Significant figures — rules for operations: Worked examples

1.1.4 · D3 · Physics › Measurement, Vectors & Kinematics › Significant figures — rules for operations

Yeh page Significant figures — rules for operations ke liye "sab kuch ek jagah" wali page hai. Solve karne se pehle, hum har tarah ki situation ka map banate hain jo in rules ko face karni padti hai. Phir har worked example us map ke ek specific box ko nail karta hai, taaki tum koi aisa case kabhi na dekho jise tumne practise na kiya ho.

The scenario matrix

Is topic ka har problem asal mein inhi cells mein se ek hota hai. Aakhri column batata hai ki neeche kaun sa example us cell ko cover karta hai.

Cell	Kya cheez ise tricky banati hai	Kaun sa rule govern karta hai	Example
A Plain multiply	baseline, fewest sig figs wins	×/÷ → sig figs	Ex 1
B Plain add	fewest decimal places wins	+/− → decimals	Ex 2
C Subtraction "digit loss"	lagbhag barabar numbers subtract karne se sig figs khatam ho jaate hain	+/− → decimals	Ex 3
D Exact / counting number	infinite sig figs — answer ko limit NAHI karna chahiye	×/÷	Ex 4
E Mixed chain (× then +)	calculation ke beech mein rules switch karo; guard digits raho	both	Ex 5
F Leading zeros / scientific notation	leading zeros significant nahi hote	counting + ×/÷	Ex 6
G Banker's-rounding edge (exactly 5)	tie-breaking to even	rounding	Ex 7
H Real-world word problem	tum decide karo kaun sa rule, units ke saath	both	Ex 8
I Zero / degenerate input	$0$ par aur limits par kya hota hai	edge case	Ex 9

Matrix ek baar padho. Ab hum ise clear karte hain, ek cell ek baar.

Ex 1 — Cell A: plain multiplication

Worked example

$3.24 \times 5.1 = ?$ Forecast: Answer mein sig figs ki sankhya pehle se guess karo, aage padhne se pehle. (Kaun sa factor sabse kamzor hai?)

Step 1 — Har factor ke sig figs count karo. $3.24$ → teeno digits non-zero hain → 3 sig figs, toh $N_{1} = 3$ . $5.1$ → 2 sig figs, toh $N_{2} = 2$ . (Yaad karo $N$ = "us factor ke sig figs ki sankhya", upar define kiya gaya hai.) Yeh step kyun? ×/÷ rule sirf relative precision ki parwah karta hai, aur sig-fig count uska proxy hai. Toh pehla kaam hamesha yeh hota hai: har factor kitne sig figs carry karta hai?

Step 2 — Raw multiply karo, saare digits raho. $3.24 \times 5.1 = 16.524$ . Yeh step kyun? Abhi kabhi mat round karo — hum ek baar end mein round karte hain. Abhi har digit rakhte hain taaki koi information pehle se na nikle.

Step 3 — $min (N_{1}, N_{2})$ apply karo. $min (N_{1}, N_{2}) = min (3, 2) = 2$ sig figs. $16.524$ ko 2 sig figs mein round karo → pehle do significant digits hain $1, 6$ ; agla digit hai $5$ … lekin $6$ ke baad $524$ clearly $> 5$ hai, toh round up: $17$ . Yeh step kyun? Answer apne sabse kamzor factor se zyada precise (relatively) nahi ho sakta.

Verify: $17/5.1 = 3.33$ , $3.24$ ke karib ✓ (order of magnitude aur leading digit match karte hain). Units wahi carry hote hain jo factors ke paas the.

Ex 2 — Cell B: plain addition

Ex 3 — Cell C: subtraction jo sig figs destroy kar deta hai

Yeh woh case hai jisme log sabse zyada gadbad karte hain, isliye ise ek figure milta hai.

Worked example

$7.556 - 7.42 = ?$ Forecast: Dono inputs 3–4 sig-fig numbers lagte hain. Guess karo kitne sig figs bachenge. (Trap!)

Step 1 — Decimal places count karo. $7.556$ → 3. $7.42$ → 2. Rule decimals ka hai, toh $min = 2$ . Yeh step kyun? Phir bhi ek additive operation hai → decimal places, sig figs nahi kabhi.

