Errors of Measurement Questions in English

(C) Density $\rho$ is given by the formula $\rho = \frac{m}{V}$.
For a product or quotient,the relative error is the sum of the relative errors of the individual measurements: $\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta V}{V}$.
Given: $m = 22.42 \; g$,$\Delta m = 0.01 \; g$,$V = 4.7 \; cc$,$\Delta V = 0.1 \; cc$.
Substituting the values:
$\frac{\Delta \rho}{\rho} = \frac{0.01}{22.42} + \frac{0.1}{4.7}$.
$\frac{\Delta \rho}{\rho} \approx 0.000446 + 0.021276 \approx 0.021722$.
To find the percentage error,multiply by $100 \%$:
Percentage error $= 0.021722 \times 100 \% \approx 2.17 \%$.
Rounding to the nearest significant value provided in the options,the maximum percentage error is approximately $2 \%$.

(D) The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta T}{T}$.
Here,the total error in time measurement is $\Delta T_{total} = \Delta T + 0.1 \, s$.
The time period $T$ is measured as $T = \frac{t}{n}$,where $t$ is the time for $n$ oscillations.
Thus,$\frac{\Delta T}{T} = \frac{\Delta T_{total}}{t} = \frac{\Delta T + 0.1}{n \times T_0}$,where $T_0 \approx 2 \, s$ for a $1 \, m$ pendulum.
Substituting values:
$(A)$ $\frac{\Delta g}{g} = \frac{5 \times 10^{-3}}{1} + 2 \left( \frac{0.2 + 0.1}{10 \times 2} \right) = 0.005 + 0.03 = 0.035$
$(B)$ $\frac{\Delta g}{g} = \frac{5 \times 10^{-3}}{1} + 2 \left( \frac{0.2 + 0.1}{20 \times 2} \right) = 0.005 + 0.015 = 0.020$
$(C)$ $\frac{\Delta g}{g} = \frac{5 \times 10^{-3}}{1} + 2 \left( \frac{0.1 + 0.1}{10 \times 2} \right) = 0.005 + 0.02 = 0.025$
$(D)$ $\frac{\Delta g}{g} = \frac{1 \times 10^{-3}}{1} + 2 \left( \frac{0.1 + 0.1}{50 \times 2} \right) = 0.001 + 0.004 = 0.005$
Since the relative error is minimum in case $(D)$,the measurement is most accurate.

(B) Given: Distance $s = (13.8 \pm 0.2) \text{ m}$ and time $t = (4.0 \pm 0.3) \text{ s}$.
Velocity $v = \frac{s}{t} = \frac{13.8}{4.0} = 3.45 \text{ m/s} \approx 3.5 \text{ m/s}$.
The relative error in velocity is given by $\frac{\Delta v}{v} = \frac{\Delta s}{s} + \frac{\Delta t}{t}$.
Substituting the values: $\frac{\Delta v}{v} = \frac{0.2}{13.8} + \frac{0.3}{4.0} = 0.0145 + 0.075 = 0.0895$.
Absolute error $\Delta v = v \times 0.0895 = 3.45 \times 0.0895 \approx 0.3087 \approx 0.31 \text{ m/s}$.
Thus,velocity $v = (3.5 \pm 0.31) \text{ m/s}$.
Percentage error = $\frac{\Delta v}{v} \times 100 = 0.0895 \times 100 = 8.95\% \approx 9\%$.

(C) The percentage error in the radius is given as $\frac{\Delta r}{r} \times 100 = 2\%$.
The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
Using the rules of propagation of errors for powers,the relative error in volume is $\frac{\Delta V}{V} = 3 \times \frac{\Delta r}{r}$.
To find the percentage error,multiply both sides by $100$:
$\frac{\Delta V}{V} \times 100 = 3 \times (\frac{\Delta r}{r} \times 100)$.
Substituting the given value:
$\frac{\Delta V}{V} \times 100 = 3 \times 2\% = 6\%$.

(B) For a body starting from rest,the distance $h$ covered in time $t$ is given by the equation of motion:
$h = \frac{1}{2}gt^2$
Rearranging for $g$,we get:
$g = \frac{2h}{t^2}$
Taking the natural logarithm on both sides:
$\ln g = \ln 2 + \ln h - 2\ln t$
Differentiating to find the relative error:
$\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2\frac{\Delta t}{t}$
For maximum permissible percentage error,we add the absolute values of the relative errors:
$\left( \frac{\Delta g}{g} \times 100 \right)_{\max} = \left( \frac{\Delta h}{h} \times 100 \right) + 2 \times \left( \frac{\Delta t}{t} \times 100 \right)$
Given that $\frac{\Delta h}{h} \times 100 = e_1$ and $\frac{\Delta t}{t} \times 100 = e_2$,the percentage error in $g$ is:
$\text{Percentage error in } g = e_1 + 2e_2$

(C) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
The relative error in the volume $V$ is given by $\frac{\Delta V}{V} = 3 \times \frac{\Delta r}{r}$.
Given the radius $r = 5.3 \; cm$ and the absolute error $\Delta r = 0.1 \; cm$.
The percentage error in volume is $\frac{\Delta V}{V} \times 100 = 3 \times \frac{\Delta r}{r} \times 100$.
Substituting the values,we get: $\left( 3 \times \frac{0.1}{5.3} \right) \times 100$.

