Density, specific gravity: Worked examples — Hinglish deep dive

2.2.2 · D3 · Physics › Fluid Mechanics › Density, specific gravity

Recall Ek symbol jis par hum zyada rely karenge:

$g$ Kuch examples mein ek floating balance use hota hai. $g$ bas gravitational acceleration hai (Earth par lagbhag $9.8 m/s^{2}$ ) — woh number jo mass ko weight mein badalta hai ( $weight = m g$ ). Neeche ke har float equation mein yeh dono sides par aata hai aur cancel ho jaata hai, isliye iska exact value yahan kabhi matter nahi karta — lekin tumhe pata hona chahiye ki yeh letter ka matlab kya hai.

Scenario matrix

Har density/SG problem in case classes mein se ek hoti hai. Hum har cell ko kam se kam ek baar cover karenge.

Cell	Case class	Kya tricky hai	Example jo ise cover karta hai
A	Plain $ρ = m / V$ , forward	bas plug in karo	Ex 1
B	Rearranged (find $m$ )	sahi symbol isolate karo	Ex 2
B'	Rearranged (find $V$ )	$V$ isolate karo, litres mein convert karo	Ex 3
C	Unit trap ( $g/cm^{3} \leftrightarrow kg/m^{3}$ )	woh sneaky $1 0^{3}$	Ex 4
C'	Imperial / non-SI unit trap ( $lb/ft^{3}$ )	ek saath do conversions	Ex 5
D	SG → float / fraction submerged	ratio, subtraction nahi	Ex 6
E	Mixture, equal volumes	arithmetic mean	Ex 7
F	Mixture, equal masses	harmonic mean, denser cheez kam jagah leti hai	Ex 8
G	Degenerate / zero input ( $V \to 0$ , ek mass $= 0$ , identical densities)	limits, "formula edge par kya karta hai?"	Ex 9
H	Real-world word problem	words → symbols mein translate karo	Ex 10
I	Exam-style twist (hidden buoyancy / hollow object)	density $\neq =$ material density jab andar air ho	Ex 11

Hum sirf wahi symbols use karte hain jo parent ne already build kiye hain: == $ρ$ (density, mass per unit volume), $m$ (mass), $V$ (volume), aur SG== (specific gravity, pure-number ratio $ρ / ρ_{water}$ jahan $ρ_{water} = 1000 kg/m^{3}$ ), plus upar recall kiya hua $g$ . Kuch bhi naya assume nahi kiya gaya.

Worked Examples

Worked example Ex 4 — Cell C: SI unit trap

Mercury ka $ρ = 13.6 g/cm^{3}$ hai. $kg/m^{3}$ mein convert karo, phir $2 L$ ka mass find karo.

Forecast: Parent note kehta hai $g/cm^{3}$ ko $1 0^{3}$ se multiply karo. Toh expect $13600 kg/m^{3}$ . Do litres ek chhoti bottle hai — lekin mercury ki, toh heavy hogi (~27 kg).

$13.6 g/cm^{3} \times 1 0^{3} = 13, 600 kg/m^{3}$ . Yeh step kyun? Factor rebuild karo: $1 g/cm^{3} = \frac{1 0 ^{- 3} kg}{1 0 ^{- 6} m ^{3}} = 1 0^{3} kg/m^{3}$ . Exponent kabhi guess mat karo — reconstruct karo.
Volume convert karo: $2 L = 2 \times 1 0^{- 3} m^{3}$ . Yeh step kyun? $1 m^{3} = 1000 L$ , toh litres ko 1000 se divide karo.
$m = ρ V = 13, 600 \times 2 \times 1 0^{- 3} = 27.2 kg$ . Yeh step kyun? Ab dono quantities SI mein hain, toh product cleanly kg mein aata hai.

Verify: ~27 kg, forecast se match karta hai ✓. Agar $1 0^{3}$ bhool jaate toh $27 g$ milta — 2 L mercury ke liye absurd, jo mistake flag karta hai.

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

Ek exam mein concrete $150 lb/ft^{3}$ quoted hai. $kg/m^{3}$ mein convert karo. Use karo $1 lb = 0.4536 kg$ aur $1 ft = 0.3048 m$ .

