, what is the total water mass in kg?
Hint: m_drop = (4/3)π r^3 · ρ_water. V_cloud = 500^3 m^3.
Answer:
$$
\boxed{
m_{\mathrm{drop}} = 7.24 \times 10^{-6}\ \mathrm{μg},\quad
m_{1\mathrm{m^3}} = 1.45\ \mathrm{g},\quad
M_{\mathrm{cloud}} \approx 1.81 \times 10^{5}\ \mathrm{kg}
}
$$
Interpretation: roughly 180 tonnes of water in a medium cloud — matching the Fermi estimate we did in §1.10 Problem 6.
Problem 16 ⭐⭐ (parallel to Halliday P40, P42)
An oxygen atom has mass 16 u (1 u = 1.660 × 10^{-27} kg). Water (H₂O) has two H atoms (each 1.0 u) and one O atom.
(a) Mass of one water molecule in kg?
(b) Number of molecules in 1 kg of water?
(c) Number of water molecules in the world’s oceans (total mass ≈ 1.4 × 10^{21} kg)?
Hint: Molecule mass = 18 u. N = m_total / m_molecule.
Answer:
$$
\boxed{
m_{\mathrm{mol}} = 2.99 \times 10^{-26}\ \mathrm{kg},\quad
N_{1\mathrm{kg}} = 3.35 \times 10^{25},\quad
N_{\mathrm{ocean}} = 4.69 \times 10^{46}
}
$$
Problem 17 ⭐⭐⭐ (parallel to Halliday P43)
A patient on a 6-month diet loses 18 kg.
(a) What is the mass loss rate in mg/s?
(b) Assuming fat has density 0.90 g/cm^3 and all of the loss was fat, what is the volume of fat lost, in L?
(c) Fat has an energy density of about 9 kcal/g. What is the equivalent total energy in kcal and MJ?
Hint:
– 6 month ≈ 1.58 × 10^7 s
– V = m/ρ
– E = m · e_specific; 1 kcal = 4184 J
Answer:
$$
\boxed{
\dot m = 1.14\ \mathrm{mg/s},\quad
V = 20.0\ \mathrm{L},\quad
E = 162{,}000\ \mathrm{kcal} \approx 678\ \mathrm{MJ}
}
$$
Note: That’s the energy content of burning ~18 L of gasoline — which is why weight loss is so hard.
Module M4 — Original scenarios (beyond Halliday)
Problem 18 ⭐⭐ (original — Iranian oil)
Oil production in Iran is reported in “barrels per day” (bpd). 1 barrel = 158.987 L. If Iran produces 3.5 × 10^6 bpd:
(a) Annual production in m^3?
(b) If crude oil density is 0.87 g/cm^3, what’s the annual mass in tonne?
(c) At $85/barrel, what’s the annual revenue in billion dollars?
Hint: 1 year = 365.25 day. 1 L = 10^{-3} m^3.
Answer:
$$
\boxed{
V_{\mathrm{year}} = 2.03 \times 10^8\ \mathrm{m^3},\quad
M = 1.77 \times 10^8\ \mathrm{tonne},\quad
\$ \approx 108.6\ \text{billion}
}
$$
Problem 19 ⭐ (original — the Persian calendar)
The Hijri Solar (Persian) calendar starts from the Prophet’s migration (year 622 CE). The mean Persian solar year is 365.2424 day (slightly more accurate than the Gregorian).
(a) Today is roughly 1405-04-13 Shamsi (2026-07-04 CE). Approximately how many days have passed since the epoch?
(b) If the Gregorian year is 365.2425 day, what’s the relative difference between the Persian and Gregorian years in ppm?
(c) After how many Persian years does this discrepancy reach 1 day?
Hint: ppm = parts per million = 10^{-6}. Difference: |T_shamsi − T_gregorian|/T_gregorian.
Answer:
$$
\boxed{
N_{\mathrm{days}} \approx 512{,}800,\quad
\Delta = 0.27\ \mathrm{ppm},\quad
\text{years to 1-day drift:}\ \approx 3.7 \times 10^6\ \mathrm{year}
}
$$
Note: The Persian calendar is accurate for millions of years — an astonishing feat of medieval astronomy by Khayyam and the Malik-Shāh court astronomers in the 11th century.
Problem 20 ⭐⭐ (original — lithium battery)
A phone lithium-ion battery typically has an energy density of 250 Wh/kg and voltage 3.7 V. Assume a new phone has 4000 mAh capacity:
(a) Total energy in Wh and J.
