Errors of Measurement Questions in English

(C) Given the formula for the time period: $T = 2\pi \sqrt{\frac{l}{g}}$.
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 $l = 100 \, cm = 1000 \, mm$ and $\Delta l = 1 \, mm$,the percentage error in $l$ is: $\frac{\Delta l}{l} \times 100 = \frac{1}{1000} \times 100 = 0.1 \%$.
For $100$ oscillations,the total time $t = 100T = 200 \, s$. The least count of the stopwatch is $\Delta t = 0.1 \, s$.
The error in the period $T$ is $\Delta T = \frac{\Delta t}{100} = \frac{0.1}{100} = 0.001 \, s$.
The percentage error in $T$ is: $\frac{\Delta T}{T} \times 100 = \frac{0.001}{2} \times 100 = 0.05 \%$.
Therefore,the percentage error in $g$ is: $\frac{\Delta g}{g} \times 100 = (0.1 \%) + 2 \times (0.05 \%) = 0.1 \% + 0.1 \% = 0.2 \%$.

(B) The formula for kinetic energy is $K.E. = \frac{1}{2}mv^2$.
The relative error in kinetic energy is given by $\frac{\Delta K.E.}{K.E.} = \frac{\Delta m}{m} + 2 \frac{\Delta v}{v}$.
To find the percentage error,we multiply by $100$:
$\text{Percentage error in } K.E. = (\% \text{ error in } m) + 2 \times (\% \text{ error in } v)$.
Given that the percentage error in mass is $2\%$ and the percentage error in speed is $3\%$,we substitute these values:
$\text{Percentage error in } K.E. = 2\% + 2 \times 3\% = 2\% + 6\% = 8\%$.
Therefore,the maximum error in the estimation of kinetic energy is $8\%$.

(D) The random error in the arithmetic mean of $n$ observations is given by $\Delta \bar{a} = \frac{\Delta a}{\sqrt{n}}$,where $\Delta a$ is the error in a single observation.
For $n_1 = 100$ observations,the random error is $x = \frac{\Delta a}{\sqrt{100}} = \frac{\Delta a}{10}$.
For $n_2 = 400$ observations,the random error is $x' = \frac{\Delta a}{\sqrt{400}} = \frac{\Delta a}{20}$.
Dividing the two expressions: $\frac{x'}{x} = \frac{\Delta a / 20}{\Delta a / 10} = \frac{10}{20} = \frac{1}{2}$.
Therefore,$x' = \frac{1}{2}x$.

(B) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
Taking the natural logarithm on both sides,we get $\ln V = \ln(\frac{4}{3} \pi) + 3 \ln r$.
Differentiating both sides,we get the relative error formula: $\frac{\Delta V}{V} = 3 \frac{\Delta r}{r}$.
To find the percentage error,multiply by $100$: $\frac{\Delta V}{V} \times 100 = 3 \times (\frac{\Delta r}{r} \times 100)$.
Given that the percentage error in the radius is $\frac{\Delta r}{r} \times 100 = 1\%$.
Therefore,the percentage error in the volume is $3 \times 1\% = 3\%$.

(C) The mean time period is given as $T = 2.00 \, s$.
The mean absolute error in the time period is given as $\Delta T = 0.05 \, s$.
Any physical quantity measured with an error is expressed in the form $(T \pm \Delta T)$.
Therefore,substituting the given values,the time period is expressed as $(2.00 \pm 0.05) \, s$.

(B) Given: Distance $S = (13.8 \pm 0.2) \ m$ and time $t = (4.0 \pm 0.3) \ s$.
Velocity $v = \frac{S}{t} = \frac{13.8}{4.0} = 3.45 \ ms^{-1}$.
The relative error in velocity is given by $\frac{\Delta v}{v} = \frac{\Delta S}{S} + \frac{\Delta t}{t}$.
Substituting the values: $\frac{\Delta v}{3.45} = \frac{0.2}{13.8} + \frac{0.3}{4.0}$.
$\frac{\Delta v}{3.45} = 0.0145 + 0.075 = 0.0895$.
$\Delta v = 3.45 \times 0.0895 \approx 0.3087 \ ms^{-1}$.
Rounding to one significant figure,$\Delta v \approx 0.3 \ ms^{-1}$.
Thus,the velocity is $(3.45 \pm 0.3) \ ms^{-1}$.