Step 2 — Raw subtract karo. $7.556 - 7.42 = 0.136$ . Yeh step kyun? Hum pehle poori precision par subtract karte hain aur rounding end tak delay karte hain, bilkul addition case ki tarah — pehle round karna us chhote difference ko corrupt kar deta jis par humein dhyan dena hai.

Step 3 — 2 decimals tak round karo. $0.136 \to 0.14$ (digit $6 > 5$ , round up). Yeh step kyun? Decimal rule wins.

Shocking part: inputs mein 4 aur 3 sig figs the, lekin answer $0.14$ mein sirf 2 sig figs hain. Do lagbhag-barabar bade numbers subtract karna leading digits cancel kar deta hai aur sirf noisy tail bachti hai — ise catastrophic cancellation kehte hain. Neeche diye figure mein dekho: do lambi bars (lavender $7.556$ aur mint $7.42$ ) almost overlap karti hain; sirf daayein wali chhoti coral sliver tumhara answer hai, aur usmein bahut kam reliable digits hain.

Figure: horizontal bar chart. Ek lavender bar length $7.556$ ki aur ek mint bar length $7.42$ ki lagbhag same point tak pahunchti hain, dikhata hai ki woh almost equal hain. $7.42$ se $7.556$ tak ek chhoti coral bar difference $0.14$ mark karti hai — woh single part jo subtraction ke baad bachta hai, sirf 2 reliable sig figs ke saath.

Verify: $7.42 + 0.14 = 7.56 \approx 7.556$ ✓ (2 decimals tak).

Ex 4 — Cell D: ek exact / counting number

Worked example Ek circle ki radius

$r = 2.5 cm$ hai (2 sig figs). Uska diameter $d = 2 r$ hai. $d$ report karo. Forecast: Kya " $2$ " answer ko… 1 sig fig tak le aata hai? Ya use akela chhod deta hai?

Step 1 — "2" ko classify karo. $d = 2 r$ mein $2$ ek definition hai, measurement nahi. Iske infinite sig figs hain. Yeh step kyun? Sirf measured numbers uncertainty carry karte hain. Ek defined multiplier perfectly exact hota hai, isliye woh kabhi "sabse kamzor" factor nahi ho sakta.

Step 2 — Multiply karo. $d = 2 \times 2.5 = 5.0 cm$ . Yeh step kyun? Hum pehle raw product compute karte hain — poori precision par, kisi bhi sig-fig rounding se pehle — taaki Step 3 mein rounding decision true value par act kare na ki prematurely-trimmed value par.

Step 3 — Sig figs = measured factors mein se fewest. Sirf $r$ measured hai → 2 sig figs → $5.0 cm$ (decimal ke baad trailing zero rakhna!) Yeh step kyun? Exact numbers $min$ mein skip kiye jaate hain. Toh answer $r$ ke 2 sig figs inherit karta hai, 1 nahi.

Verify: $5.0/2 = 2.5 = r$ ✓, aur unit cm (length) hai ✓.

Ex 5 — Cell E: mixed chain, guard digits

Worked example

$(4.2 \times 3.15) + 0.087$ compute karo. Forecast: Multiply aur add alag-alag rules follow karte hain. Har ek kahan apply karte ho?

Step 1 — Multiply karo, guard digits rakhte hue. $4.2 \times 3.15 = 13.23$ . ×-rule ke hisaab se yeh "2 sig figs chahta hai", lekin hum abhi round nahi karte — hum $13.23$ ko ek guard value ke roop mein carry karte hain. Yeh step kyun? Mid-chain round karna information throw karta hai aur error accumulate karta hai. End tak 1–2 extra digits raho.

Step 2 — Note karo ki multiply ne decimal precision kya earn ki. $4.2 \times 3.15$ : sabse kamzor factor $4.2$ mein 2 sig figs hain, toh product sirf 2 sig figs tak trustworthy hai. Product $13.23$ ka pehla sig fig tens place mein hai (" $1$ ") aur doosra units place mein (" $3$ "). "2 sig figs" ka matlab isliye hai ki aakhri reliable digit units place mein baith hai — toh pehla uncertain digit already tenths hai, deta hai 0 reliable decimal places. Yeh step kyun? $10$ se thoda zyada kisi bhi number ke liye, 2 significant digits count karne par tum units digit par pahunchte ho; decimal point ke daayein sab kuch trustworthy range se bahar hai. Isliye "2 sig figs on $13.23$ " translate hota hai "units place tak reliable = 0 decimals" — hume yeh decimal precision aage sahi se add karne ke liye chahiye.

