Dimensional analysis — checking equations, deriving relations: Exercises

1.1.3 · D4 · Physics › Measurement, Vectors & Kinematics › Dimensional analysis — checking equations, deriving relation

Level 1 — Recognition

Goal: definition se seedha "kind" padh lena.

L1.1

Density (mass per unit volume) ki dimension batao.

Recall Solution

KYA karte hain: definition par wapas jaate hain. Density $ρ = \frac{mass}{volume}$ . KYUN: dimensions hamesha defining relation se nikaalte hain, kabhi guess nahi karte.

Mass ki dimension $[M]$ hoti hai.
Volume length × length × length hota hai, isliye $[L^{3}]$ . $[ρ] = \frac{[ M ]}{[ L ^{3} ]} = M L^{- 3}$ Answer: $M L^{- 3}$ .

L1.2

Momentum ( $p = m v$ , mass times velocity) ki dimension batao.

Recall Solution

Velocity displacement over time hai, $[L T^{- 1}]$ . Mass $[M]$ se multiply karo: $[p] = [M] [L T^{- 1}] = M L T^{- 1} .$ Answer: $M L T^{- 1}$ .

L1.3

Inme se kaun si cheez force ke saath add ki ja sakti hai bina homogeneity toode? (a) pressure × area, (b) mass × velocity, (c) energy / length.

Recall Solution

Force ki dimension $[M L T^{- 2}]$ hai. Har ek check karo:

(a) pressure $= M L^{- 1} T^{- 2}$ , area $= L^{2}$ , product $= M L T^{- 2}$ ✅ force se match karta hai.
(b) $M \cdot L T^{- 1} = M L T^{- 1}$ (yeh momentum hai) ❌.
(c) energy $= M L^{2} T^{- 2}$ , $L$ se divide karo toh $M L T^{- 2}$ milta hai ✅ force se match karta hai. Answer: (a) aur (c) force ki dimension share karte hain; (b) nahi karta.

Level 2 — Application

Goal: real equations par homogeneity check chalana.

L2.1

Kya $v = u + a t$ dimensionally consistent hai? ( $v, u$ velocities, $a$ acceleration, $t$ time.)

Recall Solution

Har added/equated term match karna chahiye.

$[v] = [u] = L T^{- 1}$ .
$[a t] = (L T^{- 2}) (T) = L T^{- 1}$ ✅. Teeno $L T^{- 1}$ hain. Consistent. ✅

L2.2

Equation $E = \frac{1}{2} m v^{2} + m g h$ . Check karo ki right side ke dono terms ki dimensions barabar hain.

Recall Solution

$[\frac{1}{2} m v^{2}] = M (L T^{- 1})^{2} = M L^{2} T^{- 2}$ ( $\frac{1}{2}$ pure number hai, drop kar do).
$[m g h] = M (L T^{- 2}) (L) = M L^{2} T^{- 2}$ . Dono $M L^{2} T^{- 2}$ hain — energy ki dimension. ✅ Consistent hai (aur $[E]$ se bhi match karta hai).

L2.3

Error dhundho: ek student likhta hai $s = u t + \frac{1}{2} a t^{3}$ .

Recall Solution

$[u t] = (L T^{- 1}) (T) = L$ ✅.
$[a t^{3}] = (L T^{- 2}) (T^{3}) = L T^{1}$ ❌ — yeh length × time hai, length nahi. Dono add kiye gaye terms alag hain, isliye equation galat hai. Sahi doosra term $\frac{1}{2} a t^{2}$ hai, jo $(L T^{- 2}) (T^{2}) = L$ deta hai. ✅

Level 3 — Analysis

Goal: arguments, exponents, aur hidden dimensionless groups ke baare mein reason karna.

L3.1

Equation $y = A sin (k x - ω t)$ mein, $x$ length hai aur $t$ time hai. $k$ aur $ω$ ki dimensions nikalo.

Recall Solution

KYUN constraint hai: $sin$ ka argument dimensionless hona chahiye — tum "2 metres" ka sine nahi le sakte. Isliye bracket ke andar, $k x$ aur $ω t$ dono pure numbers hone chahiye, aur aapas mein match karne chahiye (unhe subtract kiya ja raha hai).

$[k x] = M^{0} L^{0} T^{0} \Rightarrow [k] [L] = 1 \Rightarrow [k] = L^{- 1}$ .
$[ω t] = M^{0} L^{0} T^{0} \Rightarrow [ω] [T] = 1 \Rightarrow [ω] = T^{- 1}$ . Answer: $[k] = L^{- 1}$ (per metre), $[ω] = T^{- 1}$ (per second).

L3.2

Ek quantity $F = \frac{G m _{1} m _{2}}{r ^{2}}$ se define hoti hai, jahan $F$ force hai, $m_{1}, m_{2}$ masses hain, $r$ distance hai. $G$ ki dimension nikalo.

