Physical quantities — fundamental and derived: Worked examples

1.1.1 · D3 · Physics › Measurement, Vectors & Kinematics › Physical quantities — fundamental and derived

Koi bhi symbol aane se pehle, ek reminder plain words mein. Jab hum $[X]$ likhte hain square brackets ke saath, iska matlab hai "X ki units" — bas itna hi. Toh $[speed] = m s^{- 1}$ padha jaata hai "speed ki units metres per second hain". Jab bhi units multiply ya divide hoti hain, unke exponents add ya subtract hote hain, bilkul waise jaise ordinary algebra mein letters ke saath. Wahi ek rule — units ko letters ki tarah treat karo — pura game hai.

Ek aur tool jis par hum baar baar depend karenge. Jab tum kisi quantity ko chhoti unit mein convert karte ho, tumhe uske zyada copies chahiye hoti hain, toh number baDhta hai. Likhkhar:

Scenario matrix

Har dimensional-analysis question inhi cells mein se kisi ek mein aata hai. Neeche diye gaye examples mein cell tag hai jo woh cover karte hain, aur saath milkar ye poora table fill karte hain.

Cell	Case class	Kya tricky banaata hai	Example
A	Pure multiply (unit banana)	saari base units track karo	Ex 1 (Newton)
B	Divide → negative exponent	$m / m^{2} = m^{- 1}$	Ex 2 (Pascal)
C	Quantity ki power	exponent multiply hokar aata hai	Ex 3 (area, volume)
D	Unit conversion (×1 ki chain)	kaunsa fraction upar jaata hai?	Ex 4 (density)
E	Dimensionless result	sab kuch cancel hona chahiye	Ex 5 (angle, strain)
F	Zero / degenerate input	kya unit survive karti hai jab number 0 ho?	Ex 6
G	Limiting / bahut baDi value	prefixes aur orders of magnitude	Ex 7 (prefix chain)
H	Real-world word problem	pehle words → symbols mein translate karo	Ex 8 (fuel economy)
I	Exam twist — ek unknown exponent dhundho	units match karke power solve karo	Ex 9
J	Exam trap — hidden inconsistency	galat equation ko units se pakDo	Ex 10

Cell A — multiply karke unit banana

Neeche diya gaya bar chart dikhata hai ki newton ke liye base units ke exponents kaise stack hote hain — ek visual "fingerprint" jo tum is page par har derived unit ke liye reuse karoge. Horizontal axis teen base units ke naam batata hai; vertical axis woh power hai jis par har ek raise hota hai.

deepdives/dd-physics-1.1.01-d3-s02.png

Cell B — division se negative exponent banta hai

Worked example Ex 2 ·

Pascal ki base units Statement: Pressure $P = \frac{F}{A}$ , force divided by area. Iske base units dhundho.

Forecast: metre ke exponent ka sign guess karo — kya woh positive hoga ya negative?

Upar ki units: $[F] = kg m s^{- 2}$ (Ex 1 se). Ye step kyun? $P$ ke numerator mein force hai, toh kuch bhi cancel karne se pehle hume "force" ko uske already-known base units se replace karna hoga.
Neeche ki units: area length × length hai, toh $[A] = m^{2}$ . Ye step kyun? Area bhi ek derived quantity hai; hum ise "area" nahi chhoDh sakte, humein iske base units chahiye.
Divide → like units ke exponents subtract karo. $[P] = \frac{kg m s ^{- 2}}{m ^{2}} = kg m^{1 - 2} s^{- 2} = kg m^{- 1} s^{- 2}$ Ye step kyun? $m^{1} \div m^{2} = m^{1 - 2} = m^{- 1}$ — metre neeche aa jaata hai.

Verify: $1 Pa = 1 kg m^{- 1} s^{- 2}$ . Negative exponent ka matlab hai "denominator mein metre", jo force ko area par spread karne se match karta hai. ✓

Figure newton ke fingerprint aur pascal ke fingerprint ko contrast karta hai: har base unit ke liye do bars (legend unhe alag karta hai), aur tum dekh sakte ho ki metre bar $+ 1$ (upar) se $- 1$ (neeche) flip ho jaata hai jab hum area se divide karte hain. Axes pichle figure ki tarah hi labelled hain.

deepdives/dd-physics-1.1.01-d3-s03.png

Cell C — quantity ko power par uthana

Worked example Ex 3 ·

Area aur volume ki units, aur kyun power multiply hokar aata hai Statement: Side $ℓ$ ka ek square area $ℓ^{2}$ rakhta hai; ek cube volume $ℓ^{3}$ rakhta hai. Units kya hain?

Forecast: agar length $m$ hai, toh $ℓ^{2}$ aur $ℓ^{5}$ ki units derive karne se pehle guess karo.