(C) Velocity $v$ is given by $v = \frac{d}{t}$.
For division,the relative error is the sum of the relative errors of the individual quantities: $\frac{\Delta v}{v} = \frac{\Delta d}{d} + \frac{\Delta t}{t}$.
Percentage error in velocity $= \left( \frac{\Delta d}{d} \times 100 \right) + \left( \frac{\Delta t}{t} \times 100 \right)$.
Substituting the given values: $\left( \frac{0.2}{13.8} \times 100 \right) + \left( \frac{0.3}{4.0} \times 100 \right)$.
$= 1.449 + 7.5 = 8.949 \% \approx 8.95 \%$.

(C) Percentage error is defined as the ratio of the absolute error to the measured value,multiplied by $100$.
$\text{Percentage error} = \left( \frac{\text{Measured value} - \text{Exact value}}{\text{Exact value}} \times 100 \right) \times 100\%$.
Since the numerator and the denominator have the same physical units,they cancel each other out.
Therefore,percentage error is a dimensionless quantity and has no units.

(B) The accuracy of a measurement is a measure of how close the measured value is to the true value.
Absolute error is defined as the difference between the measured value and the true value of a physical quantity.
Since absolute error directly quantifies the deviation from the true value,it is the primary indicator used to determine the accuracy of a measurement.
Therefore,the correct option is $B$.

(B) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
The relative error in volume is given by $\frac{\Delta V}{V} = 3 \frac{\Delta r}{r}$.
To find the percentage error,we multiply by $100$:
$\text{Percentage error in volume} = 3 \times \left( \frac{\Delta r}{r} \right) \times 100$.
Given $r = 5.3 \, cm$ and $\Delta r = 0.1 \, cm$,we substitute these values:
$\text{Percentage error} = 3 \times \left( \frac{0.1}{5.3} \right) \times 100$.
Thus,the correct option is $B$.

(D) Given the relation: $a = \frac{b^\alpha c^\beta}{d^\gamma e^\delta}$.
According to the theory of propagation of errors,for a quantity $a = b^\alpha c^\beta d^{-\gamma} e^{-\delta}$,the maximum relative error is given by the sum of the relative errors of the individual components multiplied by their respective powers.
The maximum percentage error is calculated as:
$\left( \frac{\Delta a}{a} \times 100 \right)_{\max} = |\alpha| \left( \frac{\Delta b}{b} \times 100 \right) + |\beta| \left( \frac{\Delta c}{c} \times 100 \right) + |-\gamma| \left( \frac{\Delta d}{d} \times 100 \right) + |-\delta| \left( \frac{\Delta e}{e} \times 100 \right)$.
Substituting the given percentage errors $b_1\%, c_1\%, d_1\%, e_1\%$:
$\left( \frac{\Delta a}{a} \times 100 \right)_{\max} = (\alpha b_1 + \beta c_1 + \gamma d_1 + \delta e_1)\%$.
Thus,the correct option is $D$.

(A) Weight in air $W_a = (5.00 \pm 0.05) \ N$.
Weight in water $W_w = (4.00 \pm 0.05) \ N$.
Loss of weight in water $W_L = W_a - W_w = (5.00 - 4.00) \pm (0.05 + 0.05) = (1.00 \pm 0.10) \ N$.
Relative density $RD = \frac{W_a}{W_L} = \frac{5.00}{1.00} = 5.0$.
The maximum permissible percentage error is given by $\frac{\Delta RD}{RD} \times 100 = \left( \frac{\Delta W_a}{W_a} + \frac{\Delta W_L}{W_L} \right) \times 100$.
Substituting the values: $\left( \frac{0.05}{5.00} + \frac{0.10}{1.00} \right) \times 100 = (0.01 + 0.10) \times 100 = 1\% + 10\% = 11\%$.
Thus,the relative density is $5.0 \pm 11\%$.

(B) Given: $V = 100 \pm 5 \, \text{V}$ and $I = 10 \pm 0.2 \, \text{A}$.
Since $R = \frac{V}{I}$,the relative error in $R$ is given by $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.
To find the percentage error,we multiply by $100$:
$\left( \frac{\Delta R}{R} \times 100 \right)_{\max} = \left( \frac{\Delta V}{V} \times 100 \right) + \left( \frac{\Delta I}{I} \times 100 \right)$.
Substituting the values:
$= \left( \frac{5}{100} \times 100 \right) + \left( \frac{0.2}{10} \times 100 \right)$.
$= 5\% + 2\% = 7\%$.
Therefore,the total percentage error in $R$ is $7\%$.