(D) The pressure $P$ on a square plate of side $l$ is given by $P = \frac{F}{A} = \frac{F}{l^2}$.
Using the formula for the propagation of errors,the relative error in $P$ is given by $\frac{\Delta P}{P} = \frac{\Delta F}{F} + 2\frac{\Delta l}{l}$.
To find the maximum percentage error,we multiply by $100$:
$\left( \frac{\Delta P}{P} \times 100 \right)_{\max} = \left( \frac{\Delta F}{F} \times 100 \right) + 2 \times \left( \frac{\Delta l}{l} \times 100 \right)$.
Given that the percentage error in force $\frac{\Delta F}{F} \times 100 = 4\%$ and the percentage error in length $\frac{\Delta l}{l} \times 100 = 2\%$.
Substituting these values,we get:
$\left( \frac{\Delta P}{P} \times 100 \right)_{\max} = 4\% + 2 \times 2\% = 4\% + 4\% = 8\%$.

(C) The formula for acceleration due to gravity is $g = 4\pi^2 \left(\frac{l}{T^2}\right)$.
Taking the relative error,we get $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}$.
Given that the percentage error in length $\left(\frac{\Delta l}{l} \times 100\right) = 1\%$ and the percentage error in time period $\left(\frac{\Delta T}{T} \times 100\right) = 3\%$.
Since errors are always additive,the percentage error in $g$ is $\frac{\Delta g}{g} \times 100 = \left(\frac{\Delta l}{l} \times 100\right) + 2 \times \left(\frac{\Delta T}{T} \times 100\right)$.
Substituting the values: $\frac{\Delta g}{g} \times 100 = 1\% + 2(3\%) = 1\% + 6\% = 7\%$.
Therefore,the percentage error in the measurement of $g$ is $7\%$.

(A) Given: $I = (2.5 \pm 0.05) \ A$ and $V = (10 \pm 0.1) \ V$.
Resistance $R = \frac{V}{I} = \frac{10}{2.5} = 4 \ \Omega$.
The relative error in resistance is given by $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.
Substituting the values: $\frac{\Delta R}{4} = \frac{0.1}{10} + \frac{0.05}{2.5}$.
$\frac{\Delta R}{4} = 0.01 + 0.02 = 0.03$.
$\Delta R = 4 \times 0.03 = 0.12 \ \Omega$.
Therefore,the resistance is $(4 \pm 0.12) \ \Omega$.

(B) According to Ohm's law,the resistance $R$ is given by $R = \frac{V}{I}$.
For a quotient,the relative error is the sum of the relative errors of the individual quantities: $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.
To find the percentage error,we multiply by $100$:
$\frac{\Delta R}{R} \times 100 = \left( \frac{\Delta V}{V} \times 100 \right) + \left( \frac{\Delta I}{I} \times 100 \right)$.
Given that the percentage error in voltage $\frac{\Delta V}{V} \times 100 = 3\%$ and the percentage error in current $\frac{\Delta I}{I} \times 100 = 3\%$.
Therefore,the percentage error in resistance is $3\% + 3\% = 6\%$.

(A) The formula for the period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$.
Squaring both sides,we get $T^2 = 4\pi^2 \frac{l}{g}$,which implies $g = 4\pi^2 \frac{l}{T^2}$.
The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2 \frac{\Delta T}{T}$.
Given values are $l = 20.0 \text{ cm}$,$\Delta l = 1 \text{ mm} = 0.1 \text{ cm}$.
Total time for $100$ oscillations is $t = 90 \text{ s}$ with a resolution $\Delta t = 1 \text{ s}$.
The period $T = \frac{t}{100} = 0.9 \text{ s}$,and the error in period $\Delta T = \frac{\Delta t}{100} = \frac{1}{100} = 0.01 \text{ s}$.
Now,the percentage error is $\frac{\Delta g}{g} \times 100 = \left( \frac{\Delta l}{l} \times 100 \right) + 2 \left( \frac{\Delta T}{T} \times 100 \right)$.
Substituting the values: $\frac{\Delta g}{g} \times 100 = \left( \frac{0.1}{20.0} \times 100 \right) + 2 \left( \frac{0.01}{0.9} \times 100 \right)$.
$= 0.5\% + 2 \times 1.11\% = 0.5\% + 2.22\% = 2.72\%$.
Rounding to the nearest integer,we get $3\%$.

(A) Step $1$: Calculate the mean time period $(T_{mean})$:
$T_{mean} = \frac{90 + 91 + 95 + 92}{4} = \frac{368}{4} = 92\;s$.
Step $2$: Calculate the absolute errors for each measurement:
$|\Delta T_1| = |92 - 90| = 2\;s$
$|\Delta T_2| = |92 - 91| = 1\;s$
$|\Delta T_3| = |92 - 95| = 3\;s$
$|\Delta T_4| = |92 - 92| = 0\;s$
Step $3$: Calculate the mean absolute error $(\Delta T_{mean})$:
$\Delta T_{mean} = \frac{2 + 1 + 3 + 0}{4} = \frac{6}{4} = 1.5\;s$.
Step $4$: Rounding off based on the least count:
The least count of the clock is $1\;s$. Therefore,the mean absolute error should be rounded to the nearest integer,which is $2\;s$.
Thus,the reported mean time is $92 \pm 2\;s$.