Recall Solution

KYA karte hain: unknown $[G]$ ke liye homogeneity equation solve karte hain. $[F] = \frac{[ G ] [ m _{1} ] [ m _{2} ]}{[ r ] ^{2}} \Rightarrow M L T^{- 2} = \frac{[ G ] M \cdot M}{L ^{2}} .$ Rearrange karo: $[G] = \frac{M L T ^{- 2} \cdot L ^{2}}{M ^{2}} = M^{- 1} L^{3} T^{- 2} .$ Answer: $M^{- 1} L^{3} T^{- 2}$ .

L3.3

Claim kiya gaya hai ki sphere par low speed mein drag force $F = 6 π η r v$ hai, jahan $η$ (viscosity) ki dimension $M L^{- 1} T^{- 1}$ hai, $r$ radius hai, $v$ speed hai. Verify karo ki equation dimensionally consistent hai.

Recall Solution

Factor $6 π$ pure number hai — drop kar do. $[η r v] = (M L^{- 1} T^{- 1}) (L) (L T^{- 1}) = M L^{- 1 + 1 + 1} T^{- 1 - 1} = M L^{1} T^{- 2} .$ Yeh exactly $[F] = M L T^{- 2}$ hai. ✅ Consistent.

Level 4 — Synthesis

Goal: $M, L, T$ ke powers match karke ek law ki form derive karna.

L4.1

Stretched string par transverse wave ki speed $v$ tension $T$ (ek force, $[M L T^{- 2}]$ ) aur mass per unit length $μ$ ( $[M L^{- 1}]$ ) par depend kar sakti hai. Derive karo ki $v$ , $T$ aur $μ$ ke saath kaise scale karta hai.

Figure — Dimensional analysis — checking equations, deriving relations

Recall Solution

Step 1 — powers ka product guess karo. Maan lo $v = k T^{a} μ^{b}$ , jahan $k$ dimensionless constant hai. KYUN product: sirf do ingredients ke saath, ek monomial hi aisi form hai jo homogeneity pin down kar sakti hai. Step 2 — dono sides ki dimensions likho. $L T^{- 1} = (M L T^{- 2})^{a} (M L^{- 1})^{b} = M^{a + b} L^{a - b} T^{- 2 a} .$ Step 3 — har base alag-alag match karo (har $M, L, T$ ek independent axis hai — figure mein teen sliders dekho):

$M : a + b = 0$
$L : a - b = 1$
$T : - 2 a = - 1 \Rightarrow a = \frac{1}{2}$ $a = \frac{1}{2}$ aur $a + b = 0$ se: $b = - \frac{1}{2}$ . (Check $L$ : $\frac{1}{2} - (- \frac{1}{2}) = 1$ ✅.) Step 4 — assemble karo. $v = k T^{1/2} μ^{- 1/2} = k \frac{T}{μ} .$ Insight: sach mein constant $k = 1$ hai, lekin dimensional analysis ise supply nahi kar sakta. Form $v \propto T / μ$ puri tarah determine ho gayi hai.

L4.2

$1 Pa = 1 N/m^{2}$ ki pressure ko CGS unit (g, cm, s) mein convert karo. $[P] = M L^{- 1} T^{- 2}$ .

Recall Solution

KYUN exponents matter karte hain: har base unit apne conversion factor se scale hota hai jo dimension $M^{1} L^{- 1} T^{- 2}$ mein us base ki power tak raise hota hai.

$M : 1 kg = 1 0^{3} g$ , power $+ 1$ → factor $1 0^{3}$ .
$L : 1 m = 1 0^{2} cm$ , power $- 1$ → factor $1 0^{- 2}$ .
$T : 1 s$ , power $- 2$ → factor $1$ . $1 Pa = 1 0^{3} \times 1 0^{- 2} \times 1 = 1 0^{1} g\cdotcm^{- 1} s^{- 2} = 10 (barye/dyne\cdotcm^{- 2}) .$ Answer: $1 Pa = 10 g\cdotcm^{- 1} s^{- 2}$ .

Level 5 — Mastery

Goal: pehchanna ki method kahan under-determine karta hai ya fail karta hai — aur precisely kyun batana.

L5.1

Propose kiya gaya hai ki vibrating water drop ki frequency $f$ us ke radius $r$ , density $ρ$ ( $[M L^{- 3}]$ ), aur surface tension $σ$ ( $[M T^{- 2}]$ , yaani force per length) par depend karti hai. Scaling $f = k r^{a} ρ^{b} σ^{c}$ derive karo.