(B) Step $1$: Calculate the mean value of the period $(T_{mean})$.
$T_{mean} = \frac{2.63 + 2.56 + 2.42 + 2.71 + 2.80}{5} = \frac{13.12}{5} = 2.624\, s \approx 2.62\, s$.
Step $2$: Calculate the absolute errors for each measurement $(|\Delta T_i| = |T_i - T_{mean}|)$.
$|\Delta T_1| = |2.63 - 2.62| = 0.01\, s$
$|\Delta T_2| = |2.56 - 2.62| = 0.06\, s$
$|\Delta T_3| = |2.42 - 2.62| = 0.20\, s$
$|\Delta T_4| = |2.71 - 2.62| = 0.09\, s$
$|\Delta T_5| = |2.80 - 2.62| = 0.18\, s$
Step $3$: Calculate the mean absolute error $(\Delta T_{mean})$.
$\Delta T_{mean} = \frac{0.01 + 0.06 + 0.20 + 0.09 + 0.18}{5} = \frac{0.54}{5} = 0.108\, s$.
Rounding to two decimal places,we get $\Delta T_{mean} = 0.11\, s$.

(C) The formula for Young's modulus is $Y = \frac{4MgL}{\pi D^2 l}$.
The relative error in $Y$ is given by $\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta g}{g} + \frac{\Delta L}{L} + 2\frac{\Delta D}{D} + \frac{\Delta l}{l}$.
Given values are $M = 3.00 \, kg$ $(\Delta M = 0.01 \, kg)$,$L = 2.820 \, m$ $(\Delta L = 0.001 \, m)$,$l = 0.087 \, cm$ $(\Delta l = 0.001 \, cm)$,$D = 0.041 \, cm$ $(\Delta D = 0.001 \, cm)$.
Substituting these values into the error formula:
$\frac{\Delta Y}{Y} \times 100 = \left( \frac{0.01}{3.00} + \frac{0.001}{2.820} + 2 \times \frac{0.001}{0.041} + \frac{0.001}{0.087} \right) \times 100$.
Note: The term $\frac{\Delta g}{g}$ is usually neglected if $g$ is taken as a constant. Calculating the values:
$\frac{\Delta Y}{Y} \times 100 \approx (0.00333 + 0.00035 + 0.04878 + 0.01149) \times 100 \approx 0.06395 \times 100 \approx 6.4\%$.
Rounding to the nearest provided option,the maximum permissible error is $6.5\%$.

(B) Given the formula for heat produced: $H = I^2Rt$.
The relative error in $H$ is given by the propagation of errors formula:
$\frac{\Delta H}{H} = 2\frac{\Delta I}{I} + \frac{\Delta R}{R} + \frac{\Delta t}{t}$.
To find the percentage error,we multiply by $100$:
$\frac{\Delta H}{H} \times 100 = \left( 2 \times \frac{\Delta I}{I} \times 100 + \frac{\Delta R}{R} \times 100 + \frac{\Delta t}{t} \times 100 \right)$.
Substituting the given percentage errors $(3\%, 4\%, 6\%)$:
$\frac{\Delta H}{H} \times 100 = (2 \times 3\% + 4\% + 6\%)$.
Calculating the result:
$\frac{\Delta H}{H} \times 100 = (6\% + 4\% + 6\%) = 16\%$.
Thus,the error in the measurement of $H$ is $\pm 16\%$.

(D) The kinetic energy $E$ of a body is given by $E = \frac{1}{2}mv^2$.
Since $m$ is constant,the relative change in kinetic energy is related to the velocity $v$ by the relation $\frac{\Delta E}{E} \approx 2 \frac{\Delta v}{v}$.
Given the percentage error in velocity is $\frac{\Delta v}{v} \times 100 = 50\%$,which means the new velocity $v' = v + 0.5v = 1.5v$.
The new kinetic energy $E'$ is $E' = \frac{1}{2}m(v')^2 = \frac{1}{2}m(1.5v)^2 = 2.25 \times (\frac{1}{2}mv^2) = 2.25E$.
The percentage error in kinetic energy is $\frac{E' - E}{E} \times 100 = \frac{2.25E - E}{E} \times 100 = 1.25 \times 100 = 125\%$.

