Exercises — Density, specific gravity

WHAT: convert grams-per-cubic-centimetre to kilograms-per-cubic-metre. WHY the factor $10^3$: $1\text{g}=10^{-3}\text{kg}$ and $1{\text{cm}}^3=(10^{-2}\text{m}{)}^3=10^{-6}{\text{m}}^3$, so the ratio scales by $\frac{10^{-3}}{10^{-6}}=10^3$. $2.7{\text{g/cm}}^3=2.7\times 1000=2700{\text{kg/m}}^3$ This is aluminium. Answer: $2700{\text{kg/m}}^3$.

The single rule is (D3): compare SG with $1$.

• $\text{SG}=0.7<1$ → less dense than water → floats.
• $\text{SG}=1.0=1$ → same density as water → neutral buoyancy, hovers at any depth.
• $\text{SG}=1.6>1$ → denser than water → sinks. No calculation of forces needed yet — SG is the verdict.

WHY multiply by 1000: rearrange (D2), $\text{SG}=\frac{\rho }{{\rho }_{\text{water}}}$, to get $\rho =\text{SG}\times {\rho }_{\text{water}}=13.6\times 1000$. ${\rho }_{\text{Hg}}=13,600{\text{kg/m}}^3$

WHAT we do: we want $V$, so isolate it in (D1), $\rho =m/V$. WHY: multiplying both sides by $V$ then dividing by $\rho$ gives $V=m/\rho$. $V=\frac{m}{\rho }=\frac{525}{10.5}=50{\text{cm}}^3$ Units are already consistent (g with g/cm³), so no conversion needed. Answer: $50{\text{cm}}^3$.

Step 1 — volume of a box. A rectangular tank has three edge lengths — length, width, height — and its volume is simply their product. Multiply the three given numbers: $V=\text{length}\times \text{width}\times \text{height}=2\times 1\times 0.5=1{\text{m}}^3$ Step 2 — apply $m=\rho V$ (the rearranged (D1)). Why this form: we know $\rho$ and $V$, want $m$, so multiply. $m=\rho V=1000\times 1=1000\text{kg}$ Answer: $1000\text{kg}$ — one cubic metre of water is a tonne. Handy fact.

Read the figure below (Fig s01). It is a 3-D sketch of the tank. The three coloured double-arrows are the three edges you multiply — the yellow ones along the floor are length ($x$-axis) and width ($y$-axis), the upright one is height ($z$-axis). The point the picture makes that the algebra hides: volume is literally how many unit cubes fit inside — the box is $2\times 1\times 0.5=1$ of the big $1{\text{m}}^3$ cube drawn faintly for scale. Bigger box ⇒ more water ⇒ more mass, in direct proportion.

WHY convert first: you cannot compare numbers in different units. Put both in ${\text{kg/m}}^3$. $X=0.92\times 1000=920{\text{kg/m}}^3,Y=940{\text{kg/m}}^3$ $940>920$, so Y is denser. (X is ice, Y is dense ice-cold something — the point is the units.)

Both parts assume volumes add exactly (A0). (a) Equal volumes → arithmetic mean. Why: with $V_1=V_2=V$, total mass $={\rho }_{1}V+{\rho }_{2}V$ over total volume $2V$ (from D1): $\rho =\frac{{\rho }_1+{\rho }_2}{2}=\frac{1000+1260}{2}=1130{\text{kg/m}}^3$ (b) Equal masses → harmonic mean. Why: with $m_1=m_2=m$, the volumes are $m/{\rho }_1$ and $m/{\rho }_2$ (rearranged D1), so $\rho =\frac{2m}{\frac{m}{{\rho }_1}+\frac{m}{{\rho }_2}}=\frac{2{\rho }_1{\rho }_2}{{\rho }_1+{\rho }_2}=\frac{2(1000)(1260)}{1000+1260}=\frac{2,520,000}{2260}\approx 1115{\text{kg/m}}^3$ Which is larger, and why: the arithmetic mean ($1130$) is larger. Reason: for equal masses, the denser liquid (glycerine) takes up less volume, so it gets "under-weighted" in the volume total — pulling the mixture density down toward the lighter liquid. The harmonic mean is always $\le$ the arithmetic mean.

