Density, specific gravity: Worked examples — Physics deep dive

Recall One symbol we will lean on:

$g$ A couple of examples use a floating balance. $g$ is just gravitational acceleration (about $9.8 m/s^{2}$ on Earth) — the number that turns a mass into a weight ( $weight = m g$ ). In every float equation below it appears on both sides and cancels, so its exact value never matters here — but you should know what the letter means.

The scenario matrix

Every density/SG problem is one of these case classes. We will hit each cell at least once.

Cell	Case class	What makes it tricky	Example that covers it
A	Plain $ρ = m / V$ , forward	just plug in	Ex 1
B	Rearranged (find $m$ )	isolate the right symbol	Ex 2
B'	Rearranged (find $V$ )	isolate $V$ , convert to litres	Ex 3
C	Unit trap ( $g/cm^{3} \leftrightarrow kg/m^{3}$ )	the sneaky $1 0^{3}$	Ex 4
C'	Imperial / non-SI unit trap ( $lb/ft^{3}$ )	two conversions at once	Ex 5
D	SG → float / fraction submerged	ratio, not subtraction	Ex 6
E	Mixture, equal volumes	arithmetic mean	Ex 7
F	Mixture, equal masses	harmonic mean, denser takes less space	Ex 8
G	Degenerate / zero input ( $V \to 0$ , one mass $= 0$ , identical densities)	limits, "what does the formula do at the edge?"	Ex 9
H	Real-world word problem	translate words → symbols	Ex 10
I	Exam-style twist (hidden buoyancy / hollow object)	density $\neq =$ material density when there's air inside	Ex 11

The only symbols we use are ones the parent already built: == $ρ$ (density, mass per unit volume), $m$ (mass), $V$ (volume), and SG== (specific gravity, the pure-number ratio $ρ / ρ_{water}$ with $ρ_{water} = 1000 kg/m^{3}$ ), plus $g$ recalled above. Nothing new is assumed.

Worked Examples

Worked example Ex 5 — Cell C′: the imperial (non-SI) unit trap

An exam quotes concrete as $150 lb/ft^{3}$ . Convert to $kg/m^{3}$ . Use $1 lb = 0.4536 kg$ and $1 ft = 0.3048 m$ .

Forecast: Concrete is a couple of times denser than water, so expect an answer around $2000 - 2500 kg/m^{3}$ .

Convert the mass unit: $1 lb = 0.4536 kg$ . Why this step? The numerator carries pounds; we must swap it for kilograms.
Convert the volume unit: $1 ft^{3} = (0.3048)^{3} m^{3} = 0.0283168 m^{3}$ . Why this step? The denominator carries cubic feet; cube the length factor (three dimensions), just like we cubed for $cm^{3}$ .
Combine: $150 \times \frac{0.4536}{0.0283168} = 150 \times 16.018 \approx 2402.8 kg/m^{3}$ . Why this step? Apply both conversions at once — pounds→kg on top, ft³→m³ on the bottom.

Verify: $\approx 2403 kg/m^{3}$ lands in the forecast band, and SG $\approx 2.4$ (denser than water, so concrete sinks) — sensible ✓. The lesson: an imperial trap is just two conversions, one for the top and one (cubed) for the bottom.

Worked example Ex 6 — Cell D: SG → fraction submerged

A wooden ball has SG $= 0.6$ . What fraction of its volume floats above the water surface?

Forecast: Lighter than water (SG<1) → floats. If 60% as dense as water, guess 60% of its volume is under and 40% above.

First, the symbol we'll introduce: let == $V_{sub}$ == stand for the submerged volume — the amount of the ball's volume that sits below the waterline (in $m^{3}$ ). It is a piece of the total volume $V$ , so $V_{sub} \leq V$ always. The whole question is really "what is the ratio $V_{sub} / V$ ?"

Floating balance: weight $=$ buoyancy $\Rightarrow ρ_{wood} V g = ρ_{w} V_{sub} g$ . Why this step? A floating object's weight (recall $g$ turns mass into weight) is exactly held up by the water it displaces — and only the submerged part $V_{sub}$ displaces water (Buoyancy and Archimedes' principle).
Cancel $g$ and $V$ : $\frac{V _{sub}}{V} = \frac{ρ _{wood}}{ρ _{w}} = SG = 0.6$ . Why this step? SG is this density ratio, so the submerged volume fraction equals SG directly.
Volume fraction above $= 1 - \frac{V _{sub}}{V} = 1 - 0.6 = 0.4 = 40%$ . Why this step? Whatever volume isn't below the waterline is above it.

