Exercises — Density, specific gravity

KYA: grams-per-cubic-centimetre ko kilograms-per-cubic-metre mein convert karo. WHY factor $10^3$: $1\text{g}=10^{-3}\text{kg}$ aur $1{\text{cm}}^3=(10^{-2}\text{m}{)}^3=10^{-6}{\text{m}}^3$, isliye ratio $\frac{10^{-3}}{10^{-6}}=10^3$ se scale hota hai. $2.7{\text{g/cm}}^3=2.7\times 1000=2700{\text{kg/m}}^3$ Ye aluminium hai. Answer: $2700{\text{kg/m}}^3$.

Ek hi rule (D3) hai: SG ko $1$ se compare karo.

• $\text{SG}=0.7<1$ → paani se kam dense → floats.
• $\text{SG}=1.0=1$ → paani jitni density → neutral buoyancy, kisi bhi depth par hover karta hai.
• $\text{SG}=1.6>1$ → paani se zyada dense → sinks. Abhi forces ka koi calculation nahi chahiye — SG hi verdict hai.

WHY 1000 se multiply karte hain: (D2) rearrange karo, $\text{SG}=\frac{\rho }{{\rho }_{\text{water}}}$, taaki $\rho =\text{SG}\times {\rho }_{\text{water}}=13.6\times 1000$ mile. ${\rho }_{\text{Hg}}=13,600{\text{kg/m}}^3$

KYA karte hain: hume $V$ chahiye, isliye ise (D1) mein isolate karo, $\rho =m/V$. WHY: dono sides ko $V$ se multiply karo phir $\rho$ se divide karo to $V=m/\rho$ milta hai. $V=\frac{m}{\rho }=\frac{525}{10.5}=50{\text{cm}}^3$ Units already consistent hain (g ke saath g/cm³), isliye koi conversion nahi chahiye. Answer: $50{\text{cm}}^3$.

Step 1 — box ka volume. Ek rectangular tank ke teen edge lengths hote hain — length, width, height — aur uska volume unka product hota hai. Diye gaye teen numbers multiply karo: $V=\text{length}\times \text{width}\times \text{height}=2\times 1\times 0.5=1{\text{m}}^3$ Step 2 — $m=\rho V$ apply karo (rearranged (D1)). Ye form kyun: hume $\rho$ aur $V$ pata hai, $m$ chahiye, isliye multiply karo. $m=\rho V=1000\times 1=1000\text{kg}$ Answer: $1000\text{kg}$ — ek cubic metre paani ek tonne hota hai. Kaam ki baat hai.

Neeche figure dekho (Fig s01). Ye tank ka ek 3-D sketch hai. Teen colored double-arrows woh teen edges hain jo tum multiply karte ho — floor ke saath wale yellow arrows length ($x$-axis) aur width ($y$-axis) hain, aur khada hua arrow height ($z$-axis) hai. Picture jo baat batata hai jo algebra chhupa leta hai: volume literally kitne unit cubes andar fit hote hain — box $2\times 1\times 0.5=1$ bada $1{\text{m}}^3$ cube hai jo halka sa draw kiya gaya hai scale ke liye. Bada box ⇒ zyada paani ⇒ zyada mass, seedhe proportion mein.

WHY pehle convert karo: alag alag units mein numbers compare nahi kar sakte. Dono ko ${\text{kg/m}}^3$ mein laao. $X=0.92\times 1000=920{\text{kg/m}}^3,Y=940{\text{kg/m}}^3$ $940>920$, isliye Y zyada dense hai. (X ice hai, Y koi dense ice-cold cheez hai — point units ka hai.)

Dono parts assume karte hain ki volumes exactly add hote hain (A0). (a) Equal volumes → arithmetic mean. Kyun: $V_1=V_2=V$ ke saath, total mass $={\rho }_{1}V+{\rho }_{2}V$ aur total volume $2V$ (D1 se): $\rho =\frac{{\rho }_1+{\rho }_2}{2}=\frac{1000+1260}{2}=1130{\text{kg/m}}^3$ (b) Equal masses → harmonic mean. Kyun: $m_1=m_2=m$ ke saath, volumes $m/{\rho }_1$ aur $m/{\rho }_2$ hain (rearranged D1), isliye $\rho =\frac{2m}{\frac{m}{{\rho }_1}+\frac{m}{{\rho }_2}}=\frac{2{\rho }_1{\rho }_2}{{\rho }_1+{\rho }_2}=\frac{2(1000)(1260)}{1000+1260}=\frac{2,520,000}{2260}\approx 1115{\text{kg/m}}^3$ Kaunsa bada hai, aur kyun: arithmetic mean ($1130$) bada hai. Wajah: equal masses ke liye, denser liquid (glycerine) kam volume leta hai, isliye volume total mein usse "under-weight" mila — mixture density ko lighter liquid ki taraf khichta hai. Harmonic mean hamesha $\le$ arithmetic mean hota hai.