Read the figure below (Fig s02). The horizontal axis is ${\rho }_1$ (density of the first liquid, in ${\text{kg/m}}^3$), sweeping from light to heavy while ${\rho }_2=1260$ is held fixed. The vertical axis is the resulting mixture density. The blue line is the equal-volume (arithmetic) result; the pink curve is the equal-mass (harmonic) result. Notice the geometric fact the picture makes obvious: the pink curve lives entirely underneath the blue line, and the two only touch where ${\rho }_1={\rho }_2=1260$ (equal densities). The yellow dots at ${\rho }_1=1000$ are exactly this problem's two answers, $1130$ and $\approx 1115$.

WHY mass is a decoy: floating depends on density, not weight. ${\rho }_{\text{log}}=600<1000={\rho }_{\text{water}}$, so it floats, no matter that it is heavy. Fraction submerged comes straight from (D4) (derived at the top of this page). Here the fluid is water, so ${\rho }_{\text{fluid}}={\rho }_{\text{water}}=1000$: $\frac{V_{\text{sub}}}{V}=\frac{{\rho }_{\text{log}}}{{\rho }_{\text{water}}}=\frac{600}{1000}=0.6$ 60% is underwater. The $300\text{kg}$ never entered — only the ratio of densities matters.

Step 1 — total mass. Why: mass is conserved. $m=200+100=300\text{g}$. Step 2 — each volume from $V=m/\rho$ (rearranged D1). Why: we need total volume; get each piece first. $V_{\text{Cu}}=\frac{200}{8.9}\approx 22.472{\text{cm}}^3,V_{\text{Sn}}=\frac{100}{7.3}\approx 13.699{\text{cm}}^3$ Step 3 — total volume (volumes add, A0). $V=22.472+13.699=36.171{\text{cm}}^3$. Step 4 — apply the definition (D1). ${\rho }_{\text{alloy}}=\frac{m}{V}=\frac{300}{36.171}\approx 8.29{\text{g/cm}}^3$ Answer: $\approx 8.29{\text{g/cm}}^3$, between $7.3$ and $8.9$ — sanity ✓.

Step 1 — outer volume of a sphere. $V_{\text{out}}=\frac{4}{3}\pi r^3$ with $r=0.05\text{m}$: $V_{\text{out}}=\frac{4}{3}\pi (0.05{)}^3=\frac{4}{3}\pi (1.25\times 10^{-4})\approx 5.236\times 10^{-4}{\text{m}}^3$ Step 2 — mass if it were solid (using $m=\rho V$, D1). Why: compare against the actual mass. $m_{\text{solid}}=\rho V_{\text{out}}=7800\times 5.236\times 10^{-4}\approx 4.08\text{kg}$ The real mass is only $2.0\text{kg}<4.08\text{kg}$ → it must be hollow. Step 3 — volume of actual metal. $V_{\text{metal}}=\frac{m}{\rho }=\frac{2.0}{7800}\approx 2.564\times 10^{-4}{\text{m}}^3$. Step 4 — cavity = outer minus metal. $V_{\text{cavity}}=V_{\text{out}}-V_{\text{metal}}\approx 5.236\times 10^{-4}-2.564\times 10^{-4}\approx 2.67\times 10^{-4}{\text{m}}^3$ Answer: hollow, cavity $\approx 2.67\times 10^{-4}{\text{m}}^3\approx 267{\text{cm}}^3$.

WHY the reference changes: in (D4) the denominator is ${\rho }_{\text{fluid}}$ — the fluid the object actually floats in. Here that fluid is seawater, so ${\rho }_{\text{fluid}}={\rho }_{\text{sea}}=1025$, not ${\rho }_{\text{water}}=1000$. First get the ice density from its SG using (D2): ${\rho }_{\text{ice}}=0.92\times 1000=920{\text{kg/m}}^3$ Now apply (D4): $\frac{V_{\text{sub}}}{V}=\frac{{\rho }_{\text{ice}}}{{\rho }_{\text{sea}}}=\frac{920}{1025}\approx 0.898$ About $89.8\%$ submerged — slightly less than the $92\%$ in pure water, because denser seawater pushes up harder, so more of the ice rides above. Physically sensible ✓.

Read the figure below (Fig s03). The horizontal blue line is the seawater surface; blue shading below it is the sea. The yellow-outlined rectangle is the ice block. The picture's job is to make the ratio (D4) visible: the pink zone is $V_{\text{sub}}$ (below the line, $89.8\%$ of the block's height) and the thin yellow zone is what rides above ($10.2\%$). The two double-arrows on the right measure those two heights so you can see that "denser fluid ⇒ shorter pink ⇒ more ice above water".