Forecast: Concrete water se kaafi zyada denser hai, toh answer around $2000 - 2500 kg/m^{3}$ expect karo.

Mass unit convert karo: $1 lb = 0.4536 kg$ . Yeh step kyun? Numerator mein pounds hain; unhe kilograms se swap karna hoga.
Volume unit convert karo: $1 ft^{3} = (0.3048)^{3} m^{3} = 0.0283168 m^{3}$ . Yeh step kyun? Denominator mein cubic feet hain; length factor ko cube karo (teen dimensions), bilkul jaise $cm^{3}$ ke liye cube kiya tha.
Combine karo: $150 \times \frac{0.4536}{0.0283168} = 150 \times 16.018 \approx 2402.8 kg/m^{3}$ . Yeh step kyun? Dono conversions ek saath apply karo — upar pounds→kg, neeche ft³→m³.

Verify: $\approx 2403 kg/m^{3}$ forecast band mein aata hai, aur SG $\approx 2.4$ (water se denser, toh concrete sink karta hai) — sensible ✓. Lesson: ek imperial trap bas do conversions hain, ek top ke liye aur ek (cubed) bottom ke liye.

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

Ek wooden ball ka SG $= 0.6$ hai. Uske volume ka kitna fraction water surface ke upar float karta hai?

Forecast: Water se lighter (SG<1) → floats. Agar water se 60% densa hai, guess karo 60% volume neeche aur 40% upar hoga.

Pehle, ek symbol introduce karte hain: == $V_{sub}$ == submerged volume ko represent karta hai — ball ke volume ka woh hissa jo waterline ke neeche baithta hai ( $m^{3}$ mein). Yeh total volume $V$ ka ek tukda hai, toh $V_{sub} \leq V$ hamesha. Poora question yeh hai ki "ratio $V_{sub} / V$ kya hai?"

Floating balance: weight $=$ buoyancy $\Rightarrow ρ_{wood} V g = ρ_{w} V_{sub} g$ . Yeh step kyun? Float karte object ka weight (yaad karo $g$ mass ko weight mein badhata hai) exactly us water se hold up hota hai jise woh displace karta hai — aur sirf submerged part $V_{sub}$ water displace karta hai (Buoyancy and Archimedes' principle).
$g$ aur $V$ cancel karo: $\frac{V _{sub}}{V} = \frac{ρ _{wood}}{ρ _{w}} = SG = 0.6$ . Yeh step kyun? SG yahi density ratio hai, toh submerged volume fraction seedha SG ke barabar hota hai.
Volume fraction above $= 1 - \frac{V _{sub}}{V} = 1 - 0.6 = 0.4 = 40%$ . Yeh step kyun? Jo volume waterline ke neeche nahi hai woh upar hai.

Verify: Volume ka $60%$ neeche $+ 40%$ upar $= 100%$ ✓, forecast se match karta hai.

Caution — volume, height nahi: Sphere ke liye submerged height diameter ka $60%$ nahi hoti, kyunki sphere beeche mein moti aur top aur bottom par patli hoti hai. Sirf volume fraction SG ke barabar hoti hai. Neeche side-view figure mein, blue-shaded submerged region ball ke volume ka $60%$ hai (2-D cross-section ke roop mein drawn, lekin volume ki tarah padho) — aur notice karo ki true waterline red dashed "60%-of-height" line se upar hai, jo prove karta hai ki dono fractions alag hain.

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

Same honey ( $ρ_{1} = 1400$ ) aur water ( $ρ_{2} = 1000$ ) ke equal masses mix karo. $ρ_{mix}$ find karo.

Forecast: Same masses — lekin lighter water zyada volume leti hai, toh woh total volume dominate karti hai, density ko 1200 ke plain average se neeche pull karti hai. Guess karo thoda 1200 se kam.

