Exercises — Physical quantities — fundamental and derived

Har ek se ek sawaal poochho: "Kya yeh saaton starter tools mein se ek hai, ya unse bana hai?"

Chaar fundamental (mass, temperature, current, luminous intensity), paanch derived — aur note karo aakhri wala, refractive index, derived hai phir bhi bilkul koi units nahi carry karta kyunki yeh do speeds ka ratio hai.

${\text{m}}^2=\text{m}\times \text{m}$ = length × length = area. Yeh length nahi ho sakti, kyunki length ke units ${\text{m}}^1$ hote hain, ${\text{m}}^2$ nahi. Metre par exponent quantity ki fingerprint hai: power 1 = length, power 2 = area, power 3 = volume.

KYA: har factor ke units substitute karo. KYUN: kyunki units bilkul waisi hi multiply hoti hain jaise quantities karti hain. $[p]=[m][v]=(\text{kg})\left({\text{m s}}^{-1}\right)=\boxed{{\text{kg m s}}^{-1}}$ Velocity hai length ÷ time $={\text{m s}}^{-1}$; mass in kg se multiply karo. Bus yahi hai — koi memorising nahi, sirf tracing.

Poora method yeh hai: conversion factors se multiply karo, jinmein se har ek 1 ke barabar hai (uske top aur bottom ek hi physical amount hain, bas alag units mein likhe hain). Kyunki har factor asal mein number 1 hai, physical density kabhi nahi badlti — sirf uska label badalta hai. Ab hum har factor aise chunte hain ki unwanted unit cancel ho aur wanted unit aaye. $1\frac{\text{g}}{{\text{cm}}^3}\times \underset{\text{KYUN: g khatam, kg laata hai}}{\underbrace{\frac{1\text{kg}}{1000\text{g}}}}\times \underset{{\text{KYUN: cm}}^3{\text{khatam, m}}^{-3}\text{laata hai}}{\underbrace{\frac{(100\text{cm}{)}^3}{(1\text{m}{)}^3}}}$

• Pehla factor — $\frac{1\text{kg}}{1000\text{g}}$ kyun? Hum chahte hain grams jaayein aur kilograms aayein. Kyunki $1\text{kg}=1000\text{g}$, yeh fraction 1 hai; $\text{g}$ ko neeche rakhne se density ke upar wala $\text{g}$ cancel hota hai.
• Doosra factor — $\frac{(100\text{cm}{)}^3}{(1\text{m}{)}^3}$ kyun? Density mein ${\text{cm}}^3$ denominator mein hai, toh cancel karne ke liye factor ke numerator mein ${\text{cm}}^3$ chahiye. Kyunki $1\text{m}=100\text{cm}$, ek cubic metre $(100\text{cm}{)}^3=10^6{\text{cm}}^3$ hai — cube zaruri hai kyunki hum ek volume convert kar rahe hain, length nahi.

Arithmetic carry karte hain: $=1\times \frac{1}{1000}\times 10^6\frac{\text{kg}}{{\text{m}}^3}=1000{\text{kg m}}^{-3}$ Paani $1000{\text{kg m}}^{-3}$ hai — same substance, bas alag label.

KYA: $G$ ke liye symbolically solve karo, phir units padho. KYUN: $G$ is equation se define hota hai, isliye uske units dimensional balance se force hote hain — dono sides match karni chahiye. $G=\frac{F{r}^2}{m_1{m}_2}$ $[G]=\frac{[F][r{]}^2}{[m_1][m_2]}=\frac{({\text{kg m s}}^{-2})({\text{m}}^2)}{(\text{kg})(\text{kg})}$ Numerator: $\text{kg}\cdot \text{m}\cdot {\text{m}}^2={\text{kg m}}^3{\text{ s}}^{-2}$. ${\text{kg}}^2$ se divide karo (kg ke exponents subtract karo: $1-2=-1$): $[G]=\boxed{{\text{m}}^3{\text{kg}}^{-1}{\text{s}}^{-2}}$

$\frac{1}{2}$ jaisa pure number koi units nahi carry karta, isliye hum ise ignore karte hain. $[\frac{1}{2}m{v}^2]=(\text{kg})({\text{m s}}^{-1}{)}^2={\text{kg m}}^2{\text{ s}}^{-2}$ $[mgh]=(\text{kg})({\text{m s}}^{-2})(\text{m})={\text{kg m}}^2{\text{ s}}^{-2}$ Dono ${\text{kg m}}^2{\text{ s}}^{-2}$ hain = joule ($\text{J}$). Match karte hain, toh dono formulas ek hi quantity describe kar sakte hain (kinetic vs potential energy).

Lekin ab torque: $[\tau ]=[r][F]=(\text{m})({\text{kg m s}}^{-2})={\text{kg m}}^2{\text{ s}}^{-2}$ — identical base units, phir bhi torque energy nahi hai. Neeche figure mein picture yeh farq visible banati hai: energy mein ek force us direction mein kaam karti hai jis taraf push kar rahi hai ($F$ aur displacement parallel), jabki torque mein force lever arm ke perpendicular kaam karti hai ($F$ aur $r$ right angles par). Same units, genuinely alag physics. Isliye hum torque ko $\text{N m}$ mein likhte hain aur energy ko $\text{J}$ mein, bhaale $\text{N m}=\text{J}$ dimensionally ho — naam woh physical meaning record karta hai jo akele units phek dete hain.