Area: $[ℓ^{2}] = (m)^{2} = m^{2}$ . Kyun? Quantity ko square karne se uski unit square hoti hai — exponent $2$ metre par land karta hai.
Volume: $[ℓ^{3}] = (m)^{3} = m^{3}$ . Ye step kyun? Volume teen baar length stack karna hai ( $ℓ \times ℓ \times ℓ$ ), toh metre teen baar khud se multiply hota hai, power $3$ milta hai.
General rule: $[ℓ^{n}] = m^{n}$ . Toh $[ℓ^{5}] = m^{5}$ . Ye step kyun? Bracket ke bahar exponent andar ke har exponent ko multiply karta hai: $(m^{1})^{n} = m^{1 \cdot n}$ .

Verify: Side $3 m$ ⇒ area $3^{2} = 9 m^{2}$ , volume $3^{3} = 27 m^{3}$ . Numbers aur units dono power track karte hain. ✓

Picture "power multiplies through" idea ko literal banata hai: ek length ek line deta hai, do ek square, teen ek cube — exponent count karta hai ki tumne metre ki kitni perpendicular copies stack ki hain.

deepdives/dd-physics-1.1.01-d3-s04.png

Cell D — unit conversion (multiply-by-one ki chain)

Worked example Ex 4 · Ek

density ko $g cm^{- 3}$ se $kg m^{- 3}$ mein convert karo Statement: Paani ki density $1 g cm^{- 3}$ hai. Ise SI derived unit $kg m^{- 3}$ mein express karo.

Forecast: SI mein paani famously $1000 kg m^{- 3}$ hai. Predict karo kyun number exactly $1000$ se jump karta hai.

Conversion factors likho, har ek 1 ke barabar. $\frac{1 kg}{1000 g} = 1$ aur $\frac{100 cm}{1 m} = 1$ . Ye step kyun? Identity $n_{1} u_{1} = n_{2} u_{2}$ se jo upar define ki gayi, $1 kg = 1000 g$ same mass hai, toh unka ratio exactly number $1$ hai. $1$ se multiply karne se quantity kabhi nahi badlti — sirf units relabel hoti hain.
Cubed length handle karo. Kyunki unit $cm^{- 3}$ hai, length factor cube hona chahiye: $(\frac{100 cm}{1 m})^{3}$ . Ye step kyun? Metre power $- 3$ par hai, toh uska conversion teen baar apply hota hai.
Sab multiply karo. $1 \frac{g}{cm ^{3}} \times \frac{1 kg}{1000 g} \times (\frac{100 cm}{1 m})^{3} = \frac{1 \times 10 0 ^{3}}{1000} \frac{kg}{m ^{3}} = \frac{1 0 ^{6}}{1 0 ^{3}} = 1000 kg m^{- 3}$ Ye step kyun? Hum factors chain karte hain taaki grams grams cancel karein aur centimetres centimetres cancel karein, sirf SI units bachti hain; baaki numbers ( $10 0^{3}$ upar, $1000$ neeche) tab final figure tak multiply hote hain.

Verify: $10 0^{3} = 1 0^{6}$ , $1 0^{3}$ se divide karo toh $1 0^{3} = 1000$ milta hai. Units bachi: $kg m^{- 3}$ . ✓ Known value se match karta hai.

Cell E — dimensionless quantity (sab kuch cancel ho jaata hai)

deepdives/dd-physics-1.1.01-d3-s01.png

Cell F — zero aur degenerate inputs

Worked example Ex 6 · Kya

zero value pe bhi unit hoti hai? Statement: Ek car rest se start karti hai: initial velocity $v_{0} = 0$ . Equation $v = v_{0} + a t$ mein, kya $0$ ki koi unit hai?

Forecast: kya " $0 m s^{- 1}$ " aur bare " $0$ " mein koi farq hai?

Check karo ki equation unit-consistent hai. Ek valid physics equation ke har term ki units same honi chahiye. Yahan har term $m s^{- 1}$ honi chahiye. Ye step kyun? Alag units ki quantities add nahi ho sakti — jaise metres ko seconds mein add karna.
Zero ko uski unit do. Toh $v_{0} = 0 m s^{- 1}$ , dimensionless $0$ nahi. Ye step kyun? Number zero hai lekin slot abhi bhi velocity measure karta hai; unit batata hai ki ye kaunsi kism ka zero hai.
Degenerate check — Ex 2 mein zero area. Agar $P = F / A$ mein area $A \to 0$ ho, toh pressure $\to \infty$ : units $Pa$ rahti hain lekin value blow up karti hai. Ye step kyun? Denominator mein zero ek genuine limit hai, value nahi — unit survive karti hai, number nahi karta.