(C) The relative error in $P$ is given by the formula: $\frac{\Delta P}{P} = 3 \frac{\Delta A}{A} + \frac{1}{2} \frac{\Delta B}{B} + 4 \frac{\Delta C}{C} + \frac{3}{2} \frac{\Delta D}{D}$.
In the expression for percentage error,the coefficient of each relative error term represents the weight of that quantity's contribution to the total error.
The coefficients are $3$ for $A$,$0.5$ for $B$,$4$ for $C$,and $1.5$ for $D$.
Since the coefficient of $\frac{\Delta C}{C}$ is the largest $(4)$,the quantity $C$ contributes the maximum percentage error to $P$.

(C) Let the length of rod $A$ be $L_A = 3.25 \pm 0.01 \,cm$ and the length of rod $B$ be $L_B = 4.19 \pm 0.01 \,cm$.
To find how much longer rod $B$ is than rod $A$,we calculate the difference $L_B - L_A$.
The difference in length is given by $\Delta L = L_B - L_A = (4.19 - 3.25) \pm (\Delta L_B + \Delta L_A)$.
Subtracting the mean values: $4.19 - 3.25 = 0.94 \,cm$.
Adding the absolute errors: $0.01 + 0.01 = 0.02 \,cm$.
Therefore,the difference is $0.94 \pm 0.02 \,cm$.

(A) Given the physical quantity $X = M^a L^b T^c$.
According to the theory of errors,if a quantity $X$ is related to measured quantities $M, L,$ and $T$ by the relation $X = M^a L^b T^c$,the relative error in $X$ is given by $\frac{\Delta X}{X} = a \left( \frac{\Delta M}{M} \right) + b \left( \frac{\Delta L}{L} \right) + c \left( \frac{\Delta T}{T} \right)$.
To find the maximum percentage error,we consider the absolute values of the coefficients.
The percentage error in $X$ is given by $\frac{\Delta X}{X} \times 100 = a \left( \frac{\Delta M}{M} \times 100 \right) + b \left( \frac{\Delta L}{L} \times 100 \right) + c \left( \frac{\Delta T}{T} \times 100 \right)$.
Substituting the given percentage errors $\alpha, \beta,$ and $\gamma$ for $M, L,$ and $T$ respectively,we get:
Maximum percentage error in $X = a\alpha + b\beta + c\gamma$.

(D) Given the relation: $A = \frac{a^2 b^3}{c \sqrt{d}}$
The percentage errors are given as:
$\frac{\Delta a}{a} \times 100 = 1\%$
$\frac{\Delta b}{b} \times 100 = 3\%$
$\frac{\Delta c}{c} \times 100 = 2\%$
$\frac{\Delta d}{d} \times 100 = 2\%$
The relative error in $A$ is given by the formula:
$\frac{\Delta A}{A} = 2 \frac{\Delta a}{a} + 3 \frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{1}{2} \frac{\Delta d}{d}$
Multiplying by $100$ to get the percentage error:
$\frac{\Delta A}{A} \times 100 = 2 \left( \frac{\Delta a}{a} \times 100 \right) + 3 \left( \frac{\Delta b}{b} \times 100 \right) + \left( \frac{\Delta c}{c} \times 100 \right) + \frac{1}{2} \left( \frac{\Delta d}{d} \times 100 \right)$
Substituting the given values:
$\frac{\Delta A}{A} \times 100 = 2(1\%) + 3(3\%) + 2\% + \frac{1}{2}(2\%)$
$\frac{\Delta A}{A} \times 100 = 2\% + 9\% + 2\% + 1\% = 14\%$
Thus,the percentage error in $A$ is $14\%$.

(D) Density $\rho$ is given by the formula $\rho = \frac{M}{V} = \frac{M}{\pi r^2 L}$.
The relative error in density is given by $\frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} + 2\frac{\Delta r}{r} + \frac{\Delta L}{L}$.
Given values are $M = 0.3 \, g, \Delta M = 0.003 \, g$; $r = 0.5 \, mm, \Delta r = 0.005 \, mm$; $L = 6 \, cm, \Delta L = 0.06 \, cm$.
Substituting these values:
$\frac{\Delta \rho}{\rho} = \frac{0.003}{0.3} + 2 \times \frac{0.005}{0.5} + \frac{0.06}{6}$.
$\frac{\Delta \rho}{\rho} = 0.01 + 2 \times 0.01 + 0.01 = 0.01 + 0.02 + 0.01 = 0.04$.
Percentage error $= \frac{\Delta \rho}{\rho} \times 100 = 0.04 \times 100 = 4\%$.

