KYA chahiye: $F_B={\rho }_f{V}_{\text{disp}}g$. KYUN $V_{\text{disp}}=V$: fully submerged matlab object apna poora volume paani ka hata deta hai. $F_B=1000\times 2.0\times 10^{-3}\times 9.8=19.6\text{N}$ Force upar ki taraf point karti hai, shallow fluid ki taraf.

KYUN hum $850$ use karte hain, $917$ nahi: buoyancy fluid ki pushing hai, isliye sirf fluid ki density $F_B$ mein jaati hai. Object ki density yeh decide karti hai ki woh float karega ya sink, push ka size nahi. $F_B=850\times 0.030\times 9.8=249.9\text{N}$

KYA karna hai: $F_B={\rho }_{f}Vg$ ko invert karo taaki $V=F_B/({\rho }_{f}g)$ mile. KYUN: push pata hai, size chahiye. $V=\frac{6.0}{1000\times 9.8}=6.12\times 10^{-4}{\text{m}}^3$

KYUN density ratio: floating equilibrium par upward push weight balance karta hai, ${\rho }_f(fV)g={\rho }_{o}Vg$. $Vg$ cancel karo: $f=\frac{{\rho }_o}{{\rho }_f}=\frac{600}{1000}=0.60$ Toh $60\%$ paani ke andar hai, $40\%$ upar dikhta hai.

KYA: "missing" weight exactly buoyant force hai, $F_B=W-W_{\text{app}}=80-52=28\text{N}$. KYUN: scale true weight minus upthrust padhta hai (Newton's equilibrium: $W_{\text{app}}+F_B=W$, dekho Newton's laws — equilibrium of forces). $V=\frac{F_B}{{\rho }_{f}g}=\frac{28}{1000\times 9.8}=2.857\times 10^{-3}{\text{m}}^3$

Step 1 — KYA hai object ki density? $V=0.20\times 0.20\times 0.30=0.012{\text{m}}^3,{\rho }_o=\frac{6.0}{0.012}=500{\text{kg/m}}^3$ Kyunki ${\rho }_o=500<1000={\rho }_f$, yeh float karta hai.

Step 2 — KYUN fraction = density ratio: floating equilibrium se $f={\rho }_o/{\rho }_f=500/1000=0.50$ milta hai. Box ka aadha submerged hai.

Step 3 — KYA height dikhti hai: vertical side $0.30\text{m}$ hai; aadha submerged matlab $0.15\text{m}$ paani ke andar hai, toh $h_{\text{above}}=0.30-0.15=0.15\text{m}$ Figure dekho: blue waterline box ko exactly mid-height par kaatti hai.

KYUN yeh subtle hai: paani rock par upar $F_B$ se push karta hai; Newton's third law se rock paani par neeche usi $F_B$ se push karta hai. Woh extra downward push scale par jaata hai. $F_B={\rho }_{f}Vg=1000\times 1.0\times 10^{-4}\times 9.8=0.98\text{N}$ Naya scale reading: $30+0.98=30.98\text{N}$ String rock ka bacha hua weight carry karti hai; scale exactly buoyant force se badh jaata hai.

Step 1 — KYA hai maximum upthrust? Sinking ki kagar par raft poori tarah submerged hai, toh woh apna poora volume displace karta hai: $F_{B,\max }={\rho }_{f}Vg=1000\times 0.50\times 9.8=4900\text{N}$ Step 2 — KYA weight support kar sakta hai? Raft ka apna weight hai: $W_{\text{raft}}={\rho }_{o}Vg=400\times 0.50\times 9.8=1960\text{N}$ Step 3 — KYUN subtract karein: extra load ka weight bacha hua buoyancy mein fit hona chahiye: $W_{\text{load}}=F_{B,\max }-W_{\text{raft}}=4900-1960=2940\text{N}$ $m_{\text{load}}=\frac{2940}{9.8}=300\text{kg}$

Boat mein (floating): ball ka weight paani ke roop mein displace hota hai. Displaced paani ka volume: $V_1=\frac{m}{{\rho }_f}=\frac{0.50}{1000}=5.0\times 10^{-4}{\text{m}}^3$ Bottom par sunk: ball ab sirf apna khud ka volume displace karti hai: $V_2=\frac{m}{{\rho }_o}=\frac{0.50}{7870}=6.35\times 10^{-5}{\text{m}}^3$ KYUN level girta hai: kyunki iron paani se dense hai, $V_2<V_1$. Sunk hone par woh kam paani hataati hai, isliye level girta hai itne equivalent se: $\Delta V=V_1-V_2=5.0\times 10^{-4}-6.35\times 10^{-5}=4.365\times 10^{-4}{\text{m}}^3$

KYUN split karein: buoyancy ab do fluids se aati hai. Maan lo fraction $x$ paani mein hai, toh $(1-x)$ oil mein hai (assume karo ki cube interface par hai, kuch bhi hawa mein nahi). Equilibrium — total upthrust weight ke barabar: ${\rho }_{\text{oil}}(1-x)Vg+{\rho }_{\text{water}}xVg={\rho }_{o}Vg$ $Vg$ cancel karo: $800(1-x)+1000x=900\Rightarrow 800+200x=900\Rightarrow x=0.50$ Toh $50\%$ paani mein hai, $50\%$ oil mein. Figure mein cube line par straddle karta hua dikhta hai, aadha dono mein.

Case ${\rho }_f\to 0$: $F_B={\rho }_{f}Vg\to 0$. Koi fluid nahi, koi push nahi — object bas girta hai. "Submerged fraction" formula $f={\rho }_o/{\rho }_f\to \infty$, jo nonsense hai: yeh signal karta hai ki object bilkul bhi float nahi kar sakta (use $f>1$ chahiye, jo impossible hai), yaani woh sink karta hai. KYUN formula break hota hai: physically $f\le 1$ zaroori hai; $f={\rho }_o/{\rho }_f>1$ matlab "apne poore volume se zyada displace karna hoga," toh woh poori tarah sink karta hai. Yeh sink condition ${\rho }_o>{\rho }_f$ se match karta hai.

Case ${\rho }_f={\rho }_o$: toh $f=1$ exactly — object neutrally buoyant hai, kisi bhi depth par poori tarah submerged float karta hai, jahan bhi rakhdo equilibrium mein. $F_B={\rho }_{f}Vg=W$ exactly.

Scale padhta hai $W_{\text{app}}=W-F_B$, jahan $F_B={\rho }_f{V}_{\text{disp}}g$. Sirf $V_{\text{disp}}$ change hota hai.

• (a) Abhi surface chhute: $V_{\text{disp}}=0\Rightarrow F_B=0$, reading $=12\text{N}$.
• (b) Aadha andar: $V_{\text{disp}}=1.0\times 10^{-3}$, $F_B=1000\times 1.0\times 10^{-3}\times 9.8=9.8\text{N}$, reading $=12-9.8=2.2\text{N}$.
• (c) Poori tarah andar: $V_{\text{disp}}=2.0\times 10^{-3}$, $F_B=19.6\text{N}$. Lekin $F_B>W$! Upthrust weight se zyada hai, toh spring neeche nahi kheench sakti — reading $12-19.6=-7.6\text{N}$ hogi, yaani ball scale ko $7.6\text{N}$ se upar ki taraf kheenchti hai. KYUN: poori tarah submerged hone par ball itni buoyant hai ki float kar sake; scale (agar dono taraf tension register kar sake) ek negative "weight" padhega, matlab string ab upar taut hai.