Equal masses → harmonic mean use karo $ρ = \frac{2 ρ _{1} ρ _{2}}{ρ _{1} + ρ _{2}}$ . Yeh step kyun? Equal mass $m$ ke saath, har volume $V_{i} = m / ρ_{i}$ hai; less dense wale ka $V_{i}$ bada hota hai, toh hum directly densities average nahi kar sakte.
$ρ = \frac{2 ( 1400 ) ( 1000 )}{1400 + 1000} = \frac{2 , 800 , 000}{2400} \approx 1166.67 kg/m^{3}$ . Yeh step kyun? Substitute karo aur divide karo.

Verify: $1166.67 < 1200$ (equal-volume answer) ✓ — exactly forecast ki tarah, kyunki low-density water extra volume occupy karti hai. Phir bhi 1000 aur 1400 ke beech ✓.

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

Test karo ki mixture formula apne edges par kya karta hai. $ρ_{1} = 1000$ throughout lo.

Forecast: Agar doosre substance ka mass ya volume zero ho jaaye, toh mixture bas substance 1 ban jaana chahiye. Aur agar dono densities equal hain, toh answer woh shared density honi chahiye — chahe koi bhi rule ho.

Ek mass $\to 0$ (maano $m_{2} = 0$ ): $ρ_{mix} = \frac{m _{1} + 0}{V _{1} + 0} = \frac{m _{1}}{V _{1}} = ρ_{1} = 1000$ . Yeh step kyun? Doosre substance ke bina, mixture substance 1 hi hai. Formula ko $ρ_{1}$ return karna chahiye.
Equal densities $ρ_{2} = ρ_{1} = 1000$ : arithmetic mean $= \frac{1000 + 1000}{2} = 1000$ ; harmonic mean $= \frac{2 ( 1000 ) ( 1000 )}{2000} = 1000$ . Yeh step kyun? Agar dono ingredients identical hain, toh koi bhi mixing rule density nahi badal sakta — dono means agree karne chahiye.
$V \to 0$ definition $ρ = \frac{m}{V}$ mein: parent note ki density ki bilkul exact definition phir se batate hain, dekho kya hota hai jab $V$ fixed mass $m$ ke saath zero ki taraf shrink hota hai: fraction $\frac{m}{V}$ bina bound ke barhta hai, toh $ρ \to \infty$ . Yeh step kyun? Finite mass ko vanishing space mein cramming karne se density blow up hoti hai — ek real physical limit (black-hole-like extreme ki taraf jaana), aur signal hai ki $V = 0$ undefined hai (zero se divide nahi kar sakte).

Verify: Dono special-case formulas $1000$ reduce ho jaate hain jab densities match karti hain ✓; zero-mass case $ρ_{1}$ return karta hai ✓. Yeh edges confirm karti hain ki formulas sensibly behave karti hain.

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

Ek jeweller ke paas ek ring hai jo pure gold (SG $= 19.3$ ) honi chahiye. Woh paati hai ki uska mass $9.65 g$ hai aur, water displacement se, uska volume $0.60 cm^{3}$ hai. Kya yeh pure gold hai?

Forecast: Pure gold ki density $g/cm^{3}$ mein uske SG ke barabar hai, yaani $19.3 g/cm^{3}$ . Dekhen ki measured density match karti hai — agar lower hai, toh ring hollow ya alloyed hai.

Measured density compute karo: $ρ = \frac{m}{V} = \frac{9.65}{0.60} \approx 16.08 g/cm^{3}$ . Yeh step kyun? Actual mass aur displaced volume se density — koi assumptions nahi.
Water ki density yaad karo. Parent note $ρ_{water} = 1000 kg/m^{3}$ deta hai, aur same water smaller unit mein $1 g/cm^{3}$ hai (yeh ek hi value hai — yaad raho $1 g/cm^{3} = 1000 kg/m^{3}$ ). Kyunki humara measurement $g/cm^{3}$ mein hai, matching form use karo $ρ_{water} = 1 g/cm^{3}$ taaki units cleanly cancel hon: $SG = \frac{ρ}{ρ _{water}} = \frac{16.08 g/cm ^{3}}{1 g/cm ^{3}} = 16.08$ . Yeh step kyun? Kyunki water exactly $1 g/cm^{3}$ hai (same substance jitna $1000 kg/m^{3}$ use kiya, bas alag units mein), isse divide karne par same number milta hai — isliye "SG g/cm³ mein density ke barabar hoti hai". Division spell out kiya taaki koi unit magic hidden na rahe.
Pure gold ke SG $= 19.3$ se compare karo: $16.08 < 19.3$ → pure gold nahi hai. Yeh step kyun? Real gold, gold se less dense nahi ho sakta; shortfall matlab alloyed ya hollow hai.

