Significant figures — rules for operations: Exercises

1.1.4 · D4 · Physics › Measurement, Vectors & Kinematics › Significant figures — rules for operations

Shuru karne se pehle, ek baat "decimal places" aur "significant figures" ka physically kya matlab hai — taaki neeche ke figures samajh aa sakein.

Figure — Significant figures — rules for operations

Upar number line dekho. Red measurement sirf tenths mark tak hi trustworthy hai — usse finer kuch bhi guess hai. Jab hum ise kisi finer number mein add karte hain, toh poora sum usi coarse red resolution ko inherit kar leta hai. Jab hum multiply karte hain, toh instead mein trustworthy digits ki count matter karti hai. Har problem ke liye yeh picture dimaag mein rakhna.

Level 1 — Recognition

Goal: ek number padhna aur batana ki woh kitni information carry karta hai.

Problem 1.1

Inme se har ek mein kitne significant figures hain? (a) $0.00405$ (b) $2.300$ (c) $4500$ (d) $6.022 \times 1 0^{23}$

Recall Solution 1.1

Hum kya karte hain: parent note se counting rules apply karte hain, digit by digit.

(a) $0.00405$ — leading zeros ( $0.00$ ) sirf place-holders hain → significant nahin. Digits $4$ , $0$ , $5$ bachte hain; beech ka $0$ non-zeros ke beech baith a hai isliye woh count hota hai. → 3 sig figs.
(b) $2.300$ — $2$ aur $3$ count hote hain; do trailing zeros decimal point ke baad aate hain, isliye woh precision ka deliberate claim hain → woh count hote hain. → 4 sig figs.
(c) $4500$ — trailing zeros bina decimal point ke ambiguous hain. 2, 3, ya 4 ho sakte hain. → ambiguous (ise fix karne ke liye $4.5 \times 1 0^{3}$ , $4.50 \times 1 0^{3}$ , etc. likho).
(d) $6.022 \times 1 0^{23}$ — scientific notation mein sirf mantissa $6.022$ count hoti hai. → 4 sig figs.

Problem 1.2

Har operation par kaun sa rule (sig figs ya decimal places) apply hoga, aur kyun? (a) $3.2 \times 4.115$ (b) $3.2 + 4.115$

Recall Solution 1.2

Hum pehle yeh kyun poochte hain: rule choose karna sabse important decision hai — agar yeh galat ho, toh har baad ka step galat hai.

(a) Multiplication → relative error propagate karta hai → significant figures count karo. $min (2, 4) = 2$ rakho.
(b) Addition → absolute error propagate karta hai → decimal places count karo. $min (1, 3) = 1$ rakho. Same do numbers, alag rule, kyunki error ka type alag hai.

Level 2 — Application

Goal: ek rule ko cleanly run karke rounded answer tak pahunchna.

Problem 2.1

$8.24 \times 0.61$ ko correct number of sig figs mein compute karo.

Recall Solution 2.1

Sig figs: $8.24 \to 3$ , $0.61 \to 2$ .
Yeh step kyun? Multiplication → $min (3, 2) = 2$ sig figs rakho.
Raw product $= 5.0264$ .
2 sig figs mein round karo → teesra digit $2 < 5$ hai, drop karo → $5.0$ . Dhyan raho trailing zero rakho — use " $5$ " karna falsely claim karega ki sirf 1 sig fig hai.

Problem 2.2

$45.76 - 2.1$ ko correct precision mein compute karo.

Recall Solution 2.2

Decimal places: $45.76 \to 2$ , $2.1 \to 1$ .
Yeh step kyun? Subtraction → $min (2, 1) = 1$ decimal place rakho.
Raw difference $= 43.66$ .
1 decimal mein round karo: cutoff ke baad digit $6 > 5$ hai → round up → $43.7$ .

Problem 2.3

$\frac{9.00}{4}$ compute karo jahan $4$ ek exact counting number hai (4 identical objects).

Recall Solution 2.3

Yeh kyun matter karta hai: ek exact number ke infinite sig figs hote hain — woh kabhi result ko limit nahin karta. Toh sirf $9.00$ (3 sig figs) count hota hai.
Raw quotient $= 2.25$ .
3 sig figs rakho → $2.25$ .