(D) Given: $V = 100\,V$,$\Delta V = 0.5\,V$,$I = 10\,A$,$\Delta I = 0.2\,A$.
Resistance $R = \frac{V}{I} = \frac{100}{10} = 10\,\Omega$.
The relative error in resistance is given by $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.
Substituting the values: $\frac{\Delta R}{10} = \frac{0.5}{100} + \frac{0.2}{10} = 0.005 + 0.02 = 0.025$.
Therefore,$\Delta R = 10 \times 0.025 = 0.25\,\Omega$.
The value of resistance is $R \pm \Delta R = 10 \pm 0.25\,\Omega$.
Since $10 \pm 0.25\,\Omega$ is not among the options $A, B,$ or $C$,the correct choice is $D$.

(B) According to Ohm's law,the resistance $R$ is given by $R = \frac{V}{I}$.
The relative error in the measurement of resistance is given by the formula $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.
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 $\frac{\Delta R}{R} \times 100 = 3\% + 3\% = 6\%$.

(B) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
Taking the relative error,we have $\frac{\Delta V}{V} = 3 \frac{\Delta r}{r}$.
To find the percentage error,multiply both sides by $100$:
$\frac{\Delta V}{V} \times 100 = 3 \times \left( \frac{\Delta r}{r} \times 100 \right)$.
Given that the percentage error in the radius is $\frac{\Delta r}{r} \times 100 = 0.2\%$.
Therefore,the percentage error in the volume is $3 \times 0.2\% = 0.6\%$.

(A) Given:
Initial temperature,$\theta_1 = (40.6 \pm 0.2)^{\circ}C$
Final temperature,$\theta_2 = (78.9 \pm 0.3)^{\circ}C$
The rise in temperature is given by $\Delta\theta = \theta_2 - \theta_1$.
$\Delta\theta = 78.9 - 40.6 = 38.3^{\circ}C$.
When subtracting two quantities,the absolute errors are added.
Error in rise in temperature,$\Delta(\Delta\theta) = \Delta\theta_1 + \Delta\theta_2$.
$\Delta(\Delta\theta) = 0.2 + 0.3 = 0.5^{\circ}C$.
Therefore,the rise in temperature with proper error limits is $(38.3 \pm 0.5)^{\circ}C$.

(A) The formula for the relative error in $X$ is given by:
$\frac{\Delta X}{X} = \frac{\Delta a}{a} + 2\frac{\Delta b}{b} + 3\frac{\Delta c}{c}$
Given percentage errors are:
$\frac{\Delta a}{a} \times 100 = \pm 1\%$
$\frac{\Delta b}{b} \times 100 = \pm 3\%$
$\frac{\Delta c}{c} \times 100 = \pm 2\%$
Substituting these values into the error formula:
$\frac{\Delta X}{X} \times 100 = \pm 1\% + 2(\pm 3\%) + 3(\pm 2\%)$
$\frac{\Delta X}{X} \times 100 = \pm 1\% \pm 6\% \pm 6\%$
To find the maximum percentage error,we add the magnitudes:
$\text{Maximum percentage error} = 1\% + 6\% + 6\% = 13\%$
Thus,the percentage error in $X$ is $\pm 13\%$.

(C) Given: Length $l = (5.7 \pm 0.1) \text{ cm}$ and Breadth $b = (3.4 \pm 0.2) \text{ cm}$.
The area $A$ is given by $A = l \times b = 5.7 \times 3.4 = 19.38 \text{ cm}^2$.
The relative error in area is given by $\frac{\Delta A}{A} = \frac{\Delta l}{l} + \frac{\Delta b}{b}$.
Substituting the values: $\frac{\Delta A}{A} = \frac{0.1}{5.7} + \frac{0.2}{3.4} = \frac{0.34 + 1.14}{19.38} = \frac{1.48}{19.38}$.
Therefore,the absolute error $\Delta A = \left( \frac{1.48}{19.38} \right) \times A = \left( \frac{1.48}{19.38} \right) \times 19.38 = 1.48 \text{ cm}^2$.
Rounding to appropriate significant figures,$\Delta A \approx 1.5 \text{ cm}^2$ and $A \approx 19.4 \text{ cm}^2$.
Thus,the area is $(19.4 \pm 1.5) \text{ cm}^2$. Note: Based on the provided options,$(19.38 \pm 1.48) \text{ cm}^2$ is the intended answer.

