Exercises — Thermal expansion — linear, area, volumetric

Volume ek 3-dimensional quantity hai, isliye yeh ==volume coefficient $\gamma$== use karta hai. Kyun: ek solid teeno directions mein phailta hai, isliye linear factor $(1+\alpha \Delta T)$ ko cube kiya jaata hai; leading correction $3\alpha \Delta T$ hoti hai. $\gamma =3\alpha .$

Length growth hai $\alpha \Delta T=0.10\%=0.0010$. Area growth hai $\beta \Delta T=2\alpha \Delta T$ kyunki ==$\beta =2\alpha$== (do directions). $\frac{\Delta A}{A_0}=2\times 0.10\%=0.20\%.$ Kyun $\times 2$: har side same factor se badhti hai, aur area = side $\times$ side, isliye fractional growth double ho jaati hai.

Hole bada ho jaata hai. Kyun: socho ki hole usi steel se bhara hua hai — woh imaginary plug bahar ki taraf $(1+\alpha \Delta T)$ factor se expand hota. Asal surrounding metal bilkul same tarah move karta hai, isliye hole ki boundary bhi bahar ki taraf move karti hai. Ek hole bilkul solid ki tarah scale up hota hai, area coefficient $\beta$ ke saath.

$\Delta T=120-20=100K$ (yeh ek difference hai, ${}^{\circ }C$ aur $K$ mein identical). $\Delta L=L_0\alpha \Delta T=2.00\times 1.7\times 10^{-5}\times 100=3.4\times 10^{-3}\text{m}=3.4\text{mm}.$ Kyun $\Delta T$ ko convert nahi karna padta: hum sirf tab absolute kelvin mein convert karte hain jab formula mein $T$ khud chahiye hota hai (gas laws); ek change scale-independent hota hai.

Area problem → $\beta =2\alpha =1.8\times 10^{-5}{K}^{-1}$, aur $\Delta T=30K$. $\Delta A=A_0\beta \Delta T=1.50\times 1.8\times 10^{-5}\times 30=8.1\times 10^{-4}{\text{m}}^2.$ Yeh $8.1{\text{cm}}^2$ hai — chhota, lekin real hai.

$\Delta V=V_0\gamma \Delta T=0.200\times 1.8\times 10^{-4}\times 50=1.8\times 10^{-3}{\text{cm}}^3.$ Kyun $\gamma$ seedha diya gaya hai ($\alpha$ nahi): liquids ki koi fixed shape nahi hoti, isliye sirf volume expansion meaningful hai — tables directly $\gamma$ list karte hain.

Hole ka diameter ek length hai, isliye yeh $\alpha$ (linear) se badhega, bhaले hi ek hole "area jaisa" dikhta ho. Hole ke aarpaar koi bhi ek line ek length hai. Humein diameter $\Delta d=5.010-5.000=0.010\text{cm}$ badhani hai. $\Delta d=d\alpha \Delta T\Rightarrow \Delta T=\frac{\Delta d}{d\alpha }=\frac{0.010}{5.000\times 1.2\times 10^{-5}}.$ $\Delta T=\frac{0.010}{6.0\times 10^{-5}}=166.7K.$ Final temperature $=20+166.7=186.7^{\circ }\text{C}\approx 187^{\circ }\text{C}$.

Liquid aur container dono expand karte hain. Overflow difference hai: liquid ko container se zyada room chahiye hota hai. Container ka volume coefficient: ${\gamma }_{\text{steel}}=3{\alpha }_{\text{steel}}=3.6\times 10^{-5}{K}^{-1}$. Effective (apparent) coefficient: ${\gamma }_{\text{app}}={\gamma }_{\text{liquid}}-{\gamma }_{\text{container}}=4.9\times 10^{-4}-3.6\times 10^{-5}=4.54\times 10^{-4}{K}^{-1}$. $\Delta V_{\text{overflow}}=V_0{\gamma }_{\text{app}}\Delta T=1000\times 4.54\times 10^{-4}\times 50=22.7{\text{cm}}^3.$ Kyun subtract karte hain: can ki cavity bhi badhti hai, jo kuch liquid growth ko "absorb" kar leti hai. Sirf excess spill hota hai.

Agar rod free hoti, toh woh strain $\frac{\Delta L}{L_0}=\alpha \Delta T$ se stretch karti. Walls isko forbid karti hain, isliye woh rod ko exactly usi strain se wapas squeeze karti hain. Stress–strain se bridge: stress $=Y\times \text{strain}$. $\text{stress}=Y\alpha \Delta T=2.0\times 10^{11}\times 1.2\times 10^{-5}\times 50.$ $\text{stress}=1.2\times 10^8\text{Pa}=120\text{MPa (compressive)}.$ Kyun $L_0$ cancel ho jaata hai: strain fractional hai, isliye length cancel ho jaati hai — thermal stress sirf material aur $\Delta T$ par depend karta hai, rod ki length par nahi.

