Dimensional analysis — checking equations, deriving relations: Worked examples

1.1.3 · D3 · Physics › Measurement, Vectors & Kinematics › Dimensional analysis — checking equations, deriving relation

The scenario matrix

Har row ko ek cell samjho — ek alag trap ya trick jo yeh topic present kar sakta hai. Hum har ek ko cover karenge.

Cell	Case class	Kya special / dangerous hai
A	Diye gaye equation ka seedha check	pass hona → sirf "disprove nahi hua"
B	Fail hua check (broken term pakdo)	ek decisive verdict
C	Derive karo ek relation, 3 unknowns = 3 equations	poora solve ho sakta hai
D	Under-determined derivation (unknowns > 3)	method toot jaata hai — batao KYU
E	Zero / degenerate exponent (ek power $0$ aati hai)	ek variable drop ho jaata hai
F	$sin / ln / e^{x}$ ka dimensionless argument	ek hidden constraint force hoti hai
G	Dimensions ke zariye unit conversion (har base ka sign & power)	exponents factors steer karte hain
H	Real-world word problem	story → dimensions mein translate karo
I	Exam twist — ek unknown constant ka dimension nikalo	symbol ko isolate karo

Nau cells, nau worked examples. Har ek ke saath woh cell label hai jo wo fill karta hai.

Cell A — ek plain check jo pass hota hai

Worked example Kya kinetic-energy–work relation

$\frac{1}{2} m v^{2} = F d$ dimensionally sahi hai? Forecast: padhne se pehle guess karo — left par energy hai, right par force×distance. Same hai ya different? Apna bet likho.

Step 1 — left term ka dimension. $[\frac{1}{2} m v^{2}] = [m] [v]^{2} = M (L T^{- 1})^{2} = M L^{2} T^{- 2} .$ Yeh step kyun? $\frac{1}{2}$ ek pure number hai ( $[M^{0} L^{0} T^{0}]$ ), isliye uska koi dimension nahi hota aur ise drop kar dete hain. $[v]$ ko square karte hain kyunki $v$ squared hai.

Step 2 — right term ka dimension. $[F d] = [F] [d] = (M L T^{- 2}) (L) = M L^{2} T^{- 2} .$ Yeh step kyun? $[F] = M L T^{- 2}$ from $F = ma$ ; ek length $[d] = L$ se multiply karo.

Step 3 — compare karo. Dono sides $M L^{2} T^{- 2}$ hain. ✅

Verify: shared dimension bilkul wahi hai jo energy ki hoti hai (parent table dekho). Pass hona matlab disprove nahi hua — iska matlab yeh nahi hai ki $\frac{1}{2}$ prove ho gaya. Cell A ke liye yahi sahi, humble conclusion hai.

Cell B — ek check jo fail hota hai (decisive)

Worked example Ek student impulse ko

$J = F / t$ likhta hai. Kya yeh sahi hai? ( $F$ = force, $t$ = time.) Forecast: impulse momentum ke change ke barabar hona chahiye, $[M L T^{- 1}]$ . Kya tumhe lagta hai $F / t$ match karega? Guess karo.

Step 1 — proposed right side ka dimension. $[F / t] = \frac{M L T ^{- 2}}{T} = M L T^{- 3} .$ Yeh step kyun? Time se divide karne par $T$ ke exponent mein se $1$ ghatta hai: $- 2 - 1 = - 3$ .

Step 2 — impulse ka dimension hona chahiye. $[J] = [Δ (m v)] = M L T^{- 1} .$ Yeh step kyun? Momentum mass×velocity hoti hai; impulse uska change hota hai, same dimension.

Step 3 — compare karo aur diagnose karo. $M L T^{- 3} \neq = M L T^{- 1}$ . $T$ ka exponent $2$ se off hai. ❌

Verify / fix: sahi relation $J = F t$ hai, jo $(M L T^{- 2}) (T) = M L T^{- 1}$ deta hai ✅. Ek failed check ek proof hai ki formula galat hai — yeh sabse strong baat hai jo dimensional analysis keh sakti hai.

Cell C — ek relation derive karo, 3 unknowns, 3 equations

Worked example Ek string par wave ki speed

$v$ tension $F$ , mass-per-length $μ$ , aur possibly length $ℓ$ par kaise depend karti hai? Forecast: kya tumhe lagta hai $ℓ$ survive karega, ya drop ho jaayega? Abhi bet karo — method bata dega.