Verify: $v = 0 m s^{- 1} + (2 m s^{- 2}) (3 s) = 6 m s^{- 1}$ . Saare terms $m s^{- 1}$ mein hain, aur zero ne apna unit-wise weight uthaya. ✓

Cell G — limiting / bahut baDi values (prefix chains)

Worked example Ex 7 · Ek

bahut baDi length: light-year se metres via prefixes Statement: Light $c = 3.0 \times 1 0^{8} m s^{- 1}$ ki speed se travel karti hai. Ek light-year (wo distance jo light $1$ saal $\approx 3.15 \times 1 0^{7} s$ mein cover karti hai) mein kitne metres hain? Prefix ke saath express karo.

Forecast: order-of-magnitude guess — kya ye $1 0^{15}$ ya $1 0^{20}$ metres ke zyada kareeb hai?

Distance = speed × time. $[d] = (m s^{- 1}) (s) = m$ — seconds cancel ho jaate hain. Ye step kyun? $s^{- 1} \times s = s^{0} = 1$ ; sirf metre bachta hai, jaisa ek distance ke liye hona chahiye.
Numbers multiply karo. $d = (3.0 \times 1 0^{8}) (3.15 \times 1 0^{7}) = 9.45 \times 1 0^{15} m$ Ye step kyun? Numbers aur powers of ten alag alag multiply hote hain: $3.0 \times 3.15 = 9.45$ mantissas handle karta hai, jabki $1 0^{8} \times 1 0^{7} = 1 0^{15}$ exponents add karta hai — wahi "exponents add karo jab multiply karo" rule jo hum units ke liye use karte hain.
Prefix lagao. $1 0^{15} = peta (P)$ , toh $d \approx 9.45 Pm$ . Ye step kyun? SI Units and Prefixes bahut baDe numbers ko scientific notation ki jagah prefix par ride karne deta hai.

Verify: $3.0 \times 3.15 = 9.45$ ; $1 0^{8} \times 1 0^{7} = 1 0^{15}$ . Toh $9.45 \times 1 0^{15} m$ . ✓ $1 0^{15}$ aur $1 0^{16}$ ke beech — peta scale.

Cell H — real-world word problem

Worked example Ex 8 ·

Fuel economy — words translate karo, phir units Statement: Ek car $100 km$ mein $6.0 litres$ fuel use karti hai. Ise $km L^{- 1}$ (kilometres per litre) mein "distance per volume" figure mein convert karo, phir base-SI $m m^{- 3} = m^{- 2}$ mein.

Forecast: "distance per volume" units $m^{- 2}$ deta hai — ek inverse area. Guess karo kyun inverse area aata hai.

Words translate karo. "6.0 L per 100 km" yaani $\frac{6.0 L}{100 km}$ hai. Hum distance per fuel chahte hain, toh invert karo: $\frac{100 km}{6.0 L} = 16.7 km L^{- 1}$ . Ye step kyun? "Distance per volume" mein distance upar hota hai; question diya gaya ratio flip karta hai.
Base SI mein. $1 km = 1 0^{3} m$ , aur $1 L (litre) = 1 0^{- 3} m^{3}$ . Ye step kyun? Ye dono baar identity $n_{1} u_{1} = n_{2} u_{2}$ disguise mein hai: $1 km$ aur $1 0^{3} m$ same distance hai, $1 L$ aur $1 0^{- 3} m^{3}$ same volume, toh inhe substitute karne se physically kuch nahi badlta — sirf har unit ko base SI mein re-express karta hai taaki agli line mein exponents combine ho sakein.
Exponents combine aur cancel karo. $16.7 \frac{km}{L} = 16.7 \times \frac{1 0 ^{3} m}{1 0 ^{- 3} m ^{3}} = 16.7 \times 1 0^{6} m m^{- 3} = 1.67 \times 1 0^{7} m^{- 2}$ Ye step kyun? $\frac{m}{m ^{3}} = m^{1 - 3} = m^{- 2}$ . Reciprocal figure — volume per distance, $m^{3} / m = m^{2}$ — literally woh cross-sectional area hai jo engine har metre travel karne par fuel "sweep up" karta hai; distance-per-volume bas uska inverse hai, isliye $m^{- 2}$ .

Verify: $100/6.0 = 16.67$ ; $1 0^{3} /1 0^{- 3} = 1 0^{6}$ ; $16.67 \times 1 0^{6} = 1.667 \times 1 0^{7} m^{- 2}$ . ✓

Cell I — exam twist: units match karke unknown exponent dhundho

Worked example Ex 9 ·

Pendulum ka period — power solve karo Statement: Ek pendulum ka period $T$ (time, unit $s$ ) length $ℓ$ (m) aur gravity $g$ ( $m s^{- 2}$ ) par $T = k ℓ^{a} g^{b}$ ke roop mein depend karta hai, jahan $k$ ek dimensionless constant hai. Dimensional Analysis use karke $a$ aur $b$ dhundho.

Forecast: $b$ ka sign guess karo — kya zyada gravity swing ko faster ya slower banati hai?