Recall Solution

Step 1 — dono sides ki dimensions. $[f] = T^{- 1}$ . $T^{- 1} = L^{a} (M L^{- 3})^{b} (M T^{- 2})^{c} = M^{b + c} L^{a - 3 b} T^{- 2 c} .$ Step 2 — bases match karo.

$M : b + c = 0$
$T : - 2 c = - 1 \Rightarrow c = \frac{1}{2} \Rightarrow b = - \frac{1}{2}$
$L : a - 3 b = 0 \Rightarrow a = 3 b = - \frac{3}{2}$ Step 3 — assemble karo. $f = k r^{- 3/2} ρ^{- 1/2} σ^{1/2} = k \frac{σ}{ρ r ^{3}} .$ Insight: teen ingredients, teen base dimensions → ek unique solution. Problem exactly determined hai.

L5.2

Ek student projectile ka full range formula, $R = \frac{v ^{2} s i n ( 2 θ )}{g},$ purely dimensional analysis se derive karne ki koshish karta hai, $R = k v^{a} g^{b}$ guess karke. Explain karo ki dimensional analysis yahan kya kar sakta hai aur kya nahi kar sakta.

Recall Solution

Jo KAR SAKTA HAI. $R = k v^{a} g^{b}$ match karo: $L = (L T^{- 1})^{a} (L T^{- 2})^{b} = L^{a + b} T^{- a - 2 b} .$

$L : a + b = 1$
$T : - a - 2 b = 0 \Rightarrow a = - 2 b$ Substitute karo: $- 2 b + b = 1 \Rightarrow b = - 1, a = 2$ . Toh $R = k \frac{v ^{2}}{g} .$ Form $R \propto v^{2} / g$ sahi hai. Jo NAHI KAR SAKTA.

Dimensionless factor $sin (2 θ)$ invisible hai — $θ$ ek angle hai (dimensionless), isliye woh kabhi base-matching mein enter nahi karta. Dimensional analysis nahi bata sakta ki $R$ launch angle par depend karta hai.
Pure number $k$ unknown hai (yahan woh $sin 2 θ$ hoga, jo 1 tak cap hai). Conclusion: dimensional analysis $v^{2} / g$ scaling deta hai lekin angular factor aur kisi bhi numeric constant ke liye blind hai. Full derivation ke liye dekho Equations of motion (kinematics).

L5.3

Fast-moving body par drag force $F$ speed $v$ , cross-sectional area $A$ ( $[L^{2}]$ ), fluid density $ρ$ ( $[M L^{- 3}]$ ), aur viscosity $η$ ( $[M L^{- 1} T^{- 1}]$ ) par depend karti hai. Explain karo ki ek single monomial $F = k v^{a} A^{b} ρ^{c} η^{d}$ under-determined kyun hai, aur free parameters ki sankhya batao.

Recall Solution

Equations aur unknowns count karo. Hamare paas 4 unknown exponents ( $a, b, c, d$ ) hain lekin sirf 3 base dimensions ( $M, L, T$ ) hain → sirf 3 equations hain. $M L T^{- 2} = (L T^{- 1})^{a} (L^{2})^{b} (M L^{- 3})^{c} (M L^{- 1} T^{- 1})^{d} .$

$M : c + d = 1$
$L : a + 2 b - 3 c - d = 1$
$T : - a - d = - 2$ Teen equations, chaar unknowns → ek free parameter ( $4 - 3 = 1$ ). Solution ek family hai, single answer nahi: physically yeh free parameter Reynolds number $Re = ρ v A / η$ hai, jo ek dimensionless group hai. Drag ban jaata hai $F = ρ v^{2} A \cdot ϕ (Re),$ jahan $ϕ$ ek unknown dimensionless function hai jo method pin down nahi kar sakta. Answer: exactly 1 parameter se under-determined; tumhe ek dimensionless group milta hai, formula nahi. Yeh exactly Buckingham Pi theorem ka content hai.

Score yourself

Recall Mastery checklist

Kisi bhi definition se dimension padh lena (L1) ::: $M, L, T$ defining ratio se. Homogeneity check chalana aur verdict sahi padh lena (L2) ::: failed = galat; passed = disprove nahi hua. $sin / ln / e^{x}$ ke argument ko dimensionless force karna (L3) ::: coefficient ki dimension milti hai. Powers match karke do-ingredient law derive karna (L4) ::: monomial → $M, L, T$ match karo → solve karo. Under-determination pehchaanna jab variables > base dimensions hoon (L5) ::: leftover count = dimensionless groups.

Connections

Units and the SI system — L1 aur L4 ke peeche unit conversions.
Significant figures and error propagation — in answers ka numeric side.
Equations of motion (kinematics) — projectile range ki full derivation (L5.2).
Buckingham Pi theorem — L5.3 ko govern karne wala general rule.
Vectors — components and addition — vector quantities ki dimensions.