(B) Density $(d)$ is given by the formula: $d = \frac{M}{V} = \frac{M}{L^3}$,where $M$ is the mass and $L$ is the length of the side of the cube.
Using the propagation of errors formula,the relative error in density is given by: $\frac{\Delta d}{d} = \frac{\Delta M}{M} + 3 \frac{\Delta L}{L}$.
Given that the relative error in mass $\frac{\Delta M}{M} = 1.5\%$ and the relative error in length $\frac{\Delta L}{L} = 1\%$.
Substituting these values into the equation: $\frac{\Delta d}{d} = 1.5\% + 3(1\%) = 1.5\% + 3\% = 4.5\%$.
Therefore,the maximum error in determining the density is $4.5\%$.

(D) The given current-voltage relation is $I = (e^{1000V/T} - 1) \text{ mA}$.
Given $I = 5 \text{ mA}$,we have $5 = e^{1000V/T} - 1$,which implies $e^{1000V/T} = 6$.
To find the error in current $(dI)$,we differentiate the expression with respect to $V$:
$dI = \frac{d}{dV} (e^{1000V/T} - 1) \cdot dV = \left( e^{1000V/T} \cdot \frac{1000}{T} \right) dV$.
Given $dV = 0.01 \text{ V}$ and $T = 300 \text{ K}$,substitute the values:
$dI = (6) \cdot \left( \frac{1000}{300} \right) \cdot (0.01)$.
$dI = 6 \cdot \left( \frac{10}{3} \right) \cdot 0.01 = 2 \cdot 10 \cdot 0.01 = 0.2 \text{ mA}$.
Thus,the error in the value of current is $0.2 \text{ mA}$.

(C) The given relation is $A = \frac{p^{1/2} q^{1/2}}{r^2 s^3}$.
The relative error in $A$ is given by the formula:
$\frac{\Delta A}{A} = \frac{1}{2} \frac{\Delta p}{p} + \frac{1}{2} \frac{\Delta q}{q} + 2 \frac{\Delta r}{r} + 3 \frac{\Delta s}{s}$.
Given percentage errors are:
$\frac{\Delta p}{p} \times 100 = 1\%$
$\frac{\Delta q}{q} \times 100 = 3\%$
$\frac{\Delta r}{r} \times 100 = 0.5\%$
$\frac{\Delta s}{s} \times 100 = 0.33\%$
Substituting these values into the expression for percentage error in $A$:
$\frac{\Delta A}{A} \times 100 = \frac{1}{2}(1\%) + \frac{1}{2}(3\%) + 2(0.5\%) + 3(0.33\%)$
$= 0.5\% + 1.5\% + 1.0\% + 0.99\%$
$= 3.99\% \approx 4\%$.

(A) The formula for resistivity is $\rho = \frac{RA}{\ell} = \frac{R \pi (d/2)^2}{\ell} = \frac{\pi R d^2}{4 \ell}$.
Taking the relative error,we have $\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta \ell}{\ell}$.
Given values: $R = 1.05, \Delta R = 0.01$; $d = 0.60, \Delta d = 0.01$; $\ell = 75.3, \Delta \ell = 0.1$.
Substituting the values: $\frac{\Delta \rho}{\rho} = \frac{0.01}{1.05} + 2 \times \frac{0.01}{0.60} + \frac{0.1}{75.3}$.
$\frac{\Delta \rho}{\rho} \approx 0.0095 + 0.0333 + 0.0013 = 0.0441$.
Rounding to the nearest provided option,the relative error is approximately $0.04$.

(B) Given the expression $Q = \frac{X^n}{Y^m}$.
Taking the natural logarithm on both sides: $\ln Q = n \ln X - m \ln Y$.
Differentiating both sides,we get: $\frac{dQ}{Q} = n \frac{dX}{X} - m \frac{dY}{Y}$.
For errors,we consider the maximum relative error,which is the sum of the absolute values of the individual relative errors:
$\frac{\Delta Q}{Q} = \pm \left( n \frac{\Delta X}{X} + m \frac{\Delta Y}{Y} \right)$.
Multiplying both sides by $Q$,we get the absolute error in $Q$:
$\Delta Q = \pm \left( n \frac{\Delta X}{X} + m \frac{\Delta Y}{Y} \right) Q$.

(A) The formula for the range $R$ of a projectile is $R = \frac{U^2 \sin(2\theta)}{g}$.
Given $U = 20 \ m/s$,$\theta = 60^{\circ}$,and $g \approx 10 \ m/s^2$.
$R = \frac{20^2 \times \sin(120^{\circ})}{10} = \frac{400 \times \frac{\sqrt{3}}{2}}{10} = 20\sqrt{3} \approx 34.64 \ m$.
The relative error in $U$ is $\frac{\Delta U}{U} = 5\% = 0.05$.
Since $R \propto U^2$,the relative error in range is $\frac{\Delta R}{R} = 2 \times \frac{\Delta U}{U} = 2 \times 0.05 = 0.10$.
Thus,$\Delta R = 0.10 \times R = 0.10 \times 34.64 = 3.464 \ m$.
The possible range lies in the interval $(R - \Delta R, R + \Delta R) = (34.64 - 3.46, 34.64 + 3.46) = (31.18 \ m, 38.10 \ m)$.
Comparing the options,$39.0 \ m$ lies outside this range.

