Errors — absolute, relative, percentage; systematic vs random: Worked examples

1.1.5 · D3 · Physics › Measurement, Vectors & Kinematics › Errors — absolute, relative, percentage; systematic vs rando

Shuru karne se pehle, teen plain-word reminders taaki koi symbol unearned na ho:

Recall Teen quantities ek hi saanss mein

Absolute error $Δ a$ — kitna door hai, same units mein (cm, g, s).
Relative error $δ_{r} = Δ a / a$ — miss as a fraction of the thing, koi units nahi.
Percentage error $δ_{%} = δ_{r} \times 100%$ — wohi fraction percent mein bola gaya.

" $\pm$ " (padho "plus or minus") ka matlab hai "true value kahin is band ke andar hai." $\sum$ (ek bada Greek S) ka matlab sirf "inhe sab jodo" hai. $\overset{a}{ˉ}$ ya $a_{mean}$ average hai.

Scenario matrix

Har error problem jo tum miloge woh in cells mein se kisi ek mein aati hai. Har row ek alag tarah ka sawaal hai; neeche ke examples us cell ke label ke saath hain jisme woh aate hain.

Cell	Situation	Trap jo test hota hai
C1	Full pipeline: mean → abs → rel → %	Error ko value ke decimals tak round karna
C2	Systematic (zero) error present	Averaging isse hata nahi sakti; tum subtract karte ho
C3	Ek hi dataset mein dono flavours	Scatter ko bias se alag karna
C4	Degenerate input: saari readings identical	Zero error ≠ "no uncertainty" (least count)
C5	Tiny quantity vs huge quantity	Same absolute error, wildly different %
C6	Limiting case: value $\to$ 0	Relative error blow up ho jaata hai
C7	Real-world word problem	Words → numbers mein translate karo
C8	Exam twist: % diya, backwards kaam karo	Pipeline ko reverse karo

Hum aathon cells ko 8 worked examples se cover karenge. Unme se do ko figures milti hain kyunki geometry se idea click karta hai.

C1 — Full pipeline

Worked example C1 · Ek length jo paanch baar measure ki gayi

Ek rod ko measure kiya gaya (cm): $12.34, 12.30, 12.36, 12.31, 12.29$ . $L = L_{mean} \pm Δ L$ report karo, aur percentage error bhi.

Forecast: readings lagbhag $0.07$ cm spread karti hain. Andaza lagao: kya % error $0.2%$ ke paas hogi ya $2%$ ke paas? (Aage padhne se pehle apna guess likho.)

Step 1 — Mean. Yeh step kyun? Average hamare single best estimate hota hai; random highs aur lows partly cancel ho jaate hain. $L_{mean} = \frac{12.34 + 12.30 + 12.36 + 12.31 + 12.29}{5} = \frac{61.60}{5} = 12.32 cm .$

Step 2 — Har reading ka absolute error. Kyun? Har $Δ L_{i} = ∣ L_{mean} - L_{i} ∣$ yeh batata hai ki woh reading kitni miss hui. $0.02, 0.02, 0.04, 0.01, 0.03 cm .$

Step 3 — Mean absolute error. Kyun? Typical miss wohi hoti hai jo hum uncertainty ke roop mein quote karte hain. $Δ L = \frac{0.02 + 0.02 + 0.04 + 0.01 + 0.03}{5} = \frac{0.12}{5} = 0.024 cm .$

Step 4 — Error ko round karo, phir value se match karo. Kyun? Error khud bhi fuzzy hoti hai, isliye isse 1 significant figure tak rakho: $Δ L \approx 0.02$ cm. Value ko same decimal place tak round karo. $L = 12.32 \pm 0.02 cm$

Step 5 — Percentage. $δ_{%} = \frac{0.024}{12.32} \times 100% = 0.195% \approx 0.2%$ .

Verify: $12.32$ min ( $12.29$ ) aur max ( $12.36$ ) ke beech mein aata hai — ek mean aisa hi hona chahiye. Aur $0.2%$ forecast band se match karta hai (readings do decimals tak agree karti hain, isliye fraction-of-a-percent error expected hai). ✓

C2 — Systematic zero error (ek figure yahan kaam aati hai)

Worked example C2 · Ek vernier caliper jo kabhi zero nahi padha

Ek vernier caliper $0.05$ cm dikhata hai jab uske jaws bilkul band hain (kuch bhi unke beech nahi). Phir tum ek ball bearing ko chaar baar measure karte ho: $2.55, 2.56, 2.55, 2.54$ cm. Sahi diameter do.

Forecast: agar tum sirf average karke report karte, toh kya tum zyada hote ya kam — aur kitna?

