Exercises — Errors — absolute, relative, percentage; systematic vs random

KAUNSA type: jaws touch karne par $0$ read hona chahiye lekin $0.03$ cm read karta hai. Har measurement $0.03$ cm se upar push hoti hai — ek direction mein consistent bias. Yeh ek systematic (zero) error hai, random nahi. KYUN hum ise fix kar sakte hain: ek known constant offset ko simply subtract kiya ja sakta hai (dekho Least Count and Vernier Calipers ki us instrument par zero errors kaise aate hain). Corrected reading: $4.57-0.03=\boxed{4.54\text{cm}}$.

Readings dono ek doosre se upar aur neeche girti hain bina kisi fixed direction ke — ek coin-flip wali nudge. Yeh random error hai. Averaging se yeh shrink hogi, kyunki highs aur lows partly cancel karte hain.

(a) $\pm$ wala number hi absolute error hai: $\Delta L=0.5$ cm ($L$ ke same units mein). (b) Relative $=\frac{0.5}{25.0}=0.02$ (dimensionless). (c) Percentage $=0.02\times 100\%=\boxed{2\%}$.

Mean (best estimate, KYUN: random scatter ka spread minimize karta hai): $t_{\text{mean}}=\frac{12.1+12.4+12.0+12.3+12.2}{5}=\frac{61.0}{5}=12.2\text{s}.$ Absolute errors $|t_{\text{mean}}-t_i|$: $0.1,0.2,0.2,0.1,0.0$. Mean absolute error: $\Delta t_{\text{mean}}=\frac{0.1+0.2+0.2+0.1+0.0}{5}=\frac{0.6}{5}=0.12\text{s}.$ Report (error ko 1 sig fig tak, value ko same decimal place tak): $\boxed{t=12.2\pm 0.1\text{s}}$.

Relative $=\frac{\Delta t_{\text{mean}}}{t_{\text{mean}}}=\frac{0.12}{12.2}=0.0098$. Percentage $=0.0098\times 100\%=\boxed{0.98\%\approx 1.0\%}$. KYUN yahan relative: yeh tumhe is timing ko directly compare karne deta hai, maano $3\%$ tak measure ki gayi mass ke saath — percentage units ko strip kar deta hai.

Mean $=\frac{0.42+0.44+0.43+0.43}{4}=\frac{1.72}{4}=0.43$ mm. Absolute errors: $0.01,0.01,0.00,0.00$ → mean $=\frac{0.02}{4}=0.005$ mm. Report: $\boxed{d=0.430\pm 0.005\text{mm}}$. Percentage $=\frac{0.005}{0.43}\times 100\%=\boxed{1.2\%}$.

(a) Constant empty-pan reading $0.15$ g har weighing ko same amount se upar push karta hai → systematic zero error. $\pm 0.02$ g ka scatter run ke baare mein random hai. (b) Uncorrected mean $=\frac{8.00+8.02+7.98}{3}=\frac{24.00}{3}=8.00$ g. Offset subtract karo: $8.00-0.15=7.85$ g. Absolute errors (constant shift se unchanged): $|8.00-8.00|=0.00,0.02,0.02$ → mean $=\frac{0.04}{3}=0.0133\approx 0.01$ g. Report: $\boxed{m=7.85\pm 0.01\text{g}}$. KYUN offset $\Delta m$ nahi badalta: har reading se same number subtract karna unhe sab ek saath slide karta hai — mean ke around unka spread unchanged rehta hai.

A ka mean $=\frac{50.31+50.29+50.30}{3}=50.30$ cm; readings tightly clustered → precise, lekin truth se $+0.30$ cm off → inaccurate (ek systematic bias). Yeh target figure ka left panel hai. B ka mean $=\frac{49.6+50.4+49.9+50.1}{4}=\frac{200.0}{4}=50.00$ cm; readings wider → less precise, lekin mean true value ke barabar → accurate. Yeh right panel hai. Averaging: B ki madad karta hai (random scatter truth tak cancel ho jaata hai) lekin A ki nahi (ek $+0.30$ bias kitne bhi repeats mein survive karta hai).

Same absolute error ka matlab same quality nahi — fairly judge karne ke liye size se divide karo. $P$: $\frac{0.1}{2.0}\times 100\%=5\%$. $Q$: $\frac{0.1}{200.0}\times 100\%=0.05\%$. $Q$ better hai; ratio $=\frac{5}{0.05}=\boxed{100\times }$ zyada precise jo cheez measure ki ja rahi hai uske fraction ke roop mein.