(C) The given relation is $x = \frac{a^2 b^3}{c \sqrt{d}}$.
The percentage error in $x$ is given by the formula:
$\frac{\Delta x}{x} \times 100 = 2 \left( \frac{\Delta a}{a} \times 100 \right) + 3 \left( \frac{\Delta b}{b} \times 100 \right) + 1 \left( \frac{\Delta c}{c} \times 100 \right) + \frac{1}{2} \left( \frac{\Delta d}{d} \times 100 \right)$.
Given percentage errors are:
$\frac{\Delta a}{a} \times 100 = 2\%$,$\frac{\Delta b}{b} \times 100 = 1\%$,$\frac{\Delta c}{c} \times 100 = 3\%$,and $\frac{\Delta d}{d} \times 100 = 4\%$.
Substituting these values into the formula:
$\frac{\Delta x}{x} \times 100 = 2(2\%) + 3(1\%) + 1(3\%) + \frac{1}{2}(4\%)$.
$\frac{\Delta x}{x} \times 100 = 4\% + 3\% + 3\% + 2\% = 12\%$.
Therefore,the percentage error in $x$ is $\pm 12\%$.

(A) Given the relation $m = \pi \tan \theta$.
Taking the differential of both sides,we get $dm = \pi \sec^2 \theta \, d\theta$.
The relative error is given by $\frac{dm}{m} = \frac{\pi \sec^2 \theta \, d\theta}{\pi \tan \theta}$.
Simplifying the expression: $\frac{dm}{m} = \frac{\sec^2 \theta}{\tan \theta} d\theta = \frac{1}{\cos^2 \theta} \cdot \frac{\cos \theta}{\sin \theta} d\theta = \frac{1}{\sin \theta \cos \theta} d\theta$.
Multiplying the numerator and denominator by $2$,we get $\frac{dm}{m} = \frac{2 d\theta}{2 \sin \theta \cos \theta} = \frac{2 d\theta}{\sin 2\theta}$.
For the percentage error to be minimum,the denominator $\sin 2\theta$ must be maximum.
The maximum value of $\sin 2\theta$ is $1$,which occurs when $2\theta = 90^\circ$.
Therefore,$\theta = 45^\circ$.

(A) Given $Z = x^2y - xy^2$,$x = 3.0 \pm 0.1$,and $y = 2.0 \pm 0.1$.
First,calculate the nominal value of $Z$:
$Z = (3.0)^2(2.0) - (3.0)(2.0)^2 = (9.0)(2.0) - (3.0)(4.0) = 18.0 - 12.0 = 6.0$.
Next,calculate the absolute error $\delta Z$ using the partial differentiation method:
$\delta Z = \left| \frac{\partial Z}{\partial x} \right| \delta x + \left| \frac{\partial Z}{\partial y} \right| \delta y$.
$\frac{\partial Z}{\partial x} = 2xy - y^2 = 2(3.0)(2.0) - (2.0)^2 = 12.0 - 4.0 = 8.0$.
$\frac{\partial Z}{\partial y} = x^2 - 2xy = (3.0)^2 - 2(3.0)(2.0) = 9.0 - 12.0 = -3.0$.
Now,substitute the values into the error formula:
$\delta Z = |8.0| \times 0.1 + |-3.0| \times 0.1 = 0.8 + 0.3 = 1.1$.
Therefore,the value of $Z$ is $6.0 \pm 1.1$.

(A) The resistance of a rod is given by $R = \frac{\rho \ell}{A} = \frac{\rho \ell}{\pi r^2} = \frac{4 \rho \ell}{\pi D^2}$,where $\ell$ is length and $D$ is diameter.
Taking the relative error: $\frac{\Delta R}{R} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta D}{D}$ (since $\rho$ is exact,$\Delta \rho = 0$).
Given: $\ell = 20 \ cm = 200 \ mm$,$\Delta \ell = 1 \ mm$.
Diameter $D = 4 \ mm$. The least count of the screw gauge is $LC = \frac{\text{pitch}}{\text{number of divisions}} = \frac{1 \ mm}{50} = 0.02 \ mm$. Thus,$\Delta D = 0.02 \ mm$.
Percentage error = $\left( \frac{\Delta \ell}{\ell} + 2 \frac{\Delta D}{D} \right) \times 100\%$.
Percentage error = $\left( \frac{1}{200} + 2 \times \frac{0.02}{4} \right) \times 100\% = (0.005 + 0.01) \times 100\% = 0.015 \times 100\% = 1.5\%$.

(C) Given the relation $x = a - b$.
The absolute error in $x$,denoted by $\Delta x$,is given by the sum of the absolute errors in $a$ and $b$ because errors always add up in subtraction as well.
Therefore,$\Delta x = \Delta a + \Delta b$.
The percentage error is defined as $\frac{\Delta x}{x} \times 100\%$.
Substituting the values of $\Delta x$ and $x$,we get:
Percentage error $= \left( \frac{\Delta a + \Delta b}{a - b} \right) \times 100\% = \left( \frac{\Delta a}{a - b} + \frac{\Delta b}{a - b} \right) \times 100\%$.