Period $T=2\pi \sqrt{L/g}$. Ek small fractional length change se fractional period change aadha hota hai: $\frac{\Delta T_{\text{period}}}{T}=\frac{1}{2}\frac{\Delta L}{L}=\frac{1}{2}\alpha \Delta T_{\text{temp}}.$ Kyun $\frac{1}{2}$: $T\propto \sqrt{L}$, aur square root koi bhi small fractional change ko aadha kar deta hai ($\sqrt{1+x}\approx 1+\frac{x}{2}$ se). $\frac{\Delta T_{\text{period}}}{T}=\frac{1}{2}\times 1.9\times 10^{-5}\times 15=1.425\times 10^{-4}.$ Lamba period → clock slow chalti hai. Seconds lost per day (86400 s): $86400\times 1.425\times 10^{-4}=12.3\text{s lost per day}.$

Yahan $\alpha \Delta T=4.5\times 10^{-6}\times 3000=0.0135$. Linear form: $L=1.000\times (1+0.0135)=1.013500\text{m}$. Exact form: $L=1.000\times e^{0.0135}=1.013591\text{m}$. Difference $\approx 9.1\times 10^{-5}\text{m}=0.091\text{mm}$, yaani total expansion ka lagbhag $0.0090\%$. Kyun itna chhota hai extreme $\Delta T$ par bhi: chhoda hua term $\frac{1}{2}(\alpha \Delta T{)}^2=\frac{1}{2}(0.0135{)}^2\approx 9.1\times 10^{-5}$ hai. $3000K$ par bhi $\alpha \Delta T$ bahut chhota hai, isliye linear form $\sim 0.01\%$ tak safe hai. Approximation tabhi fail hoti hai jab $\alpha \Delta T$, $1$ ke paas pahunche — jo koi bhi solid survive nahi karta.

$\gamma <0$ ke saath, formula $\Delta V=V_0\gamma \Delta T$ mein $\Delta T=1-2=-1K$ hai (cooling). $\Delta V=V_0\times (\text{negative})\times (\text{negative})=\text{positive}.$ Toh $2^{\circ }\text{C}$ se $1^{\circ }\text{C}$ tak cool karne se paani expand karta hai — normal ke ulta. Kyun matter karta hai: paani $4^{\circ }\text{C}$ par sabse dense hota hai. Jab ek pond thanda hota hai, $4^{\circ }\text{C}$ ke paas ka paani neeche doob jaata hai; thanda, halka paani (0–4°C) upar tairta hai aur wahan freeze hota hai. Ice surface par banti hai, jo neeche ke liquid ko insulate karti hai taaki machhliyan survive kar sakein. Yeh is chapter mein ek aisa aur akela jagah hai jahan ek coefficient negative hai.

Dono ke liye same $\Delta T$ hai, lekin brass ka $\alpha$ zyada hai, isliye brass layer steel layer se zyada lambi ho jaati hai. Ek taraf lamba strip aur doosri taraf chhota strip hone se ek curve force hota hai: lamba (brass) side bahar ki taraf bow karta hai, chhota (steel) side arc ke andar hota hai. Kyun: dono layers ko bonded rakhte hue jab ek zyada lamba hota hai, toh strip ko aise wrap karna padta hai ki lambi layer bada radius trace kare. Yahi exactly woh mechanism hai jisse thermostat switch circuit trip karta hai.

Solid: expansion ek tiny fractional effect hai jo material ki fixed atomic spacing par built hai; natural variable change $\Delta T$ hai, aur $\gamma =3\alpha \sim 10^{-5}{K}^{-1}$ — $K$ ya ${}^{\circ }C$ dono mein same rahega kyunki yeh ek difference multiply karta hai. Gas: Gas laws ke mutabiq, constant pressure par $V\propto T$ (absolute). Yahan $V$ seedha $T$ ke proportional hai, isliye absolute $T$ double karne se $V$ double ho jaata hai — ek bahut bada effect (${\gamma }_{\text{gas}}=1/T\approx 3.4\times 10^{-3}{K}^{-1}$ at $293K$, solid se ~100× zyada). Kyunki $V$, $T=0K$ par $0$ hit karta hai, tum zero shift nahi kar sakte: tumhe kelvin use karni hi padegi, sirf difference nahi. Bridge: solids spacing barely change karte hain (asymmetric potential, chhota effect); gas molecules free-flying hote hain (Kinetic theory of gases), isliye unka volume seedha absolute temperature track karta hai.

Liquid ke volume ke liye kaun sa coefficient? ::: $\gamma$ (liquids ke liye sirf volume defined hota hai). Hole ka diameter kaun se coefficient se badhega? ::: $\alpha$ (diameter ek length hai). Filled container ko heat karne par overflow ke liye kaun sa coefficient use hoga? ::: ${\gamma }_{\text{liquid}}-{\gamma }_{\text{container}}$. Clamped rod mein thermal stress uski length par depend karta hai? ::: Nahi — strain fractional hota hai, isliye $L_0$ cancel ho jaata hai; stress $=Y\alpha \Delta T$. Pendulum error mein $\frac{1}{2}$ factor kyun aata hai? ::: Kyunki $T\propto \sqrt{L}$, isliye fractional period change, fractional length change ka aadha hota hai. $L_0(1+\alpha \Delta T)$ kab fail hota hai? ::: Sirf tab jab $\alpha \Delta T$, $1$ ke paas pahunche — real solids ke liye kabhi nahi.