(A) The measured temperature is $T = 37.8 ^\circ C$.
The least count of the thermometer is the absolute error,$\Delta T = 0.2 ^\circ C$.
Thus,the temperature with error is expressed as $(T \pm \Delta T) = (37.8 \pm 0.2) ^\circ C$.
The percentage error is calculated as:
$\text{Percentage Error} = \left( \frac{\Delta T}{T} \times 100 \right) \%$
$= \left( \frac{0.2}{37.8} \times 100 \right) \%$
$= \left( \frac{20}{37.8} \right) \% \approx 0.529 \% \approx 0.5 \%$.

(B) Let $X = ABC$.
Taking the natural logarithm on both sides,we get $\ln X = \ln A + \ln B + \ln C$.
Differentiating both sides,we get $\frac{dX}{X} = \frac{dA}{A} + \frac{dB}{B} + \frac{dC}{C}$.
For small errors,this can be written as $\frac{\Delta X}{X} = \frac{\Delta A}{A} + \frac{\Delta B}{B} + \frac{\Delta C}{C}$.
Given that $a, b, c$ are the percentage errors,we have $a = \frac{\Delta A}{A} \times 100$,$b = \frac{\Delta B}{B} \times 100$,and $c = \frac{\Delta C}{C} \times 100$.
Therefore,the percentage error in $X = ABC$ is $\frac{\Delta X}{X} \times 100 = (\frac{\Delta A}{A} \times 100) + (\frac{\Delta B}{B} \times 100) + (\frac{\Delta C}{C} \times 100) = a + b + c$.

(B) The angular momentum $(J)$ of a ring about its geometrical axis is given by $J = I \omega$.
Since the moment of inertia of a ring is $I = M R^2$,we have $J = M R^2 \omega$.
The relative error in $J$ is given by $\frac{\Delta J}{J} = \frac{\Delta M}{M} + 2 \frac{\Delta R}{R} + \frac{\Delta \omega}{\omega}$.
To find the maximum percentage error,we multiply by $100$:
$\frac{\Delta J}{J} \times 100 = \left( \frac{\Delta M}{M} \times 100 \right) + 2 \left( \frac{\Delta R}{R} \times 100 \right) + \left( \frac{\Delta \omega}{\omega} \times 100 \right)$.
Given $\frac{\Delta M}{M} \times 100 = 2 \%$,$\frac{\Delta R}{R} \times 100 = 1 \%$,and $\frac{\Delta \omega}{\omega} \times 100 = 1 \%$.
Substituting these values:
$\frac{\Delta J}{J} \times 100 = 2 \% + 2(1 \%) + 1 \% = 2 \% + 2 \% + 1 \% = 5 \%$.

(C) The least count of the stopwatch,which represents the absolute error in the measurement,is $\Delta t = \frac{1}{5} \ s = 0.2 \ s$.
The measured time for $20$ oscillations is $t = 25 \ s$.
The percentage error is calculated using the formula: $\text{Percentage Error} = \left( \frac{\Delta t}{t} \right) \times 100$.
Substituting the values: $\text{Percentage Error} = \left( \frac{0.2}{25} \right) \times 100$.
$\text{Percentage Error} = 0.008 \times 100 = 0.8 \%$.

(A) The thickness of the wall $(t)$ is given by the difference between the outer radius $(r_1)$ and the inner radius $(r_2)$.
$t = r_1 - r_2 = 4.23 \text{ cm} - 3.89 \text{ cm} = 0.34 \text{ cm}$.
When subtracting two physical quantities,the absolute errors are added.
$\Delta t = \Delta r_1 + \Delta r_2 = 0.01 \text{ cm} + 0.01 \text{ cm} = 0.02 \text{ cm}$.
Therefore,the thickness of the wall is $(t \pm \Delta t) = (0.34 \pm 0.02) \text{ cm}$.

(C) The circumference $C$ of a circle is given by the formula $C = \pi D$,where $D$ is the diameter.
Taking the relative error,we have $\frac{\Delta C}{C} = \frac{\Delta D}{D}$.
Given that the percentage error in the diameter is $\frac{\Delta D}{D} \times 100\% = 4\%$.
Therefore,the percentage error in the circumference is $\frac{\Delta C}{C} \times 100\% = \frac{\Delta D}{D} \times 100\% = 4\%$.
Thus,the error in the circumference is $4\%$.