(B) The formula for kinetic energy is $E = \frac{1}{2}mv^2$.
The relative error in kinetic energy is given by $\frac{\Delta E}{E} = \frac{\Delta m}{m} + 2 \frac{\Delta v}{v}$.
To find the percentage error,multiply by $100$:
$\left( \frac{\Delta E}{E} \times 100 \right) = \left( \frac{\Delta m}{m} \times 100 \right) + 2 \times \left( \frac{\Delta v}{v} \times 100 \right)$.
Given that the percentage error in mass is $3\%$ and the percentage error in speed is $2\%$,we substitute these values:
$\text{Percentage error in } E = 3\% + 2 \times (2\%) = 3\% + 4\% = 7\%$.
Therefore,the error in kinetic energy is $7\%$.

(B) The moment of inertia of a disc about an axis tangent to it is given by the parallel axis theorem: $I = I_{cm} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
However,for a disc,the moment of inertia about a tangent in its plane is $I = \frac{5}{4}MR^2$.
Using the relative error formula for $I = k M^a R^b$,we have $\frac{\Delta I}{I} = a \frac{\Delta M}{M} + b \frac{\Delta R}{R}$.
Here,$a = 1$ and $b = 2$.
Given $\frac{\Delta M}{M} = 1\%$ and $\frac{\Delta R}{R} = 1.5\%$.
Therefore,$\frac{\Delta I}{I} \times 100 = (1 \times 1\%) + (2 \times 1.5\%) = 1\% + 3\% = 4\%$.

(A) Given the relation $P = \frac{A^3}{B^{5/2}}$.
Taking the natural logarithm on both sides,we get $\ln P = 3 \ln A - \frac{5}{2} \ln B$.
Differentiating both sides,we obtain $\frac{dP}{P} = 3 \frac{dA}{A} - \frac{5}{2} \frac{dB}{B}$.
For maximum relative error,we consider the absolute values of the terms,so $\frac{\Delta P}{P} = \pm \left( 3 \frac{\Delta A}{A} + \frac{5}{2} \frac{\Delta B}{B} \right)$.
Multiplying both sides by $P$,we get the absolute error $\Delta P = \pm \left( 3 \frac{\Delta A}{A} + \frac{5}{2} \frac{\Delta B}{B} \right) P$.

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
Taking the natural logarithm on both sides,we get $\ln T = \ln(2\pi) + \frac{1}{2} \ln l - \frac{1}{2} \ln g$.
Differentiating to find the relative error,we have $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l} + \frac{1}{2} \frac{\Delta g}{g}$.
Given that the percentage error in $l$ is $\frac{\Delta l}{l} \times 100 = 2\%$ and in $g$ is $\frac{\Delta g}{g} \times 100 = 4\%$.
Substituting these values,the percentage error in $T$ is $\frac{\Delta T}{T} \times 100 = \frac{1}{2} (2\%) + \frac{1}{2} (4\%) = 1\% + 2\% = 3\%$.

(C) $1$. Accuracy refers to how close the average of the measured values is to the true value $(T)$.
$2$. Precision refers to the consistency or repeatability of the measurements,indicated by the spread or width of the distribution curve.
$3$. $A$ narrow peak indicates high precision (low variation),while a broad,wide peak indicates low precision (high variation or imprecise).
$4$. In Figure $A$,the peak is narrow (precise) and centered at $T$ (accurate).
$5$. In Figure $B$,the peak is narrow (precise) but shifted away from $T$ (inaccurate).
$6$. In Figure $C$,the peak is broad (imprecise) and centered at $T$ (accurate).
$7$. In Figure $D$,the peak is broad (imprecise) and shifted away from $T$ (inaccurate).
$8$. Therefore,Figure $C$ represents a measurement that is imprecise (broad peak) but accurate (centered at $T$).

(D) The equivalent resistance $R_{eq}$ for two resistors in parallel is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$.
First,calculate $R_{eq}$: $R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2 \Omega$.
Differentiating the expression $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$,we get $-\frac{dR_{eq}}{R_{eq}^2} = -\frac{dR_1}{R_1^2} - \frac{dR_2}{R_2^2}$,which simplifies to $\frac{dR_{eq}}{R_{eq}^2} = \frac{dR_1}{R_1^2} + \frac{dR_2}{R_2^2}$.
Multiplying by $R_{eq}$,we get $\frac{dR_{eq}}{R_{eq}} = \frac{R_{eq}}{R_1} \left( \frac{dR_1}{R_1} \right) + \frac{R_{eq}}{R_2} \left( \frac{dR_2}{R_2} \right)$.
Given $\frac{dR_1}{R_1} = 1\%$ and $\frac{dR_2}{R_2} = 2\%$,we substitute the values:
$\frac{dR_{eq}}{R_{eq}} = \frac{2}{3} (1\%) + \frac{2}{6} (2\%) = \frac{2}{3}\% + \frac{2}{3}\% = \frac{4}{3}\%$.
$\frac{dR_{eq}}{R_{eq}} = 1.33\%$.