Verify: $60%$ of the volume below $+ 40%$ above $= 100%$ ✓, matches forecast.

Caution — volume, not height: For a sphere the submerged height is not $60%$ of the diameter, because a sphere is fat in the middle and thin at top and bottom. Only the volume fraction equals SG. In the side-view figure below, the blue-shaded submerged region is $60%$ of the ball's volume (drawn as a 2-D cross-section, but read it as volume) — and notice the true waterline sits higher than the red dashed "60%-of-height" line, proving the two fractions are different.

Worked example Ex 8 — Cell F: mixture, equal masses (harmonic mean)

Mix equal masses of the same honey ( $ρ_{1} = 1400$ ) and water ( $ρ_{2} = 1000$ ). Find $ρ_{mix}$ .

Forecast: Same masses — but the lighter water takes up more volume, so it dominates the total volume, pulling the density below the plain average of 1200. Guess a bit under 1200.

Equal masses → use harmonic mean $ρ = \frac{2 ρ _{1} ρ _{2}}{ρ _{1} + ρ _{2}}$ . Why this step? With equal mass $m$ , each volume is $V_{i} = m / ρ_{i}$ ; the less dense one has a bigger $V_{i}$ , so we can't average densities directly.
$ρ = \frac{2 ( 1400 ) ( 1000 )}{1400 + 1000} = \frac{2 , 800 , 000}{2400} \approx 1166.67 kg/m^{3}$ . Why this step? Substitute and divide.

Verify: $1166.67 < 1200$ (the equal-volume answer) ✓ — exactly as forecast, because the low-density water occupies extra volume. Still between 1000 and 1400 ✓.

Worked example Ex 9 — Cell G: degenerate & limiting cases

Test what the mixture formula does at its edges. Take $ρ_{1} = 1000$ throughout.

Forecast: If the second substance's mass or volume goes to zero, the mixture should just become substance 1. And if both densities are equal, the answer should equal that shared density — no matter which rule.

One mass $\to 0$ (say $m_{2} = 0$ ): $ρ_{mix} = \frac{m _{1} + 0}{V _{1} + 0} = \frac{m _{1}}{V _{1}} = ρ_{1} = 1000$ . Why this step? With no second substance, the mixture is substance 1. The formula must return $ρ_{1}$ .
Equal densities $ρ_{2} = ρ_{1} = 1000$ : arithmetic mean $= \frac{1000 + 1000}{2} = 1000$ ; harmonic mean $= \frac{2 ( 1000 ) ( 1000 )}{2000} = 1000$ . Why this step? If both ingredients are identical, no mixing rule can change the density — both means must agree.
$V \to 0$ in the definition $ρ = \frac{m}{V}$ : restating the very definition of density from the parent note, watch what happens as $V$ shrinks toward zero with fixed mass $m$ : the fraction $\frac{m}{V}$ grows without bound, so $ρ \to \infty$ . Why this step? Cramming finite mass into vanishing space makes density blow up — a real physical limit (approaching a black-hole-like extreme), and a signal that $V = 0$ is undefined (you can't divide by zero).

Verify: Both special-case formulas reduce to $1000$ when densities match ✓; the zero-mass case returns $ρ_{1}$ ✓. These edges confirm the formulas behave sanely.

Worked example Ex 10 — Cell H: real-world word problem

A jeweller has a ring that should be pure gold (SG $= 19.3$ ). She finds its mass is $9.65 g$ and, by water displacement, its volume is $0.60 cm^{3}$ . Is it pure gold?

Forecast: Pure gold's density in $g/cm^{3}$ equals its SG, i.e. $19.3 g/cm^{3}$ . Let's see if the measured density matches — if it's lower, the ring is hollow or alloyed.