Setup. Powers ka product guess karo: $v = k F^{a} μ^{b} ℓ^{c}, k dimensionless .$ Dimensions: $[v] = L T^{- 1}$ , $[F] = M L T^{- 2}$ , $[μ] = M L^{- 1}$ (mass per length), $[ℓ] = L$ .

Step 1 — right side expand karo. $L T^{- 1} = (M L T^{- 2})^{a} (M L^{- 1})^{b} (L)^{c} = M^{a + b} L^{a - b + c} T^{- 2 a} .$ Yeh step kyun? Har base ko alag-alag collect karo — $M, L, T$ mein se har ek ek independent axis hai, isliye unke exponents ek axis ek time par match karne chahiye.

Step 2 — har base ki powers match karo.

$M : a + b = 0$
$L : a - b + c = 1$
$T : - 2 a = - 1 \Rightarrow a = \frac{1}{2}$

Step 3 — solve karo. $M$ se: $b = - a = - \frac{1}{2}$ . $L$ mein substitute karo: $\frac{1}{2} - (- \frac{1}{2}) + c = 1 \Rightarrow 1 + c = 1 \Rightarrow c = 0$ .

Step 4 — assemble karo. $v = k F^{1/2} μ^{- 1/2} ℓ^{0} = k \frac{F}{μ} .$

Verify: $[F / μ] = \frac{M L T ^{- 2}}{M L ^{- 1}} = L^{2} T^{- 2} = L T^{- 1} = [v]$ ✅. Aur $c = 0$ kehta hai length drop ho jaati hai — tumhara forecast ab settle ho gaya. (True value $k = 1$ .)

Cell D — under-determined: equations se zyada unknowns

Worked example Ek sphere par drag force

$F$ speed $v$ , radius $r$ , fluid density $ρ$ , aur viscosity $η$ par depend kar sakti hai. Ise derive karne ki koshish karo. Forecast: hamare paas char unknown powers hain lekin sirf teen base dimensions hain. Kya tumhe lagta hai ek clean answer niklega? Guess yes/no.

Setup. $F = k v^{a} r^{b} ρ^{c} η^{d}$ with $[F] = M L T^{- 2}$ , $[v] = L T^{- 1}$ , $[r] = L$ , $[ρ] = M L^{- 3}$ , $[η] = M L^{- 1} T^{- 1}$ .

Step 1 — powers match karo. $M L T^{- 2} = (L T^{- 1})^{a} (L)^{b} (M L^{- 3})^{c} (M L^{- 1} T^{- 1})^{d} .$

$M : c + d = 1$
$L : a + b - 3 c - d = 1$
$T : - a - d = - 2 \Rightarrow a + d = 2$

Yeh step kyun? Same recipe as Cell C — lekin unknowns gino.

Step 2 — gino. Char unknowns $a, b, c, d$ ; sirf teen equations. Ek free parameter bach jaata hai — maan lo hum $d$ ko undetermined chhodte hain.

Step 3 — $d$ ke terms mein express karo. Equations se: $a = 2 - d$ , $c = 1 - d$ , aur $b = 1 - a + 3 c + d = 1 - (2 - d) + 3 (1 - d) + d = 2 - d$ . $F = k v^{2 - d} r^{2 - d} ρ^{1 - d} η^{d} = k (ρ v^{2} r^{2}) (\frac{η}{ρ v r})^{d} .$

Verify / lesson: bacha hua group $\frac{η}{ρ v r}$ dimensionless hai (yeh $1/ Reynolds number$ hai). Dimensional analysis $d$ fix nahi kar sakti — uske liye experiment ya Buckingham Pi theorem chahiye, jo exactly is "ek extra variable per ek free dimensionless group" ko formalise karta hai. Yeh woh boundary hai jiske baare mein parent note ne warn kiya tha.

Cell E — ek power zero aati hai (ek variable gayab ho jaata hai)

Worked example Simple-pendulum period dobara derive karo aur dikhao ki mass sach mein drop ho jaati hai.

Forecast: yeh tumne pehle dekha hai; yahan hum zero exponent ko spotlight karte hain. Kaun sa base equation mass ko bahar dhakelta hai?

Setup. $T = k ℓ^{a} m^{b} g^{c}$ , with $[ℓ] = L$ , $[m] = M$ , $[g] = L T^{- 2}$ , $[T] = T$ .

Step 1 — match karo. $T^{1} = M^{b} L^{a + c} T^{- 2 c} .$

$M : b = 0$ ← mass equation
$L : a + c = 0$
$T : - 2 c = 1 \Rightarrow c = - \frac{1}{2}, a = \frac{1}{2}$

Step 2 — zero padhho. $b = 0$ isliye aata hai kyunki koi bhi doosri quantity mass carry nahi karti: $ℓ$ aur $g$ mass-free hain, isliye $M^{b}$ ko left par $M^{0}$ ke barabar banane ka ek hi tarika hai: $b = 0$ .