Neeche figure dekho (Fig s02). Horizontal axis ${\rho }_1$ hai (pehle liquid ki density, ${\text{kg/m}}^3$ mein), light se heavy ki taraf jaata hai jabki ${\rho }_2=1260$ fixed hai. Vertical axis resulting mixture density hai. Blue line equal-volume (arithmetic) result hai; pink curve equal-mass (harmonic) result hai. Geometric fact jo picture clearly dikhata hai: pink curve poori tarah blue line ke neeche hai, aur dono sirf wahan milte hain jahan ${\rho }_1={\rho }_2=1260$ (equal densities) ho. ${\rho }_1=1000$ par yellow dots exactly is problem ke do answers hain, $1130$ aur $\approx 1115$.

WHY mass ek decoy hai: floating density par depend karta hai, weight par nahi. ${\rho }_{\text{log}}=600<1000={\rho }_{\text{water}}$, isliye ye floats karta hai, chahe bhaari hi kyun na ho. Submerged fraction seedhe (D4) se aata hai (is page ke upar derive kiya gaya hai). Yahan fluid paani hai, isliye ${\rho }_{\text{fluid}}={\rho }_{\text{water}}=1000$: $\frac{V_{\text{sub}}}{V}=\frac{{\rho }_{\text{log}}}{{\rho }_{\text{water}}}=\frac{600}{1000}=0.6$ 60% paani ke andar hai. $300\text{kg}$ kabhi use nahi hua — sirf densities ka ratio matter karta hai.

Step 1 — total mass. Kyun: mass conserved hota hai. $m=200+100=300\text{g}$. Step 2 — har cheez ka volume $V=m/\rho$ se (rearranged D1). Kyun: total volume chahiye; pehle har piece nikalo. $V_{\text{Cu}}=\frac{200}{8.9}\approx 22.472{\text{cm}}^3,V_{\text{Sn}}=\frac{100}{7.3}\approx 13.699{\text{cm}}^3$ Step 3 — total volume (volumes add, A0). $V=22.472+13.699=36.171{\text{cm}}^3$. Step 4 — definition apply karo (D1). ${\rho }_{\text{alloy}}=\frac{m}{V}=\frac{300}{36.171}\approx 8.29{\text{g/cm}}^3$ Answer: $\approx 8.29{\text{g/cm}}^3$, $7.3$ aur $8.9$ ke beech — sanity ✓.

Step 1 — sphere ka outer volume. $V_{\text{out}}=\frac{4}{3}\pi r^3$ jahan $r=0.05\text{m}$: $V_{\text{out}}=\frac{4}{3}\pi (0.05{)}^3=\frac{4}{3}\pi (1.25\times 10^{-4})\approx 5.236\times 10^{-4}{\text{m}}^3$ Step 2 — agar solid hota toh mass ($m=\rho V$ use karke, D1). Kyun: actual mass se compare karo. $m_{\text{solid}}=\rho V_{\text{out}}=7800\times 5.236\times 10^{-4}\approx 4.08\text{kg}$ Real mass sirf $2.0\text{kg}<4.08\text{kg}$ hai → ye zaroor hollow hai. Step 3 — actual metal ka volume. $V_{\text{metal}}=\frac{m}{\rho }=\frac{2.0}{7800}\approx 2.564\times 10^{-4}{\text{m}}^3$. 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 hai, cavity $\approx 2.67\times 10^{-4}{\text{m}}^3\approx 267{\text{cm}}^3$.

WHY reference badalta hai: (D4) mein denominator ${\rho }_{\text{fluid}}$ hai — woh fluid jismein object actually float karta hai. Yahan woh fluid seawater hai, isliye ${\rho }_{\text{fluid}}={\rho }_{\text{sea}}=1025$, ${\rho }_{\text{water}}=1000$ nahi. Pehle SG se ice ki density (D2) use karke nikalo: ${\rho }_{\text{ice}}=0.92\times 1000=920{\text{kg/m}}^3$ Ab (D4) apply karo: $\frac{V_{\text{sub}}}{V}=\frac{{\rho }_{\text{ice}}}{{\rho }_{\text{sea}}}=\frac{920}{1025}\approx 0.898$ Lagbhag $89.8\%$ submerged hai — pure water ke $92\%$ se thoda kam, kyunki denser seawater zyada push karta hai upar, isliye zyada ice surface ke upar rehti hai. Physically samajh aata hai ✓.