Verify: Uska SG $16.08$ hai, clearly $19.3$ se neeche ✓. Ek relative-density measurement yeh ek nazar mein confirm kar deti.

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

Ek sealed steel box (steel $ρ = 8000$ ) ka outer volume $0.002 m^{3}$ hai lekin total mass sirf $1.2 kg$ hai (andar mostly empty hai). Kya yeh water mein float karta hai? Uski average density kya hai?

Forecast: Trap! Log kehte hain "steel sinks, SG≈8." Lekin floating depend karta hai average density par (mass ÷ total outer volume), material ki density par nahi. Agar mostly air hai, average density 1000 se kam ho sakti hai.

Average density $\overset{ρ}{ˉ} = \frac{m}{V _{outer}} = \frac{1.2}{0.002} = 600 kg/m^{3}$ . Yeh step kyun? Buoyancy care karta hai ki poori shape kitna water displace karti hai, toh outer volume use karo — steel ki density nahi.
Water se compare karo: $600 < 1000$ . Yeh step kyun? Average density water se kam matlab object float karta hai.
Volume fraction submerged $= \frac{600}{1000} = 0.6 = 60%$ . Yeh step kyun? Same floating balance jaise Ex 6 mein, average density use karke (phir $g$ cancel ho jaata hai).

Verify: $600 < 1000$ → floats, uske volume ka $60%$ neeche ✓. Exactly isliye steel ships float karti hain steel ke SG 8 ke bawajood — hollow shape average density ko lower karti hai. Compare karo ek solid steel block se jiska $\overset{ρ}{ˉ} = 8000$ hai aur woh sink karta hai.

Recall Poore matrix ka one-line summary

Jab bhi koi density problem mile, order mein teen questions poochho: (1) Main kya solve kar raha hoon — $ρ$ , $m$ , ya $V$ ? (rearrange karo) (2) Kya units already SI hain, ya koi hidden $\times 1 0^{3}$ hai (g/cm³) ya do conversions (imperial)? (convert karo) (3) Kya yeh ek mixture hai (equal volume → average, equal mass → harmonic) ya float question hai (compare average density to 1000; submerged volume fraction = SG)?

Active Recall — answers cover karo

Hollow object ke float karne ke liye kaun sa volume use karte hain?

Outer (total) volume, jo average density deta hai, material ki density nahi.

Equal masses of 1400 aur 1000 → mixture density?

Harmonic mean

\approx 1166.67 kg/m^{3}

Equal volumes of 1400 aur 1000 → mixture density?

Arithmetic mean

= 1200 kg/m^{3}

Jab volume

V \to 0

fixed mass ke saath, density kya karti hai?

Bina bound ke barhti hai (

ρ \to \infty

);

V = 0

undefined hai.

SG 0.6 wale floating sphere ke liye, kya submerged HEIGHT diameter ka 60% hai?

Nahi — sirf volume fraction SG ke barabar hoti hai; height fraction alag hoti hai kyunki sphere ka cross-section vary karta hai.

800 kg/m³ par 780 kg oil ka volume kya hai?

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

g

ka matlab kya hai, aur float problems mein uski value kyun matter nahi karti?

Gravitational acceleration; yeh weight = buoyancy ke dono sides se cancel ho jaata hai.

Connections

Density, specific gravity (index 2.2.2) — parent: definitions aur mixture-mean derivations jo yahan use hue.
Buoyancy and Archimedes' principle — Ex 6 aur Ex 11 mein float/submerged-fraction logic.
Relative density measurement (hydrometer) — Ex 10 ka purity check lab mein kaise hota hai.
Pressure in fluids — density seedha $P = ρ g h$ mein jaati hai.
Continuity equation — incompressible flow ke liye constant $ρ$ assume karta hai.
Pascal's law

Worked examples — Density, specific gravity