(b) Cell mass (excluding casing) in g.
(c) If a laptop needs 70 Wh, what’s the corresponding cell mass?
Hint:
– E = V · Q (Ah); watch mAh → Ah
– m = E / e_specific
Answer:
$$
\boxed{
E_{\mathrm{phone}} = 14.8\ \mathrm{Wh} = 5.33 \times 10^4\ \mathrm{J},\quad
m_{\mathrm{phone\ cell}} = 59.2\ \mathrm{g},\quad
m_{\mathrm{laptop\ cell}} = 280\ \mathrm{g}
}
$$
Problem 21 ⭐⭐⭐ (original — Bitcoin hashrate)
The Bitcoin network in 2026 has a hashrate of about 700 EH/s (exahashes per second). 1 EH = 10^{18} H.
(a) Hashrate in H/s and ZH/day (zettahash per day, 1 ZH = 10^{21} H).
(b) If each hash operation consumes about 5 nJ (best ASICs), what’s the total network power in MW and GW?
(c) That power equals how many 1 GW nuclear plants?
Hint:
– 1 day = 86400 s
– P = (H/s) · (energy/hash)
– For MW: 1 MW = 10^6 W.
Answer:
$$
\boxed{
\begin{aligned}
&\text{hash rate:}\ 7 \times 10^{20}\ \mathrm{H/s} = 60.5\ \mathrm{ZH/day} \
&P \approx 3.5\ \mathrm{GW} = 3500\ \mathrm{MW} \
&\text{≈ 3.5 nuclear plants}
\end{aligned}
}
$$
Note: That’s about 0.1% of the world’s total electricity use — hence the sustainability debate around proof-of-work networks.
Problem 22 ⭐⭐⭐ (original — earthquake energy)
The Richter scale (moment magnitude) has a logarithmic relationship with earthquake energy:
$$
\log_{10}(E) = 1.5\,M + 4.8
\qquad (E\ \text{in joules})
$$
(a) Energy of an M=6.0 quake (moderate, e.g. Varzaqan 2012) in J and TJ?
(b) Energy of an M=8.0 quake (major, e.g. Bam 2003 or Kobe 1995)?
(c) Ratio of M=8.0 to M=6.0 energy — i.e. each unit of magnitude adds how much energy?
Hint:
– E = 10^{1.5M + 4.8}
– For the ratio: E(M=8)/E(M=6) = 10^{1.5(8) − 1.5(6)} = 10^3 — what do you see?
Answer:
$$
\boxed{
\begin{aligned}
&M{=}6.0:\ E = 6.31 \times 10^{13}\ \mathrm{J} = 63.1\ \mathrm{TJ} \
&M{=}8.0:\ E = 6.31 \times 10^{16}\ \mathrm{J} = 63.1\ \mathrm{PJ} \
&\text{Ratio:}\ 10^3 = 1000\ (\text{per unit}\ M \approx 32×)
\end{aligned}
}
$$
Interpretation: An M=8 quake releases as much energy as a thousand M=6 quakes — this is why big quakes are so devastating. Each unit of magnitude adds roughly 10^{1.5} ≈ 32× energy (not 10×!).
Short on time? Solve these six
If you have limited time, choose these six “signature” problems — they exercise all four tools at least once:
- Problem 1 (Moon geometry) — simple conversion with powers
- Problem 4 (Tehran rain) — density, volume, percentage
- Problem 10 (Earth day slowdown) — linear rate of change, big/small numbers
- Problem 15 (cumulus cloud) — combining scientific notation and three-step chains
- Problem 18 (Iranian oil) — economic chain conversion
- Problem 22 (earthquake) — logarithmic relations, energy ratios
What’s next
After this section:
– If you solved all 22 correctly, the Chapter 1 tools are yours — move to Chapter 2 (kinematics) without worry.
– If you still hit errors, revisit §1.10 and identify which step (statement/tool/dim-check/sig-fig) tripped you up.
– If you want full worked solutions for any of these, tell us — we’ll add a “solutions appendix” to the chapter.
📚 See also: Halliday Vol 1, Ch 1 — Module 1.1–1.3 problems for scenario variety.
📖 Open reference: OpenStax University Physics Vol 1 — Chapter 1 Problems (with solutions).
📖 Physics Stack Exchange — for tougher problems (but resist looking at the full solution before attempting yourself).
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