Neeche figure dekho (Fig s03). Horizontal blue line seawater ki surface hai; neeche blue shading samudra hai. Yellow-outlined rectangle ice block hai. Picture ka kaam (D4) ke ratio ko visible banana hai: pink zone $V_{\text{sub}}$ hai (line ke neeche, block ki height ka $89.8\%$) aur patla yellow zone woh hai jo upar ride karta hai ($10.2\%$). Right side ke do double-arrows un dono heights ko measure karte hain taaki tum dekh sako ki "denser fluid ⇒ chhota pink ⇒ zyada ice paani ke upar".

(D4) se reason karo: ek bead jo exactly interface par ruki hai woh oil se denser hai (warna oil mein upar float karti, oil ke relative $\text{SG}<1$) aur paani se kam dense hai (warna neeche sink karti). Boundary par dono ko touch karte hue equilibrium mein hai, isliye ${\rho }_{\text{oil}}\le {\rho }_{\text{bead}}\le {\rho }_{\text{water}}\Rightarrow 800\le {\rho }_{\text{bead}}\le 1000{\text{kg/m}}^3$ General rule: do stacked liquids ke interface par baithne wale object ki density dono liquid densities ke beech hoti hai. Ye poora density column ka principle hai — objects daalo aur unki density wahan se padho jahaan woh ruk jaayein.

KYA dikhana hai: $\frac{2{\rho }_1{\rho }_2}{{\rho }_1+{\rho }_2}\le \frac{{\rho }_1+{\rho }_2}{2}$. Step 1 — cross-multiply. Dono denominators, ${\rho }_1+{\rho }_2$ aur $2$, positive hain, isliye cross multiply karne par inequality direction same rehti hai. Hume milta hai: $2\cdot 2{\rho }_1{\rho }_2\le ({\rho }_1+{\rho }_2{)}^2⟺4{\rho }_1{\rho }_2\le ({\rho }_1+{\rho }_2{)}^2$ Step 2 — right-hand side expand karo. $({\rho }_1+{\rho }_2{)}^2={\rho }_1^2+2{\rho }_1{\rho }_2+{\rho }_2^2$ use karke, claim ban jaata hai: $4{\rho }_1{\rho }_2\le {\rho }_1^2+2{\rho }_1{\rho }_2+{\rho }_2^2$ Step 3 — sab ek taraf le jao. Dono sides se $4{\rho }_1{\rho }_2$ subtract karo: $0\le {\rho }_1^2-2{\rho }_1{\rho }_2+{\rho }_2^2$ Step 4 — perfect square pehchano. Right side factor hota hai: ${\rho }_1^2-2{\rho }_1{\rho }_2+{\rho }_2^2=({\rho }_1-{\rho }_2{)}^2$ isliye claim reduce ho jaata hai $0\le ({\rho }_1-{\rho }_2{)}^2$ par. Step 5 — conclude karo. Kisi bhi real number ka square hamesha $\ge 0$ hota hai, isliye inequality sare positive ${\rho }_1,{\rho }_2$ ke liye sahi hai. Equality sirf tab hoti hai jab $({\rho }_1-{\rho }_2{)}^2=0$, yaani ${\rho }_1={\rho }_2$. Proved. Physical meaning: equal masses se mix karne par density hamesha equal-volume mix se $\le$ hoti hai — bilkul L3·1 ke numbers se match karta hai.

WHY buoyancy set karta hai: ek floating hydrometer ka weight uske displaced liquid ke weight ke barabar hota hai — wahi equilibrium jo (D4) derive karne mein use kiya tha. Weight ko mass times $g$ likhte hain: $m_{\text{hydro}}g={\rho }_{\text{liquid}}{V}_{\text{sub}}g$ $g$ cancel karo — ye dono sides ko multiply karta hai, isliye drop ho jaata hai, aur float ka mass $=$ displaced liquid ka mass bachta hai: ${\rho }_{\text{liquid}}=\frac{m_{\text{hydro}}}{V_{\text{sub}}}$ Plug in karo (consistent units, grams aur cm³, jahan $V_{\text{sub}}=28{\text{cm}}^3$): ${\rho }_{\text{liquid}}=\frac{30}{28}\approx 1.071{\text{g/cm}}^3$ SI mein convert karo — $\times 1000$ rebuild karo: $1\text{g}=10^{-3}\text{kg}$, $1{\text{cm}}^3=10^{-6}{\text{m}}^3$, ratio $10^3$ se scale hota hai: $1.071{\text{g/cm}}^3=1.071\times 1000\approx 1071{\text{kg/m}}^3$ Answer: $\approx 1071{\text{kg/m}}^3$ (SG $\approx 1.07$). Paani se denser → hydrometer isme paani ki tulna mein upar float karta hai. ✓

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. $920/1025\approx 0.898$, yaani lagbhag $89.8\%$ submerged.
5. $\text{SG}=1$ → density paani ke barabar → neutral buoyancy, kisi bhi depth par hover karta hai; $g$ weight aur buoyancy dono ko multiply karta hai isliye cancel ho jaata hai (jaise D4 derivation mein).