Area $A=L\times W=50.0{\text{cm}}^2$ (ek product → relative errors add hote hain). $\frac{\Delta A}{A}=\frac{\Delta L}{L}+\frac{\Delta W}{W}=\frac{0.1}{10.0}+\frac{0.1}{5.0}=0.01+0.02=0.03$. Percentage $=\boxed{3\%}$, aur $\Delta A=0.03\times 50.0=1.5{\text{cm}}^2$, toh $A=50.0\pm 1.5{\text{cm}}^2$. KYUN relatives add hote hain, absolutes nahi: product ke liye har factor result ko proportionally stretch karta hai, isliye har ek ka fractional wobble through feed hota hai.

$g$ mein $L$ power $1$ par aur $T$ power $2$ par (denominator mein) hai. $\pi$ exact hai, kuch contribute nahi karta. $\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T}=\frac{0.01}{1.00}+2\times \frac{0.02}{2.00}=0.01+0.02=\mathrm{0.03.}$ Percentage in $g=\boxed{3\%}$. KYUN $T$ par factor $2$ hai: $T$ squared appear karta hai, toh $T$ mein $1\%$ wobble $g$ mein $2\%$ wobble ban jaati hai. Squaring sensitivity ko double kar deta hai.

Value $=12.5-12.0=0.5$ cm. Difference ke liye absolute errors add hote hain: $\Delta =0.1+0.1=0.2$ cm. Percentage $=\frac{0.2}{0.5}\times 100\%=\boxed{40\%}$. Comment: do almost-equal numbers subtract karna dangerous hai — result chhota hota hai lekin errors pile up ho jaate hain, percentage error explode ho jaata hai. Jab ho sake "large near-equal quantities ka difference" avoid karo.

Diameter: mean $=\frac{2.00+2.02+1.98}{3}=2.00$ cm; abs errors $0.00,0.02,0.02$ → mean $=\frac{0.04}{3}=0.01\overline{3}$ cm, jo hum $1$ sig fig tak $\Delta d=0.01$ cm quote karte hain. $\frac{\Delta d}{d}=\frac{0.01}{2.00}=0.0050$ (aur $\Delta r/r=\Delta d/d$ kyunki $r=d/2$ top aur bottom same scale karta hai). Length: mean $=5.00$ cm; abs errors $0.00,0.02,0.02$ → mean $=\frac{0.04}{3}=0.01\overline{3}\to \Delta \ell =0.01$ cm; $\frac{\Delta \ell }{\ell }=\frac{0.01}{5.00}=0.0020$. Mass: $\frac{\Delta m}{m}=\frac{0.5}{50.0}=0.010$. Propagate $\rho =\frac{m}{\pi r^2\ell }$ → $r$ ki power $2$ hai, $\ell$ ki power $1$, $m$ ki power $1$: $\frac{\Delta \rho }{\rho }=\frac{\Delta m}{m}+2\frac{\Delta d}{d}+\frac{\Delta \ell }{\ell }=0.010+2(0.0050)+0.0020=\mathrm{0.0220.}$ Percentage $\approx \boxed{2.2\%}$. Central value: $r=1.00$ cm, $V=\pi (1.00{)}^2(5.00)=15.708{\text{cm}}^3$, $\rho =\frac{50.0}{15.708}=3.183{\text{g/cm}}^3$. $\Delta \rho =0.022\times 3.183=0.070$, toh $\rho =3.18\pm 0.07{\text{g/cm}}^3$. Rounding note: har mean absolute error propagation mein feed karne se pehle $1$ sig fig tak round ki gayi ($0.01$ cm) — L2 rule se match karta hai — shuruaat se ant tak consistent.

Har term ka contribution dekho $\frac{\Delta \rho }{\rho }=0.0220$ mein: mass $0.010$, diameter $2(0.0050)=0.0100$, length $0.0020$. Mass aur diameter $0.0100$ par tie karte hain — lekin diameter ka contribution uski power $2$ se inflate hota hai, toh $\Delta d$ ko half karna raw fractional error ki har unit per twice as much remove karta hai. $\Delta d$ ko half karne par uska contribution $0.0100$ se $0.0050$ ho jaata hai, jo deta hai $\frac{\Delta \rho }{\rho }=0.010+0.0050+0.0020=0.0170\approx 1.7\%.$ $\Delta m$ ko half karne par mass ka $0.010$ se $0.005$ ho jaata, jo same $0.0170$ yahan deta — lekin diameter woh hai jiska leverage uski power ke saath badhta hai, toh kisi bhi cylinder mein jahan $r$ ka fractional error tiny nahi hai, diameter woh term hai jise dekhna chahiye. Neeche contribution bars dekho.