(B) For resistors connected in series,the equivalent resistance $R_s$ is the sum of individual resistances:
$R_s = R_1 + R_2 = 300 \,\Omega + 500 \,\Omega = 800 \,\Omega$
The error in the sum of two quantities is the sum of their absolute errors:
$\Delta R_s = \Delta R_1 + \Delta R_2 = 3 \,\Omega + 4 \,\Omega = 7 \,\Omega$
Therefore,the equivalent resistance with its uncertainty is expressed as:
$R_s \pm \Delta R_s = (800 \pm 7) \,\Omega$

(B) Given: Distance $s = (13.8 \pm 0.2) \ m$ and time $t = (4 \pm 0.3) \ s$.
Velocity $v = \frac{s}{t} = \frac{13.8}{4} = 3.45 \ ms^{-1}$. Rounding to one decimal place,$v = 3.5 \ ms^{-1}$.
The relative error in velocity is given by $\frac{\Delta v}{v} = \frac{\Delta s}{s} + \frac{\Delta t}{t}$.
Substituting the values: $\frac{\Delta v}{3.45} = \frac{0.2}{13.8} + \frac{0.3}{4}$.
$\frac{\Delta v}{3.45} = 0.0145 + 0.075 = 0.0895$.
$\Delta v = 0.0895 \times 3.45 \approx 0.3087 \ ms^{-1}$.
Rounding to one significant figure,$\Delta v \approx 0.3 \ ms^{-1}$.
Thus,the velocity is $(3.5 \pm 0.3) \ ms^{-1}$.

(A) The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$.
Squaring both sides and rearranging for $g$,we get $g = \frac{4\pi^2 l}{T^2}$.
The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}$.
Given: $l = 100 \, cm$,$\Delta l = 0.1 \, cm$,$T_{total} = 50 \, s$,$\Delta T_{total} = 0.1 \, s$.
Note that $T$ is the time period of one vibration,$T = \frac{T_{total}}{25} = \frac{50}{25} = 2 \, s$.
The error in $T$ is $\Delta T = \frac{\Delta T_{total}}{25} = \frac{0.1}{25} = 0.004 \, s$.
Substituting these values: $\frac{\Delta g}{g} = \frac{0.1}{100} + 2 \left( \frac{0.004}{2} \right) = 0.001 + 0.004 = 0.005$.
Percentage error = $0.005 \times 100 = 0.5 \%$.

(C) The surface area of a sphere is given by $S = 4\pi r^2$. The relative error in the surface area is $\frac{\Delta S}{S} = 2 \frac{\Delta r}{r} = \alpha$.
From this,the relative error in the radius is $\frac{\Delta r}{r} = \frac{\alpha}{2}$.
The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$. The relative error in the volume is $\frac{\Delta V}{V} = 3 \frac{\Delta r}{r}$.
Substituting the value of $\frac{\Delta r}{r}$,we get $\frac{\Delta V}{V} = 3 \times \frac{\alpha}{2} = \frac{3}{2}\alpha$.

(D) Given the formula $A = \frac{P^3 Q^2}{R^{1/2} S}$.
To find the maximum percentage error in $A$,we use the formula for propagation of errors:
$\frac{\Delta A}{A} \times 100 = 3 \left( \frac{\Delta P}{P} \times 100 \right) + 2 \left( \frac{\Delta Q}{Q} \times 100 \right) + \frac{1}{2} \left( \frac{\Delta R}{R} \times 100 \right) + 1 \left( \frac{\Delta S}{S} \times 100 \right)$.
Substituting the given percentage errors:
$\text{Max } \% \text{ error in } A = 3(0.5\%) + 2(1\%) + 0.5(3\%) + 1(1.5\%)$.
$= 1.5\% + 2.0\% + 1.5\% + 1.5\%$.
$= 6.5\%$.

(C) Given the relation: $p = a^{1/2} b^2 c^3 d^{-4}$.
The formula for the maximum relative error is given by:
$\frac{\Delta p}{p} = \frac{1}{2} \frac{\Delta a}{a} + 2 \frac{\Delta b}{b} + 3 \frac{\Delta c}{c} + 4 \frac{\Delta d}{d}$.
Substituting the given percentage errors:
$\frac{\Delta p}{p} \times 100 = \frac{1}{2}(2\%) + 2(1\%) + 3(3\%) + 4(5\%)$.
Calculating the values:
$= 1\% + 2\% + 9\% + 20\% = 32\%$.
Therefore,the relative error in $p$ is $32\%$.

(B) The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$.
Squaring both sides,we get $T^2 = 4\pi^2 \frac{L}{g}$,which implies $g = 4\pi^2 \frac{L}{T^2}$.
The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$.
Given: $L = 20.0$ cm,$\Delta L = 1$ mm = $0.1$ cm.
$T_{total} = 90.0$ s,$\Delta T_{total} = 1$ s. Since $T = \frac{T_{total}}{100}$,$\Delta T = \frac{\Delta T_{total}}{100} = \frac{1}{100} = 0.01$ s.
Substituting the values:
$\frac{\Delta g}{g} = \frac{0.1}{20.0} + 2 \times \frac{1}{90} = 0.005 + 0.0222 = 0.0272$.
Percentage error = $0.0272 \times 100 = 2.72 \% \approx 2.7 \%$.