Step 1 — Flavour identify karo. Kyun? Jaws ke beech kuch na hone ke bawajood ek fixed reading aana ek zero error hai: wahi $+ 0.05$ cm har reading mein "baked in" hai. Yeh ek systematic error hai — averaging isse kabhi nahi hata sakti.

Step 2 — Raw readings ka mean. Kyun? Pehle yeh best estimate lo ki instrument ne kya report kiya. $d_{raw} = \frac{2.55 + 2.56 + 2.55 + 2.54}{4} = \frac{10.20}{4} = 2.55 cm .$

Step 3 — Known offset subtract karo. Kyun? Jab ek systematic error identify aur measure ho jaati hai, tum usse exactly correct karte ho. Figure dekho: blue "true" scale, red "instrument" scale se $0.05$ cm fixed shift par hai har jagah. $d_{true} = 2.55 - 0.05 = 2.50 cm .$

Step 4 — Random error bacha rehta hai, aur hum usse abhi bhi report karte hain. Mean se absolute errors: $0.00, 0.01, 0.00, 0.01$ ; mean abs $= 0.005 \approx 0.01$ cm. $d = 2.50 \pm 0.01 cm$

Verify: correction har reading ko exactly same amount se neeche shift karta hai, isliye scatter (aur us wajah se $\pm 0.01$ ) unchanged rehta hai — ek bias correct karna kabhi random spread nahi badalta. ✓

C3 — Ek hi dataset mein dono flavours

Worked example C3 · Ek stopwatch jo lag karta hai aur ek kaanpta haath

Ek stopwatch ke baare mein pata hai ki woh har baar $0.10$ s late start karta hai (bias). Ek fixed interval ki paanch timings: $8.4, 8.6, 8.3, 8.7, 8.5$ s. Corrected best value aur uski random uncertainty nikalo.

Forecast: yahan kaun sa number systematic hai aur kaun sa random $\pm$ ? Compute karne se pehle unhe naam do.

Step 1 — Dono effects alag karo. Kyun? Inse alag tarah se lada jaata hai: bias subtract karo, scatter average se door karo.

$- 0.10$ s ka hamesha-late start systematic hai (har reading $0.10$ s zyada chhoti dikhti hai).
$8.3$ – $8.7$ mein spread random hai.

Step 2 — Raw readings ka mean. $t_{raw} = \frac{8.4 + 8.6 + 8.3 + 8.7 + 8.5}{5} = \frac{42.5}{5} = 8.50 s .$

Step 3 — Bias correct karo. Kyun? Clock late start hoti hai, isliye woh kam count karti hai; sahi interval zyada lamba hai. $0.10$ s jodo. $t_{true} = 8.50 + 0.10 = 8.60 s .$

Step 4 — Random uncertainty. $8.50$ se abs errors: $0.10, 0.10, 0.20, 0.20, 0.00$ ; mean abs $= 0.60/5 = 0.12$ s. $t = 8.60 \pm 0.12 s$

Verify: bias ne centre move kiya ( $8.50 \to 8.60$ ) lekin width ( $\pm 0.12$ ) untouched chhaodi — exactly wohi jo ek systematic shift karta hai. Units poore time seconds mein hain. ✓

C4 — Degenerate input: saari readings identical

Worked example C4 · Jab teri readings perfectly agree karti hain

Ek length ko ek ruler se jo least count $0.1$ cm ka hai, teen baar padha jaata hai aur har baar $7.5, 7.5, 7.5$ cm aata hai. Kya uncertainty zero hai?

Forecast: mean-absolute-error formula $0$ dega. Kya honest error isliye $0$ hai? Guess karo haan/nahi.

Step 1 — Formula apply karo. Kyun? Dekhne ke liye ki woh apne aap kya kehta hai. Mean $= 7.5$ ; har $∣7.5 - 7.5∣ = 0$ , isliye $Δ a_{mean} = 0$ .

Step 2 — Reality check. Kyun? Zero scatter ka matlab zero uncertainty nahi hota. Tum abhi bhi ruler ke smallest marks ke beech nahi padh sakte. Instrument ka least count ek floor set karta hai.

Step 3 — Least-count rule use karo. Kyun? Jab statistical scatter truth ko under-report kare, uncertainty kam se kam least count hoti hai (aksar uski half li jaati hai, lekin safe exam convention full least count hai). $L = 7.5 \pm 0.1 cm$

Verify: reported $\pm 0.1$ least count ke barabar hai — yeh legitimately kabhi us se neeche nahi ja sakta jo instrument resolve kar sakta hai. ✓ (Dekho Least Count and Vernier Calipers.)