Level 3 — Analysis

Goal: pressure mein sahi rule choose karna, aur precision ka behavior dekhna.

Problem 3.1

$12.11 + 0.3 + 1.012$ compute karo aur phir us sum ko $2.5$ se multiply karo. Har answer ko correct precision mein do, sirf bilkul end mein rounding karo.

Recall Solution 3.1

Step 1 (addition): raw sum $= 13.422$ . Terms mein sabse kam decimal places $0.3$ ke hain jisme 1 decimal hai. Guard-digit trick: abhi $13.4$ mein round mat karo — step 2 mein poora $13.422$ carry karo. Step 2 (multiplication): $13.422 \times 2.5 = 33.555$ .

Final answer mein kitne sig figs ho sakte hain? Multiplication hume $min (sig figs)$ tak limit karta hai. Yahan $2.5$ ke 2 sig figs hain, aur $13.422$ (carried) ek aisa number represent karta hai jiski precision-limiting factor 1 decimal place ≈ 3 sig figs thi. Toh 2 sig figs rakho.
$33.555$ ko 2 sig figs mein round karo → $34$ . Agar tumne prematurely $13.4$ mein round kar diya hota, toh $13.4 \times 2.5 = 33.5 \to 34$ milta — yahan same hai, lekin early rounding safe guarantee nahin hai, isliye kabhi uss par rely mat karo.

Problem 3.2

Near-equal numbers ka subtraction: $5.004 - 4.999$ compute karo aur surviving precision par comment karo.

Recall Solution 3.2

Dono ke 3 decimal places hain, isliye 3 decimal places rakho.
Raw $= 0.005$ .
Answer $= 0.005$ , jiske sirf 1 sig fig hai!
Yeh kaisa dikhta hai: do numbers jinhe 4 sig figs tak jaana jaata tha, subtract hone par sirf 1 sig fig wala result dete hain. Yeh catastrophic cancellation hai — reliable leading digits cancel ho jaate hain, uncertain tail expose ho jaata hai. Decimal-place rule ne ise automatically handle kiya; inputs par sig-fig counting tumhe mislead karti.

Level 4 — Synthesis

Goal: ek multi-step physical formula mein rules combine karna.

Problem 4.1

Ek rectangular plate ki length $L = 12.5 cm$ aur width $W = 3.20 cm$ hai. Area aur perimeter dono ko correct precision mein compute karo.

Recall Solution 4.1

Area (multiplication):

$A = L \times W = 12.5 \times 3.20 = 40.0$ (raw $= 40.00$ ).
Sig figs: $12.5 \to 3$ , $3.20 \to 3$ → 3 rakho → $A = 40.0 cm^{2}$ .

Perimeter (addition, with an exact multiplier):

$P = 2 (L + W)$ . $2$ exact hai — infinite sig figs.
Inner sum $L + W = 12.5 + 3.20 = 15.70$ ; sabse kam decimals $12.5$ ke hain (1 decimal) → precision 1 decimal tak limited → $15.7$ .
Exact $2$ se multiply karo: $P = 2 \times 15.7 = 31.4$ (exact $2$ precision reduce nahin karta).
$P = 31.4 cm$ . Dhyan do: ek problem ke andar do alag rules — area ke liye ×, perimeter ke inner step ke liye +.

Problem 4.2

Density: $m = 24.0 g$ mass ka ek sample $V = 5.1 cm^{3}$ volume occupy karta hai. Density $ρ = m / V$ nikalo.

Recall Solution 4.2

Division → sig figs. $24.0 \to 3$ , $5.1 \to 2$ → $min (3, 2) = 2$ rakho.
Raw $ρ = 24.0/5.1 = 4.7058 \dots g/cm^{3}$ .
2 sig figs mein round karo: teesra digit $0 < 5$ → drop karo → $ρ = 4.7 g/cm^{3}$ .

Level 5 — Mastery

Goal: poora reasoning — rules choose karna, guard digits, exact numbers, aur cancellation sab ek saath.