Compute measured density: $ρ = \frac{m}{V} = \frac{9.65}{0.60} \approx 16.08 g/cm^{3}$ . Why this step? Density from the actual mass and displaced volume — no assumptions.
Recall water's density. The parent note gives $ρ_{water} = 1000 kg/m^{3}$ , and the same water expressed in the smaller unit is $1 g/cm^{3}$ (these are one and the same value — remember $1 g/cm^{3} = 1000 kg/m^{3}$ ). Since our measurement is in $g/cm^{3}$ , use the matching form $ρ_{water} = 1 g/cm^{3}$ so the units cancel cleanly: $SG = \frac{ρ}{ρ _{water}} = \frac{16.08 g/cm ^{3}}{1 g/cm ^{3}} = 16.08$ . Why this step? Because water is exactly $1 g/cm^{3}$ (the very same substance as the $1000 kg/m^{3}$ used earlier, just in different units), dividing by it leaves the same number — that's why "SG equals density in g/cm³". We spell out the division so no unit magic is hidden.
Compare to pure gold's SG $= 19.3$ : $16.08 < 19.3$ → not pure gold. Why this step? Real gold can't be less dense than gold; the shortfall means it's alloyed or hollow.

Verify: Its SG is $16.08$ , clearly below $19.3$ ✓. A relative-density measurement would confirm this at a glance.

Worked example Ex 11 — Cell I: exam twist (hollow object)

A sealed steel box (steel $ρ = 8000$ ) has outer volume $0.002 m^{3}$ but total mass only $1.2 kg$ (it's mostly empty inside). Does it float in water? What's its average density?

Forecast: Trap! People say "steel sinks, SG≈8." But floating depends on average density (mass ÷ total outer volume), not the material's. If it's mostly air, average density could be under 1000.

Average density $\overset{ρ}{ˉ} = \frac{m}{V _{outer}} = \frac{1.2}{0.002} = 600 kg/m^{3}$ . Why this step? Buoyancy cares about how much water the whole shape displaces, so use outer volume — not steel's density.
Compare to water: $600 < 1000$ . Why this step? Average density below water's means the object floats.
Volume fraction submerged $= \frac{600}{1000} = 0.6 = 60%$ . Why this step? Same floating balance as Ex 6, using average density (again $g$ cancels).

Verify: $600 < 1000$ → floats, $60%$ of its volume under ✓. This is exactly why steel ships float despite steel's SG of 8 — hollow shape lowers the average density. Contrast with a solid steel block whose $\overset{ρ}{ˉ} = 8000$ and sinks.

Recall One-line summary of the whole matrix

Whenever you meet a density problem, ask three questions in order: (1) What am I solving for — $ρ$ , $m$ , or $V$ ? (rearrange) (2) Are units already SI, or is there a hidden $\times 1 0^{3}$ (g/cm³) or two conversions (imperial)? (convert) (3) Is it a mixture (equal volume → average, equal mass → harmonic) or a float question (compare average density to 1000; submerged volume fraction = SG)?

Active Recall — cover the answers

Which volume do you use for floating a hollow object?

The outer (total) volume, giving average density, not the material's density.

Equal masses of 1400 and 1000 → mixture density?

Harmonic mean

\approx 1166.67 kg/m^{3}

Equal volumes of 1400 and 1000 → mixture density?

Arithmetic mean

= 1200 kg/m^{3}

As volume

V \to 0

with fixed mass, density does what?

Grows without bound (

ρ \to \infty

);

V = 0

is undefined.

For a floating sphere with SG 0.6, is the submerged HEIGHT 60% of the diameter?

No — only the volume fraction equals SG; the height fraction differs because a sphere's cross-section varies.

780 kg of oil at 800 kg/m³ occupies what volume?

V = 780/800 = 0.975 m^{3} = 975 L

What does

g

mean, and why doesn't its value matter in float problems?

Gravitational acceleration; it cancels from both sides of weight = buoyancy.

Connections

Density, specific gravity (index 2.2.2) — parent: definitions and the mixture-mean derivations used here.
Buoyancy and Archimedes' principle — the float/submerged-fraction logic in Ex 6 and Ex 11.
Relative density measurement (hydrometer) — how Ex 10's purity check is done in a lab.
Pressure in fluids — density feeds directly into $P = ρ g h$ .
Continuity equation — assumes constant $ρ$ for incompressible flow.
Pascal's law

Worked examples — Density, specific gravity