Step 3 — Add karo, phir decimal-place rule apply karo. $13.23 + 0.087 = 13.317$ . Decimal places rakhe jaayenge = $min (0, 3) = 0$ → whole number tak round karo → $13$ . Yeh step kyun? Sum apne worst term se zyada precise nahi ho sakta, jo hai multiply ka units-level result.

Verify: chhota $0.087$ barely $13.23$ ko nudge karta hai; final $13$ agree karta hai " $\approx 4.2 \times 3.15 \approx 13$ " ke saath ✓.

Ex 6 — Cell F: leading zeros & scientific notation

Worked example

$0.00680 \div 0.4 = ?$ Forecast: $0.00680$ mein actually kitne sig figs hain? (Leading zeros decoys hain.)

Step 1 — Sig figs dhyaan se count karo. $0.00680$ : leading zeros ( $0.00$ ) placeholders hain → nahi significant. Digits $6, 8, 0$ hain — decimal ke baad trailing zero count hota hai → 3 sig figs ( $N_{1} = 3$ ). $0.4$ → 1 sig fig ( $N_{2} = 1$ ). Yeh step kyun? Leading zeros sirf decimal point ki position mark karte hain; woh koi measured information carry nahi karte. Final zero, however, precision ka ek deliberate claim hai.

Step 2 — Raw divide karo. $0.00680/0.4 = 0.017$ . Yeh step kyun? Hum division poori precision par carry out karte hain rounding se pehle, taaki Step 3 mein sig-fig cut exact quotient par apply ho na ki pre-rounded stand-in par.

Step 3 — $min = 1$ sig fig apply karo. $min (N_{1}, N_{2}) = min (3, 1) = 1$ . $0.017$ ko 1 sig fig mein round karo → $0.02$ , better written $2 \times 1 0^{- 2}$ . Yeh step kyun? Single-sig-fig $0.4$ weak link hai; Scientific notation akele sig fig ko unambiguous banata hai.

Verify: $0.02 \times 0.4 = 0.008 \approx 0.0068$ ✓ (same order of magnitude, 1-sig-fig level).

Ex 7 — Cell G: exactly-5 tie (banker's rounding)

Worked example Har ek ko bataye gaye precision tak round karo: (a)

$2.45$ ko 1 decimal tak, (b) $8.750$ ko 2 sig figs tak. Forecast: Cutoff par akele $5$ ke saath, kya tum hamesha round up karte ho? (Nahi.)

Step 1 — Tie dhundho. (a) $2.45$ ko tenths ke baad cut karna: dropped part exactly " $5$ " hai uske baad kuch nahi. (b) $8.750$ ko do sig figs ke baad cut karna: dropped part exactly " $50$ " = exactly half. Yeh step kyun? Trailing exact-5 wahi case hai jahan "round up" bahut saare measurements mein systematic upward bias create kar sakta hai.

Step 2 — Nearest EVEN tak round karo. (a) tenths digit $4$ hai (even) → use rakho → $2.4$ . (b) doosra sig-fig digit $7$ hai (odd) → even $8$ par bump karo → $8.8$ . Yeh step kyun? Banker's rounding kept digit ko even hone par force karta hai, toh upward aur downward rounds average par cancel ho jaate hain.

Verify: dono kept digits ( $4$ aur $8$ ) even hain ✓ — yahi correct banker's round ka defining check hai.

Ex 8 — Cell H: real-world word problem

Worked example Ek runner

$d = 400. m$ (3 sig figs, trailing dot ka matlab hai zero real hai) $t = 62.3 s$ mein cover karti hai. Uski average speed correct sig figs ke saath report karo. Forecast: Kya speed multiply/divide (sig figs) ya add/subtract (decimals) situation hai?

Step 1 — Rule choose karo. Speed $= d / t$ → division → sig-fig rule use karo. Yeh step kyun? Pehle hamesha poochho "multiplicative ya additive?" Division = relative errors add hote hain = sig figs count karo.