C5 — Same absolute error, opposite worlds (ek figure)

Worked example C5 · Ek baal par aur ek sadak par ek millimetre

Ek caliper same absolute error $Δ = 0.001$ m se do kamon mein miss karta hai: (a) $0.002$ m diameter ka ek baal; (b) $2.000$ m lambi ek sadak. Percentage errors compare karo.

Forecast: wahi $0.001$ m — kya dono % errors barabar hongi, ya wildly different?

Step 1 — Baal ke liye percentage. Value se kyun divide karte hain? Relative error poochti hai "miss cheez ke comparison mein kitni badi hai?" $δ_{%}^{hair} = \frac{0.001}{0.002} \times 100% = 50%.$

Step 2 — Sadak ke liye percentage. $δ_{%}^{road} = \frac{0.001}{2.000} \times 100% = 0.05%.$

Step 3 — Picture padho. Kyun? Green bar (sadak) $1$ mm error ko mushkil se notice karta hai; red bar (baal) aadha galat hai. Identical absolute error → percentage mein $1000 \times$ fark.

Verify: percentages ka ratio $= 50/0.05 = 1000$ , jo dono lengths ke ratio ke barabar hai $2.000/0.002 = 1000$ . Consistent. ✓

C6 — Limiting case: value zero ki taraf jaati hai

Worked example C6 · Kuch almost weightless cheez weighing karna

Ek balance ki ek fixed random uncertainty hai $Δ m = 0.01$ g. Percentage error calculate karo jab measured mass ghatti jaaye: $m = 10.00$ g, phir $0.10$ g, phir $0.02$ g.

Forecast: jab $m \to 0$ aur $Δ m$ fixed ho, toh kya $δ_{%}$ settle hoti hai, ya infinity ki taraf bhaag jaati hai?

Step 1 — Teen percentages. Kyun? Trend dekhne ke liye. $δ_{%} = \frac{0.01}{m} \times 100%$ . $m = 10.00 : 0.1%, m = 0.10 : 10%, m = 0.02 : 50%.$

Step 2 — Limit interpret karo. Kyun? Kyunki $m$ denominator mein hai, $m$ ko $Δ m$ fixed rakh ke ghataane se $δ_{%} \to \infty$ ho jaata hai. Chhhoti quantities ko fractionally measure karna mushkil hota hai chahe absolute error tiny ho.

Step 3 — Practical lesson. Kyun? Ek chhoti quantity ko acche percentage tak pin down karne ke liye, tumhe ek finer instrument chahiye (chhota $Δ m$ ) — ya bahut saari ek saath measure karo aur divide karo (100 daane measure karo, ek daane ka mass $1/100$ error ke saath pao).

Verify: $0.01/0.02 = 0.5 = 50%$ ; sequence $0.1%, 10%, 50%$ strictly increase karta hai jab $m$ girta hai — " $δ_{%} \to \infty$ as $m \to 0$ " se match karta hai. ✓

C7 — Real-world word problem

Worked example C7 · Fuel per 100 km

Ek car ka trip meter distance $d = 240 \pm 2$ km padhta hai aur fuel gauge $V = 16.0 \pm 0.5$ L. Fuel-per-distance figure $V / d$ mein percentage error kya hai?

Forecast: dono inputs mein se kaun zyada error contribute karta hai, $d$ ya $V$ — bada number ya chhota?

Step 1 — Har input ka percentage error. Kyun? Quotient ke liye, percentage errors add hote hain (dekho Combination of Errors). $δ_{%} (d) = \frac{2}{240} \times 100% = 0.83%, δ_{%} (V) = \frac{0.5}{16.0} \times 100% = 3.13%.$

Step 2 — Ratio $V / d$ ke liye combine karo. Add kyun? Product ya quotient mein, fractional errors addition se combine hote hain; yahi woh rule hai jisse uncertainties propagate hoti hain. $δ_{%} (V / d) = 0.83% + 3.13% = 3.96% \approx 4.0%.$

Verify: fuel measurement dominate karta hai ( $3.13%$ vs $0.83%$ ) — forecast se match karta hai ki chhoti-precision input (fuel) rule karta hai. Final $4.0%$ kisi bhi single input se bada hai, jaise ki positives jodhne se hona chahiye. ✓

C8 — Exam twist: percentage se backwards kaam karo

Recall Jaane se pehle self-test

Inme se har ek kaun si cell hai?

"Paanchon readings $3.20$ mm aayi." ::: C4 (degenerate → least count sets the floor).
"Ammeter circuit open hone par $0.2$ A padhta hai." ::: C2 (systematic zero error → subtract).
"Area mein error $=$ do sides ke % errors ka sum." ::: C7-style (combination, percentages add).
"Wohi $\pm 0.1$ mm ek $0.5$ mm wire par aur ek $500$ mm rod par." ::: C5 (fractional error enormously alag hoti hai).