(A) Step $1$: Calculate the mean value of resistance $\bar{R}$.
$\bar{R} = \frac{4.12 + 4.08 + 4.22 + 4.14}{4} = \frac{16.56}{4} = 4.14 \; \Omega$.
Step $2$: Calculate the absolute errors in each measurement.
$\Delta R_1 = |4.14 - 4.12| = 0.02 \; \Omega$
$\Delta R_2 = |4.14 - 4.08| = 0.06 \; \Omega$
$\Delta R_3 = |4.14 - 4.22| = 0.08 \; \Omega$
$\Delta R_4 = |4.14 - 4.14| = 0.00 \; \Omega$
Step $3$: Calculate the mean absolute error $\Delta \bar{R}$.
$\Delta \bar{R} = \frac{0.02 + 0.06 + 0.08 + 0.00}{4} = \frac{0.16}{4} = 0.04 \; \Omega$.
Step $4$: Calculate the relative error.
Relative error $= \frac{\Delta \bar{R}}{\bar{R}} = \frac{0.04}{4.14} \approx 0.00966 \approx 0.0096$.

(A) The formula for the moment of inertia $(I)$ of a ring about its geometric axis is given by $I = M R^2$.
To find the maximum percentage error,we use the rule for propagation of errors in products and powers:
$\frac{\Delta I}{I} \times 100 = \frac{\Delta M}{M} \times 100 + 2 \left( \frac{\Delta R}{R} \times 100 \right)$.
Given:
Percentage error in mass $\frac{\Delta M}{M} \times 100 = 2 \%$.
Percentage error in radius $\frac{\Delta R}{R} \times 100 = 1 \%$.
Substituting these values into the equation:
$\frac{\Delta I}{I} \times 100 = 2 \% + 2(1 \%) = 2 \% + 2 \% = 4 \%$.
Thus,the maximum percentage error in the measurement of the moment of inertia is $4 \%$.

(C) The relative error or the standard error in a measurement is inversely proportional to the square root of the number of observations,$n$,if the errors are random.
However,in the context of the mean absolute error,the error is often expressed as $\Delta x_{m} = \frac{\sum |\Delta x|}{n}$.
For a large number of observations,the statistical error (standard deviation of the mean) is given by $\sigma_{m} = \frac{\sigma}{\sqrt{n}}$.
When the number of observations $n$ increases from $100$ to $400$,the factor $n$ increases by a factor of $4$.
Since the error is inversely proportional to $\sqrt{n}$,the new error becomes $\frac{1}{\sqrt{4}} = \frac{1}{2}$ of the original error.
Thus,the possible error is halved.

(C) The height $h$ of the bridge is given by the equation of motion: $h = \frac{1}{2}gt^2$.
Given $g = 9.8\;m/s^2$ and $t = 2\;s$,the height is $h = \frac{1}{2} \times 9.8 \times (2)^2 = 19.6\;m$.
To find the error in height $\Delta h$,we differentiate the equation $h = \frac{1}{2}gt^2$ with respect to $t$:
$dh = gt \cdot dt$.
Substituting the given values,where $dt = \Delta t = 0.1\;s$:
$\Delta h = 9.8 \times 2 \times 0.1 = 1.96\;m$.
Thus,the error in the estimation of the height is $1.96\;m$.

(B) Let the initial temperature be $T_1 = 16 \pm 0.6 \ ^\circ C$ and the final temperature be $T_2 = 56 \pm 0.3 \ ^\circ C$.
The rise in temperature is given by $\Delta T = T_2 - T_1$.
For subtraction,the absolute errors are added: $\Delta T = (56 - 16) \pm (0.3 + 0.6) \ ^\circ C$.
Therefore,$\Delta T = 40 \pm 0.9 \ ^\circ C$.

(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}} = 2\pi l^{1/2} g^{-1/2}$.
Using the formula for propagation of errors,the maximum relative error in $T$ is given by:
$\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l} + \frac{1}{2} \frac{\Delta g}{g}$.
To find the maximum percentage error,multiply by $100$:
$\left( \frac{\Delta T}{T} \times 100 \right) \% = \frac{1}{2} \left( \frac{\Delta l}{l} \times 100 \right) \% + \frac{1}{2} \left( \frac{\Delta g}{g} \times 100 \right) \%$.
Given $\left( \frac{\Delta l}{l} \times 100 \right) \% = 2\%$ and $\left( \frac{\Delta g}{g} \times 100 \right) \% = 4\%$.
Substituting these values:
$\text{Max } \% \text{ error in } T = \frac{1}{2} (2\%) + \frac{1}{2} (4\%) = 1\% + 2\% = 3\%$.
Thus,the maximum percentage error in the time period is $\pm 3\%$.