(C) The formula for heat produced in an electric circuit is given by $H = I^2Rt.$
To find the maximum relative error in $H,$ we use the formula for propagation of errors:
$\frac{\Delta H}{H} = 2\left(\frac{\Delta I}{I}\right) + \frac{\Delta R}{R} + \frac{\Delta t}{t}.$
Given percentage errors are $\frac{\Delta I}{I} \times 100 = 2\%,$ $\frac{\Delta R}{R} \times 100 = 1\%,$ and $\frac{\Delta t}{t} \times 100 = 1\%.$
Substituting these values into the error equation:
$\frac{\Delta H}{H} \times 100 = 2(2\%) + 1\% + 1\% = 4\% + 1\% + 1\% = 6\%.$
Thus,the maximum possible error in the total heat produced is $6\%.$

(C) The formula for acceleration due to gravity is $g = \frac{4\pi^2 L}{T^2}$.
Taking the relative error, we have $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}$.
Given:
$L = 55.0\,cm$, $\Delta L = 1\,mm = 0.1\,cm$.
Time for $20$ oscillations is $t = 30\,s$, so $T = \frac{30}{20} = 1.5\,s$.
The least count of the watch is $1\,s$, so the error in total time is $\Delta t = 1\,s$. Thus, $\Delta T = \frac{\Delta t}{20} = \frac{1}{20} = 0.05\,s$.
Now, calculate the percentage error:
$\frac{\Delta g}{g} \times 100 = \left( \frac{\Delta L}{L} + 2\frac{\Delta T}{T} \right) \times 100\%$
$= \left( \frac{0.1}{55.0} + 2 \times \frac{0.05}{1.5} \right) \times 100\%$
$= \left( 0.001818 + 0.06666 \right) \times 100\%$
$= 0.06848 \times 100\% \approx 6.8\%$.

(D) The density $\rho$ of a cube is given by $\rho = \frac{m}{L^3}$,where $m$ is the mass and $L$ is the edge length.
Given: $m = 10.00 \, kg$,$\Delta m = 0.10 \, kg$,$L = 0.10 \, m$,$\Delta L = 0.01 \, m$.
The measured density is $\rho = \frac{10.00}{(0.10)^3} = \frac{10.00}{0.001} = 10000 \, kg/m^3$.
Since the relative error in $L$ is $\frac{\Delta L}{L} = \frac{0.01}{0.10} = 0.1$,which is not small,we calculate the maximum and minimum possible density values.
$\rho_{max} = \frac{m + \Delta m}{(L - \Delta L)^3} = \frac{10.10}{(0.09)^3} = \frac{10.10}{0.000729} \approx 13854.6 \, kg/m^3$.
$\rho_{min} = \frac{m - \Delta m}{(L + \Delta L)^3} = \frac{9.90}{(0.11)^3} = \frac{9.90}{0.001331} \approx 7438.0 \, kg/m^3$.
The absolute error in density is $\Delta \rho = \frac{\rho_{max} - \rho_{min}}{2} = \frac{13854.6 - 7438.0}{2} = \frac{6416.6}{2} = 3208.3 \, kg/m^3$.
Since this value is not among the given options,the correct choice is $D$.

(D) The density $\rho$ of a cube is given by the formula $\rho = \frac{M}{V} = \frac{M}{L^3}$,where $M$ is the mass and $L$ is the length of the side of the cube.
Using the formula for relative error,the maximum percentage error in density is given by:
$\frac{\Delta \rho}{\rho} \times 100 = \frac{\Delta M}{M} \times 100 + 3 \times \left( \frac{\Delta L}{L} \times 100 \right)$
Given that the maximum error in mass $\frac{\Delta M}{M} \times 100 = 3\%$ and the maximum error in length $\frac{\Delta L}{L} \times 100 = 2\%$.
Substituting these values into the equation:
$\text{Percentage error in } \rho = 3\% + 3 \times (2\%) = 3\% + 6\% = 9\%$.
Therefore,the maximum error in the measurement of the density of the cube is $9\%$.

(A) Given temperatures are $t_1 = 20^{\circ}C \pm 0.5^{\circ}C$ and $t_2 = 50^{\circ}C \pm 0.5^{\circ}C$.
The temperature difference is calculated as $\Delta t = t_2 - t_1 = 50^{\circ}C - 20^{\circ}C = 30^{\circ}C$.
When subtracting two quantities,the absolute errors are added.
The error in the difference is $\Delta t = \Delta t_1 + \Delta t_2 = 0.5^{\circ}C + 0.5^{\circ}C = 1^{\circ}C$.
Thus,the temperature difference is $30^{\circ}C \pm 1^{\circ}C$.

(D) Given the physical quantity $A = P^2/Q^3$.
To find the maximum percentage error,we use the formula for relative error in powers:
$\frac{\Delta A}{A} = 2 \left( \frac{\Delta P}{P} \right) + 3 \left( \frac{\Delta Q}{Q} \right)$.
Multiplying by $100$ to convert to percentage error:
$\left( \frac{\Delta A}{A} \times 100 \right) = 2 \left( \frac{\Delta P}{P} \times 100 \right) + 3 \left( \frac{\Delta Q}{Q} \times 100 \right)$.
Given that the percentage error in $P$ is $x$ and in $Q$ is $y$,we substitute these values:
$\text{Percentage error in } A = 2x + 3y$.

(A) The volume of a sphere is given by the formula $V = \frac{4}{3}\pi r^3$.
Taking the logarithmic derivative,we get $\frac{\Delta V}{V} = 3 \frac{\Delta r}{r}$.
Given that the percentage error in volume is $\frac{\Delta V}{V} \times 100 = 6\%$.
Substituting this into the equation: $6\% = 3 \times (\frac{\Delta r}{r} \times 100)$.
Therefore,the percentage error in radius is $\frac{\Delta r}{r} \times 100 = \frac{6\%}{3} = 2\%$.