Route 1 — pressure: $[P]=\frac{[F]}{[\text{area}]}=\frac{{\text{kg m s}}^{-2}}{{\text{m}}^2}={\text{kg m}}^{-1}{\text{ s}}^{-2}$ Route 2 — energy density: $[u]=\frac{[\text{energy}]}{[\text{volume}]}=\frac{{\text{kg m}}^2{\text{ s}}^{-2}}{{\text{m}}^3}={\text{kg m}}^{-1}{\text{ s}}^{-2}$ Metre ke exponents subtract karo: $2-3=-1$. Dono ${\text{kg m}}^{-1}{\text{ s}}^{-2}$ = ek pascal ($\text{Pa}$) dete hain. Do alag physical stories, ek unit — aur yahan, torque-vs-energy ke unlike, yeh sach mein ek hi underlying cheez describe karte hain.

(a) $P=\frac{180000\text{J}}{12\text{s}}=15000\text{W}$. Base units: $\frac{[\text{J}]}{[\text{s}]}=\frac{{\text{kg m}}^2{\text{ s}}^{-2}}{\text{s}}=\boxed{{\text{kg m}}^2{\text{ s}}^{-3}}$ (second par $-2-1=-3$ add karo). Base units ka yeh dher bilkul wahi hai jo naam watt ($\text{W}$) abbreviate karta hai. (b) Conversion factor se multiply karo (jo 1 ke barabar hai, kyunki $746\text{W}$ aur $1\text{hp}$ same power hain): $15000\text{W}\times \frac{1\text{hp}}{746\text{W}}=20.1\text{hp}(\text{3 s.f.})$ (Dekho Errors and Significant Figures yeh jaanne ke liye ki hum 3 significant figures par kyun rukते hain.)

Strategy: units ko pehchane jaane wale chunks mein toro. L4 se yaad karo ki power ${\text{kg m}}^2{\text{ s}}^{-3}$ hai (watt, $\text{W}$). Toh ${\text{kg m}}^2{\text{ s}}^{-3}{\text{ A}}^{-2}=\frac{[\text{power}]}{{\text{A}}^2}=\frac{[P]}{[I{]}^2}.$ Formula $P=I^{2}R$ (resistance $R$ wale resistor mein dissipate hui heat) $R=P/I^2$ mein rearrange hota hai. Toh $Q$ hai electrical resistance, unit hai ohm ($\Omega$). Har base-unit fingerprint ek real quantity ki taraf wapas point karti hai agar tum ise dhairye se decompose karo.

(a) Rearrange karo: $\rho =\frac{RA}{\ell }$. L5·Q1 se $[R]=\Omega ={\text{kg m}}^2{\text{ s}}^{-3}{\text{ A}}^{-2}$ use karte hain: $[\rho ]=\frac{[R][A]}{[\ell ]}=\frac{({\text{kg m}}^2{\text{ s}}^{-3}{\text{ A}}^{-2})({\text{m}}^2)}{\text{m}}={\text{kg m}}^3{\text{s}}^{-3}{\text{A}}^{-2}$ Metre exponent: $2+2-1=3$. Shorthand mein yeh $\Omega \text{m}$ hai — diye gaye units se match karta hai. ✓

(b) Pehle area. Radius convert karo: $r=0.50\text{mm}=0.50\times 10^{-3}\text{m}=5.0\times 10^{-4}\text{m}$. $A=\pi r^2=\pi (5.0\times 10^{-4}{)}^2=\pi \times 2.5\times 10^{-7}=7.854\times 10^{-7}{\text{m}}^2$ Ab resistance: $R=\frac{\rho \ell }{A}=\frac{(1.7\times 10^{-8})(2.0)}{7.854\times 10^{-7}}=\boxed{0.0433\Omega }(\text{3 s.f.})$ (c) Jaise $r\to 0$, area $A=\pi r^2\to 0$, aur $R=\rho \ell /A\to \infty$. Neeche figure $R$ ko $r$ ke against plot karta hai: curve upar rocket karta hai jaise radius zero ki taraf sirakti hai. Physically sensible: infinitely thin wire charge ke liye infinitely narrow pipe hai — yeh current ko choke karti hai, toh resistance blows up. Amber dot hamara worked point $r=0.50\text{mm}$ mark karta hai, curve ke gentle part par baitha; baayein $r=0$ ki taraf slide karo aur tum wall chad jaate ho. Formula ka limiting behaviour intuition se match karta hai, yeh ek achha sign hai ki shape sahi hai.

1. Momentum ke base units, aur woh formula jisse aaye.
2. Density convert karte waqt length-conversion factor ko cube kyun karna padta hai?
3. Ek cheez batao jo dimensional analysis kar sakta hai aur ek jo nahi kar sakta.
4. $G$ (gravitation constant) ke base units.
5. Physically explain karo ki wire ka radius $\to 0$ hone par $R\to \infty$ kyun hota hai.
6. Do quantities batao jo base units ${\text{kg m}}^2{\text{ s}}^{-2}$ share karte hain phir bhi physically alag hain.
7. Teen dimensionless derived quantities ke naam batao aur kaho ki unke units kyun nahi hote.