Step 3 — assemble karo. $T = k \frac{ℓ}{g}, m^{0} = 1 (mass vanish ho gayi) .$

Verify: $[ℓ / g] = L / (L T^{- 2}) = T^{2} = T$ ✅. Ek degenerate/zero exponent koi error nahi hai — yeh method ka yeh batana hai ki koi dependence exist nahi karti.

Cell F — ek function ka dimensionless argument

Worked example Ek pulse

$y = A sin (k x - ω t)$ ki tarah travel karti hai. $k$ aur $ω$ ke dimensions nikalo, aur confirm karo ki argument pure hai. Forecast: $x$ ek length hai aur $t$ ek time hai. $k$ aur $ω$ kya carry karte hain taaki $k x$ aur $ω t$ pure numbers hon? Guess karo.

Step 1 — governing rule. $sin$ ka argument dimensionless hona chahiye (parent corollary). Isliye andar ka har piece $[M^{0} L^{0} T^{0}]$ hona chahiye: $[k x] = 1 and [ω t] = 1.$ Yeh step kyun? $sin$ apne argument ki powers ka ek sum hai; tum ek length ko uske cube se add nahi kar sakte — argument ek pure number hona chahiye.

Step 2 — $k$ solve karo. $[k] [L] = 1 \Rightarrow [k] = L^{- 1}$ (per metre — ek wave number).

Step 3 — $ω$ solve karo. $[ω] [T] = 1 \Rightarrow [ω] = T^{- 1}$ (per second — ek angular frequency).

Step 4 — amplitude $A$ ka dimension. Kyunki $sin (\cdot)$ dimensionless hai aur $y$ ek displacement hai, $[A] = [y] = L$ .

Verify: $[k x - ω t] = (L^{- 1}) (L) - (T^{- 1}) (T) = 1 - 1$ , dono pure numbers hain — subtraction legal hai (same kind). ✅ Dhyan do: homogeneity bracket ke andar bhi hai: dono terms dimensionless hone chahiye, sirf unka difference nahi.

Cell G — unit conversion jahan har power factor steer karti hai

Worked example

$1 joule$ ko ergs mein express karo (CGS energy unit, $g\cdotcm^{2} \cdots^{- 2}$ ). Forecast: energy $[M L^{2} T^{- 2}]$ hai. Mass $1 0^{3}$ se scale hoti hai, length $1 0^{2}$ se — lekin length squared hai. Total power of ten kya hogi? Step 3 se pehle guess karo.

Step 1 — dimension recipe drive karta hai. $[E] = M^{1} L^{2} T^{- 2}$ . Har base unit ko (new-per-old factor) se replace karte hain jo dimension mein uski power tak raise hoti hai. $1 J = (1 kg)^{1} (1 m)^{2} (1 s)^{- 2} .$ Yeh step kyun? Exponents $1, 2, - 2$ seedhe $[E]$ se copy hote hain; dimensions track karne ka poora point yahi hai.

Step 2 — conversion factors substitute karo. $1 kg = 1 0^{3} g, 1 m = 1 0^{2} cm, 1 s = 1 s .$ $1 J = (1 0^{3} g)^{1} (1 0^{2} cm)^{2} (1 s)^{- 2} .$

Step 3 — ten ki powers combine karo. $1 0^{3} \times 1 0^{2 \cdot 2} = 1 0^{3} \times 1 0^{4} = 1 0^{7} .$ $1 J = 1 0^{7} g\cdotcm^{2} \cdots^{- 2} = 1 0^{7} erg .$

Verify: length ka exponent $2$ hona hi woh cheez hai jisne $1 0^{2}$ ko $1 0^{4}$ banaya — exactly woh trap jise forecast ne flag kiya tha. Agar tum square bhool jaate to $1 0^{5}$ milta, $100 \times$ galat. General scaling rule ke liye dekho Units and the SI system.

Cell H — ek real-world word problem

Worked example Mass

$m$ ki ek raindrop hawa se girte hue ek terminal speed tak pahunchti hai jahan drag gravity ko balance karta hai. Drag force ki magnitude $F = c v^{2}$ hai kisi constant $c$ ke liye jiska $[c] = M L^{- 1}$ hai. Estimate karo ki terminal speed $v_{t}$ $m$ aur $g$ ke saath kaise scale karti hai. Forecast: bhaari drops tez girte hain ya dheere? Zyada gravity — tez ya dheere? Dono trends guess karo.