(B) According to Mayer's relation,the gas constant $R$ is given by $R = C_P - C_V$.
Substituting the given values: $R = 12.28 - 3.97 = 8.31 \text{ units}$.
For the error in $R$,we use the rule for the subtraction of physical quantities: $\Delta R = \Delta C_P + \Delta C_V$.
Substituting the error values: $\Delta R = 0.2 + 0.3 = 0.5 \text{ units}$.
Therefore,the value of the gas constant is $R = (8.31 \pm 0.5) \text{ units}$.

(B) Given the expression $Q = \frac{X^n}{Y^m}$.
Taking the natural logarithm on both sides: $\ln Q = n \ln X - m \ln Y$.
Differentiating both sides,we get the expression for relative error: $\frac{dQ}{Q} = n \frac{dX}{X} - m \frac{dY}{Y}$.
For maximum absolute error,we consider the sum of the magnitudes of the relative errors: $\frac{\Delta Q}{Q} = \pm \left( n \frac{\Delta X}{X} + m \frac{\Delta Y}{Y} \right)$.
Therefore,the absolute error $\Delta Q$ is given by: $\Delta Q = \pm \left( n \frac{\Delta X}{X} + m \frac{\Delta Y}{Y} \right) Q$.

(A) Given: Mass $M = 10.000\,g$,Volume $V = 10.00\,cm^3$.
The uncertainty in mass $\Delta M = 0.001\,g$ and in volume $\Delta V = 0.01\,cm^3$.
Density $\rho = \frac{M}{V} = \frac{10.000}{10.00} = 1.000\,g\,cm^{-3}$.
The relative error in density is given by $\frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} + \frac{\Delta V}{V}$.
Substituting the values: $\frac{\Delta \rho}{\rho} = \frac{0.001}{10.000} + \frac{0.01}{10.00} = 0.0001 + 0.001 = 0.0011$.
Therefore,the absolute error $\Delta \rho = 0.0011 \times \rho = 0.0011 \times 1.000 = 0.0011\,g\,cm^{-3}$.

(C) Density $\rho$ is defined as $\rho = \frac{m}{V}$.
For a quotient,the relative error is the sum of the relative errors of the individual quantities: $\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta V}{V}$.
Given: $m = 5.00 \ kg$,$\Delta m = 0.05 \ kg$,$V = 1.00 \ m^3$,$\Delta V = 0.05 \ m^3$.
The percentage error in density is given by: $\frac{\Delta \rho}{\rho} \times 100 = \left( \frac{\Delta m}{m} + \frac{\Delta V}{V} \right) \times 100$.
Substituting the values: $\left( \frac{0.05}{5.00} + \frac{0.05}{1.00} \right) \times 100 = (0.01 + 0.05) \times 100 = 0.06 \times 100 = 6 \%$.
Therefore,the maximum possible percentage error in density is $6 \%$.

(D) The formula for the time period of a simple pendulum is $T = 2 \pi \sqrt{\frac{l}{g}}$.
Squaring both sides,we get $T^2 = 4 \pi^2 \frac{l}{g}$.
Rearranging for $g$,we get $g = 4 \pi^2 \frac{l}{T^2}$.
The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2 \frac{\Delta T}{T}$.
To find the percentage error,multiply by $100$:
$\frac{\Delta g}{g} \times 100 = \left( \frac{\Delta l}{l} \times 100 \right) + 2 \left( \frac{\Delta T}{T} \times 100 \right)$.
Given $\frac{\Delta l}{l} \times 100 = 2 \%$ and $\frac{\Delta T}{T} \times 100 = 2 \%$.
Substituting these values: $\frac{\Delta g}{g} \times 100 = 2 \% + 2(2 \%) = 2 \% + 4 \% = 6 \%$.

(C) The given physical quantity is $X = \frac{2k^3l^2}{m\sqrt{n}}$.
The relative error in $X$ is given by the formula: $\frac{\Delta X}{X} = 3\frac{\Delta k}{k} + 2\frac{\Delta l}{l} + \frac{\Delta m}{m} + \frac{1}{2}\frac{\Delta n}{n}$.
Given percentage errors are $\frac{\Delta k}{k} \times 100 = 1\%$,$\frac{\Delta l}{l} \times 100 = 2\%$,$\frac{\Delta m}{m} \times 100 = 3\%$,and $\frac{\Delta n}{n} \times 100 = 4\%$.
Substituting these values into the expression for percentage error:
$\frac{\Delta X}{X} \times 100 = 3(1\%) + 2(2\%) + 3\% + \frac{1}{2}(4\%)$.
Calculating the result: $3 + 4 + 3 + 2 = 12\%$.
Therefore,the value of $X$ is uncertain by $12\%$.

(D) Let the length of one rod be $L = (11.05 \pm 0.05) \ cm$.
For two rods,the total length $L_{total} = L + L = 2 \times (11.05 \pm 0.05) \ cm$.
The mean value is $2 \times 11.05 = 22.10 \ cm$.
The absolute error adds up: $\Delta L_{total} = 0.05 + 0.05 = 0.10 \ cm$.
According to the rules of significant figures in addition/subtraction,the result should have the same number of decimal places as the measurement with the least number of decimal places.
Here,$11.05$ has two decimal places,so the result $22.10$ is appropriate.
Thus,the total length is $(22.10 \pm 0.10) \ cm$.