Step 2 — Sig figs count karo. $400.$ → decimal point teeno zeros/digits ko count karwata hai → 3 sig figs ( $N_{1} = 3$ ). $62.3$ → 3 sig figs ( $N_{2} = 3$ ). Yeh step kyun? Likha hua decimal point author ka signal hai ki woh trailing zeros measured hain, placeholders nahi (dekho Scientific notation).

Step 3 — Raw compute karo, phir round karo. $400./62.3 = 6.42054 \dots m/s$ . $min (N_{1}, N_{2}) = min (3, 3) = 3$ → $6.42 m/s$ . Yeh step kyun? Dono factors 3 sig figs dete hain, toh answer 3 rakhta hai.

Verify: $6.42 \times 62.3 = 400.0 \approx 400.$ m ✓; units $m / s$ = speed ✓ (Dimensional analysis).

Ex 9 — Cell I: zero aur degenerate inputs

Worked example (a)

$0 \times 9.87 = ?$ — kitne sig figs? (b) $d / t$ mein kya hota hai jab $t \to 0$ ? Forecast: Kya "0" ke sig figs hote hain? Kya ×-rule edges par bhi make sense karta hai?

Step 1 — Exact zero handle karo. Yahan $0$ ek pure zero hai (jaise "koi displacement nahi"), toh yeh exact hai — infinite sig figs, counting number ki tarah. Tab $0 \times 9.87 = 0$ , aur answer exactly $0$ hai, " $0.00$ " nahi. Yeh step kyun? Ek pure zero koi measurement nahi hai jisme uncertain last digit ho; yeh ek exact count hai, toh yeh kabhi sig figs limit nahi karta (Cell D logic). Agar input ki jagah ek measured $0.0$ hota, toh hum uski decimal-place information Ex 2 ki tarah rakhte.

Step 2 — Degenerate/limiting case $t \to 0$ . $v = d / t$ mein, jab $d > 0$ fixed ke saath $t \to 0^{+}$ , $v \to + \infty$ . Koi bhi sig-fig rule (near-)zero se division ko rescue nahi kar sakta — relative error $Δ t / t$ blow up ho jaata hai. Yeh step kyun? Sig-fig bookkeeping assume karta hai ki har factor ki relative error chhoti hai (< kuch %). Zero denominator ke paas woh assumption fail ho jaati hai, toh rule meaningful rehna band kar deta hai — tumhe measurement ko unresolved report karna hoga, fake precise answer nahi dena.

Verify: $lim_{t \to 0^{+}} 1/ t = + \infty$ (ratio bina bound ke badhta hai) ✓; aur $0 \times 9.87 = 0$ exactly ✓. Yeh Error propagation — relative vs absolute se connect hota hai: apni khud ki uncertainty ke paas wale value se divide karna relative error explode kar deta hai.

Recall Recall check — ek line har ek

×/÷ answer kaun sa count rakhta hai? ::: Fewest significant figures. +/− answer kaun sa count rakhta hai? ::: Fewest decimal places. Near-equal numbers subtract karna sig figs kyun kho deta hai? ::: Leading digits cancel ho jaate hain, sirf noisy tail bachti hai (catastrophic cancellation). $d = 2 r$ mein "2" ke kitne sig figs hain? ::: Infinite — yeh exact hai. $8.750$ ko 2 sig figs (banker's) mein round karo ::: $8.8$ (nearest even). $2.45$ ko 1 decimal (banker's) mein round karo ::: $2.4$ (nearest even). $400./62.3$ correct sig figs mein ::: $6.42 m/s$ . Pure $0$ aur measured $0.0$ mein kya difference hai? ::: Pure $0$ exact hai (infinite sig figs); measured $0.0$ decimal-place uncertainty carry karta hai.

Connections

Significant figures — rules for operations — parent rules jinhe yeh examples exercise karte hain.
Error propagation — relative vs absolute — kyun cancellation aur division-by-small error blow up karte hain.
Scientific notation — trailing zeros ko disambiguate karta hai (Ex 6, Ex 8).
Measurement & uncertainty — "uncertain last digit" ka physical meaning.
Orders of magnitude & estimation — answers ko sanity-check karna (har "Verify" mein use hota hai).
Dimensional analysis — Ex 8 mein units check.