Step 1 — physics ko ek balance mein translate karo. Terminal speed par net force zero hoti hai, isliye weight = drag: $m g = c v_{t}^{2} .$ Yeh step kyun? "Terminal" matlab koi aur acceleration nahi — dono forces size mein equal hain, isliye unke dimensions pehle se match karne chahiye ( $[m g] = M L T^{- 2}$ , $[c v^{2}] = (M L^{- 1}) (L T^{- 1})^{2} = M L T^{- 2}$ ✅ — homogeneity solve karne se pehle setup confirm karti hai).

Step 2 — $v_{t}$ solve karo. $v_{t} = \frac{m g}{c} .$

Step 3 — scaling padhho. $v_{t} \propto m$ aur $v_{t} \propto g$ . Zyada mass → tez; zyada gravity → tez. Apne forecast se compare karo.

Verify: $[m g / c] = \frac{M \cdot L T ^{- 2}}{M L ^{- 1}} = L^{2} T^{- 2} = L T^{- 1} = [v]$ ✅. Terminal speed tak pahunchne se pehle kya hota hai, uske liye dekho Equations of motion (kinematics).

Cell I — exam twist: ek unknown constant ka dimension nikalo

Worked example Real gas ke liye van der Waals equation mein ek term

$\frac{a}{V ^{2}}$ hai aur ise ek pressure $P$ mein add kiya jaata hai. $V$ ek volume hai, toh $[a]$ nikalo. Forecast: $a$ divided by volume-squared ek pressure hona chahiye. Toh $a$ pressure × volume² carry karta hai. Woh dimension kya hai? Guess karo.

Step 1 — sum par homogeneity use karo. Kyunki $\frac{a}{V ^{2}}$ ko $P$ mein add kiya jaata hai, uska $P$ jaisa dimension hona chahiye: $[\frac{a}{V ^{2}}] = [P] = M L^{- 1} T^{- 2} .$ Yeh step kyun? Tum sirf like ko like mein add kar sakte ho — yeh ek rule poora answer pin kar deta hai.

Step 2 — $[a]$ isolate karo. Dono sides ko $[V^{2}] = (L^{3})^{2} = L^{6}$ se multiply karo: $[a] = [P] [V^{2}] = (M L^{- 1} T^{- 2}) (L^{6}) = M L^{5} T^{- 2} .$

Verify: wapas daal ke dekho — $\frac{[ a ]}{[ V ^{2} ]} = \frac{M L ^{5} T ^{- 2}}{L ^{6}} = M L^{- 1} T^{- 2} = [P]$ ✅. Har "constant ka dimension nikalo" twist ke liye trick yeh hai: jo constant jahan bhi baitha hai, uska combination us term ka dimension banana chahiye jiske saath woh exist karta hai.

Figures — do derivations ki geometry

Wave-on-a-string result (Cell C) aur exponent-matching machine (Cell C/D) ko pictures mein dekhna worth hai.

Figure — Dimensional analysis — checking equations, deriving relations

Upar ki figure teen base "axes" ( $M$ , $L$ , $T$ ) ko bookkeeping columns ki tarah dikhati hai: derivation ka matlab hai har column ko alag-alag balance karna.

Doosri figure dikhati hai ki terminal speed (Cell H) $m$ ki tarah kaise badhti hai — ek curve jo steep se shuru hoti hai aur flat ho jaati hai, square root ka visual signature.

Recap — har cell ke liye ek line

Recall Kaun sa move kaun sa cell solve karta hai?

Check that passes (A) ::: dono sides compute karo, sirf "disprove nahi hua" conclude karo. Check that fails (B) ::: ek bhi mismatched exponent prove karta hai ki equation galat hai. Solvable derivation (C) ::: 3 base equations, 3 unknown powers → unique law up to constant $k$ . Under-determined (D) ::: unknowns > 3 ek dimensionless group free chhodta hai; experiment / Buckingham Pi chahiye. Zero exponent (E) ::: koi carrier nahi hone par woh power $0$ ho jaati hai; variable drop ho jaata hai. Dimensionless argument (F) ::: $sin / ln / e^{x}$ ke andar har piece pure number hona chahiye, $[k], [ω]$ fix karta hai. Unit conversion (G) ::: har base factor ko uske dimension ke exponent tak raise karke scale karo. Word problem (H) ::: physics ko force/energy balance mein translate karo, phir powers match karo. Unknown constant (I) ::: constant ka dimension wahi hai jo uske term ko uske neighbour se match karaaye.