Dono sides ki units likho. Left: $[T] = s = m^{0} s^{1}$ . Right: $[ℓ^{a} g^{b}] = (m)^{a} (m s^{- 2})^{b} = m^{a + b} s^{- 2 b}$ . Ye step kyun? Ek valid equation mein dono sides ki units identical honi chahiye, letter by letter.
Har base unit ka exponent match karo.
- Metre: $a + b = 0$ .
- Second: $- 2 b = 1 \Rightarrow b = - \frac{1}{2}$ . Ye step kyun? Do unknowns ke liye do equations chahiye; har base unit exactly ek equation contribute karta hai kyunki uska exponent left aur right par agree karna chahiye.
Do linked equations solve karo. "Second" equation already akele $b = - \frac{1}{2}$ deta hai. Ise "metre" equation $a + b = 0$ mein substitute karo: $a = - b = + \frac{1}{2}$ . Ye step kyun? Hum inhe sequence mein solve karte hain — second-unit equation mein sirf $b$ hai, toh woh pehle $b$ pin karta hai; phir metre equation, jisme ab sirf unknown $a$ hai, finish karna trivial hai. Ye do linked linear equations ke liye standard back-substitution hai.

Result: $a = + \frac{1}{2}, b = - \frac{1}{2}$ , toh $T = k ℓ^{1/2} g^{- 1/2} = k \frac{ℓ}{g} .$

Verify: Ye true pendulum formula $T = 2 π ℓ / g$ se match karta hai (jahan $k = 2 π$ , jo dimensional analysis supply nahi kar sakta). $b < 0$ : baDa $g$ ⇒ chhota $T$ ⇒ faster swing, intuition se match karta hai. ✓

Cell J — exam trap: inconsistent equation pakDo

Worked example Ex 10 · Kaunsi equation

dimensionally impossible hai? Statement: Do students distance $s$ (unit m) ke formulas propose karte hain: (i) $s = v t + \frac{1}{2} a t^{2}$ , aur (ii) $s = v t^{2} + \frac{1}{2} a t$ . Sirf ek sahi ho sakta hai — sirf units se decide karo.

Forecast: check karne se pehle guess karo kaunsa fail hota hai, sirf $t$ ki powers dekh kar.

(i) mein har term ki units. $v t = (m s^{- 1}) (s) = m$ . $a t^{2} = (m s^{- 2}) (s^{2}) = m$ . Dono metres. ✓ Ye step kyun? Add karne ke liye matching units chahiye; dono terms metres dete hain, $s$ se consistent.
(ii) mein har term ki units. $v t^{2} = (m s^{- 1}) (s^{2}) = m s$ . $a t = (m s^{- 2}) (s) = m s^{- 1}$ . Ye step kyun? Koi bhi term metre nahi hai, aur woh ek dusre se bhi match nahi karte — tum $m s$ ko $m s^{- 1}$ mein add nahi kar sakte.
Conclude karo. Equation (ii) dimensionally impossible hai; (i) sahi kinematic equation hai. Ye step kyun? Formula tabhi valid hai jab har added term left-hand side ( $m$ ) ki units share kare; kyunki (ii) ke terms na metres hain na ek dusre ke barabar, woh kisi distance ko describe nahi kar sakta, toh hum ise discard karte hain.

Verify: (i) mein har term $m$ hai; (ii) mein terms $m s$ aur $m s^{- 1}$ hain, koi bhi $m$ ke barabar nahi. Dimensional analysis (ii) reject karta hai. ✓ (Woh formulas ko bahar rule kar sakta hai, dimensionless constants confirm nahi — $\frac{1}{2}$ usse invisible hai.)

Recall

Recall Ek-line self-tests

Base SI mein pressure ki units ::: kg m⁻¹ s⁻² Lengths ko divide karne se dimensionless angle kyun milta hai? ::: m ÷ m = m⁰ = 1, koi unit nahi bachi v = v₀ + at mein v₀ = 0 ke saath, 0 kaunsi unit carry karta hai? ::: m s⁻¹ (slot abhi bhi velocity measure karta hai) g cm⁻³ ko kg m⁻³ mein convert karne se number multiply hota hai ::: 1000 se T = k ℓᵃ gᵇ ke liye, exponents hain ::: a = +½, b = −½ Which fails units: s = vt + ½at² or s = vt² + ½at? ::: doosra (vt² + ½at)

Connections

Dimensional Analysis — Ex 9 aur Ex 10 pure dimensional-analysis technique hain.
SI Units and Prefixes — Ex 4, 7, 8 mein multiply-by-one conversions.
Newton's Second Law — Ex 1 mein newton ka source.
Errors and Significant Figures — yahan har numeric answer consistently rounded hai.
Vectors — yahan velocity aur force vectors hain; unki units scalars hain, bilkul same tarah handle hoti hain.