Reason it out with (D4): a bead that rests exactly at the interface is denser than oil (else it floats up into oil, $\text{SG}<1$ relative to oil) and less dense than water (else it sinks through). At the boundary it is in equilibrium touching both, so ${\rho }_{\text{oil}}\le {\rho }_{\text{bead}}\le {\rho }_{\text{water}}\Rightarrow 800\le {\rho }_{\text{bead}}\le 1000{\text{kg/m}}^3$ General rule: an object sitting at an interface between two stacked liquids has a density between the two liquid densities. This is the whole principle of a density column — drop objects in and read their density from where they stop.

WHAT to show: $\frac{2{\rho }_1{\rho }_2}{{\rho }_1+{\rho }_2}\le \frac{{\rho }_1+{\rho }_2}{2}$. Step 1 — cross-multiply. Both denominators, ${\rho }_1+{\rho }_2$ and $2$, are positive, so multiplying across keeps the inequality direction. We get $2\cdot 2{\rho }_1{\rho }_2\le ({\rho }_1+{\rho }_2{)}^2⟺4{\rho }_1{\rho }_2\le ({\rho }_1+{\rho }_2{)}^2$ Step 2 — expand the right-hand side. Using $({\rho }_1+{\rho }_2{)}^2={\rho }_1^2+2{\rho }_1{\rho }_2+{\rho }_2^2$, the claim becomes $4{\rho }_1{\rho }_2\le {\rho }_1^2+2{\rho }_1{\rho }_2+{\rho }_2^2$ Step 3 — move everything to one side. Subtract $4{\rho }_1{\rho }_2$ from both sides: $0\le {\rho }_1^2-2{\rho }_1{\rho }_2+{\rho }_2^2$ Step 4 — recognise the perfect square. The right side factors: ${\rho }_1^2-2{\rho }_1{\rho }_2+{\rho }_2^2=({\rho }_1-{\rho }_2{)}^2$ so the claim reduces to $0\le ({\rho }_1-{\rho }_2{)}^2$. Step 5 — conclude. A real number squared is always $\ge 0$, so the inequality holds for all positive ${\rho }_1,{\rho }_2$. Equality happens only when $({\rho }_1-{\rho }_2{)}^2=0$, i.e. ${\rho }_1={\rho }_2$. Proved. Physical meaning: mixing by equal masses always gives a density $\le$ the equal-volume mix — exactly matching the numbers in L3·1.

WHY buoyancy sets it: a floating hydrometer's weight equals the weight of liquid it displaces — the same equilibrium we used to derive (D4). Writing weight as mass times $g$: $m_{\text{hydro}}g={\rho }_{\text{liquid}}{V}_{\text{sub}}g$ Cancel $g$ — it multiplies both sides, so it drops out, leaving mass of float $=$ mass of displaced liquid: ${\rho }_{\text{liquid}}=\frac{m_{\text{hydro}}}{V_{\text{sub}}}$ Plug in (consistent units, grams and cm³, where $V_{\text{sub}}=28{\text{cm}}^3$): ${\rho }_{\text{liquid}}=\frac{30}{28}\approx 1.071{\text{g/cm}}^3$ Convert to SI — rebuild the $\times 1000$: $1\text{g}=10^{-3}\text{kg}$, $1{\text{cm}}^3=10^{-6}{\text{m}}^3$, ratio scales by $10^3$: $1.071{\text{g/cm}}^3=1.071\times 1000\approx 1071{\text{kg/m}}^3$ Answer: $\approx 1071{\text{kg/m}}^3$ (SG $\approx 1.07$). Denser than water → the hydrometer floats higher in it than in water. ✓

1. $19.3\times 1000=19,300{\text{kg/m}}^3$.
2. $V=525/10.5=50{\text{cm}}^3$.
3. Equal masses → harmonic mean $\frac{2{\rho }_1{\rho }_2}{{\rho }_1+{\rho }_2}\approx 1115{\text{kg/m}}^3$.
4. $920/1025\approx 0.898$, i.e. about $89.8\%$ submerged.
5. $\text{SG}=1$ → density equals water's → neutral buoyancy, hovers at any depth; $g$ multiplies both weight and buoyancy so it cancels (as in the D4 derivation).