Problem 5.1

Ek car $d = 1000.0 m$ (5 sig figs) $t = 20.1 s$ mein cover karti hai. Phir woh ek doosra stretch cover karti hai, marker $x_{1} = 850.42 m$ se $x_{2} = 851.00 m$ tak, $t_{2} = 0.30 s$ mein. Har stretch ka average speed correct precision mein nikalo, aur batao kaun sa speed kam trustworthy hai aur kyun.

Recall Solution 5.1

Stretch 1: $v_{1} = d / t = 1000.0/20.1$ .

Division → sig figs. $1000.0 \to 5$ , $20.1 \to 3$ → 3 rakho.
Raw $= 49.751 \dots$ → 3 sig figs → cutoff digit $5$ hai aur baad mein $1$ hai → round up → $v_{1} = 49.8 m/s$ .

Stretch 2 distance (pehle subtraction!): $Δ x = 851.00 - 850.42$ .

Subtraction → decimal places. Dono ke 2 decimals hain → 2 decimals rakho.
Raw $= 0.58 m$ — lekin iske sirf 2 sig figs hain (catastrophic cancellation: 5-sig-fig inputs, 2-sig-fig output).

Stretch 2 speed: $v_{2} = Δ x / t_{2} = 0.58/0.30$ .

Division → sig figs. $0.58 \to 2$ , $0.30 \to 2$ → 2 rakho.
Raw $= 1.9333 \dots$ → 2 sig figs → $v_{2} = 1.9 m/s$ .

Kaun sa kam trustworthy hai? $v_{2}$ . Bhaale hi uski input positions 5 sig figs tak jaani jaati thin, subtraction ne us precision ka zyaadatar hissa destroy kar diya, sirf 2 sig figs bachaye — isliye derived speed $v_{1}$ se kaafi coarse hai. Lesson: subtraction of near-equal numbers mein precision conserved nahin hoti.

Problem 5.2

Tum ek pendulum ka period $n = 10$ swings (ek exact count) timing karke measure karte ho, stopwatch reading $t = 18.46 s$ hai. Period $T = t / n$ hai. Phir theory kehti hai $g = \frac{4 π ^{2} L}{T ^{2}}$ , jahan $L = 0.9930 m$ hai. $T$ aur phir $g$ ko correct precision mein nikalo. $π$ ko exact maano.

Recall Solution 5.2

Period: $T = t / n = 18.46/10$ .

$n = 10$ exact hai (infinite sig figs) → sirf $18.46$ (4 sig figs) hume limit karta hai.
$T = 1.846 s$ (4 sig figs rakho) — guard value ke roop mein carry karo.

$g = 4 π^{2} L / T^{2}$ compute karo (sab multiplications/divisions; $4$ aur $π$ exact hain):

Sig-fig-limiting factors: $L = 0.9930$ (4 sig figs), $T = 1.846$ (4 sig figs) → final answer 4 sig figs rakhega.
$T^{2} = 1.84 6^{2} = 3.40772$ (guard digits rakhe).
$4 π^{2} = 39.4784 \dots$ (exact, kaafi digits carry karo).
$g = 39.4784 \times 0.9930/3.40772 = 39.2001/3.40772 = 11.503 \dots m/s^{2}$ .
4 sig figs mein round karo → $g = 11.50 m/s^{2}$ . (Physically zyaada hai kyunki real pendulum mein aur errors hote hain — lekin sig-fig bookkeeping exactly yahi hai.)

Recall Is poore page ka ek-line summary

Pehle operation type ( $\times \div \Rightarrow$ sig figs, $+ - \Rightarrow$ decimals) ::: guard digits carry karo ::: exact numbers ignore karo ::: end mein ek baar round karo ::: subtraction se precision wipout hone se savdhan raho.

Connections

Error propagation — relative vs absolute — dono rules ke peeche ka rigorous "why".
Measurement & uncertainty — input uncertainties kahan se aati hain.
Scientific notation — Problem 1.1(c) mein ambiguous $4500$ ko humne kaise fix kiya.
Orders of magnitude & estimation — jab 2 sig figs bhi zaroorat se zyaada ho.
Dimensional analysis — $g$ ki units check karta hai; sig figs uski precision check karte hain.