(A) Step $1$: Calculate the mean value of the observations.
Mean value $a_{mean} = \frac{3.29 + 3.28 + 3.29 + 3.31 + 3.28 + 3.27 + 3.29 + 3.30}{8} = \frac{26.31}{8} = 3.28875 \, cm \approx 3.29 \, cm$.
Step $2$: Calculate the absolute error for the fourth observation $(a_4 = 3.31 \, cm)$.
Absolute error $\Delta a_4 = a_4 - a_{mean} = 3.31 - 3.29 = 0.02 \, cm$.
Step $3$: Calculate the absolute error for the eighth observation $(a_8 = 3.30 \, cm)$.
Absolute error $\Delta a_8 = a_8 - a_{mean} = 3.30 - 3.29 = 0.01 \, cm$.
Thus,the absolute errors are $0.02 \, cm$ and $0.01 \, cm$.

(A) Random errors in experimental measurements are caused by unknown and unpredictable changes in the experiment.
Systematic errors in experimental observations usually arise from the measuring instruments themselves,such as calibration issues or zero errors.
$A$ zero error is a type of systematic error because it is consistent and can be corrected by applying a known offset to the measured values.
Therefore,the zero error of an instrument introduces systematic errors.

(A) For resistors connected in series,the equivalent resistance is $R = R_1 + R_2$.
$R = 100 \,\Omega + 200 \,\Omega = 300 \,\Omega$.
The maximum absolute error in the series combination is the sum of the individual absolute errors: $\Delta R = \Delta R_1 + \Delta R_2 = 3 \,\Omega + 4 \,\Omega = 7 \,\Omega$.
Thus,the equivalent resistance is $(300 \pm 7) \,\Omega$.
The percentage error is given by $\frac{\Delta R}{R} \times 100 \%$.
Percentage error $= \frac{7}{300} \times 100 \% = \frac{7}{3} \% \approx 2.33 \%$.
Rounding to one decimal place,we get $2.3 \%$.

(C) The radius $r$ of a circle is related to its diameter $D$ by the formula $r = \frac{D}{2}$.
Taking the derivative or considering relative errors,we have $dr = \frac{1}{2} dD$.
The relative error in the radius is given by $\frac{dr}{r} = \frac{\frac{1}{2} dD}{\frac{1}{2} D} = \frac{dD}{D}$.
Since the relative error in the diameter $\frac{dD}{D} \times 100 = 4\%$,the relative error in the radius is also $\frac{dr}{r} \times 100 = 4\%$.
Therefore,the error in the radius is $4\%$.

(C) The volume of a cylinder is given by $V = \pi r^2 l$,where $r$ is the radius and $l$ is the length.
Given: $l = 5.0 \; cm$,$\Delta l = 0.1 \; cm$,$d = 2.00 \; cm$,$\Delta d = 0.01 \; cm$.
The radius $r = d/2 = 1.00 \; cm$. The error in radius is $\Delta r = \Delta d / 2 = 0.005 \; cm$.
The relative error in volume is given by $\frac{\Delta V}{V} = 2 \frac{\Delta r}{r} + \frac{\Delta l}{l}$.
Percentage error = $\left( 2 \frac{\Delta r}{r} + \frac{\Delta l}{l} \right) \times 100$.
Substituting the values: $\left( 2 \times \frac{0.005}{1.00} + \frac{0.1}{5.0} \right) \times 100$.
$= (2 \times 0.5 \% + 2 \%) = 1 \% + 2 \% = 3 \%$.

(A) Let $m_1 = (10.1 \pm 0.1) \, g$ be the mass of the empty beaker.
Let $m_2 = (17.3 \pm 0.1) \, g$ be the mass of the beaker filled with liquid.
The mass of the liquid $m$ is given by $m = m_2 - m_1$.
$m = 17.3 - 10.1 = 7.2 \, g$.
When two quantities are subtracted,the absolute errors are added.
Therefore,the absolute error in the mass of the liquid is $\Delta m = \Delta m_1 + \Delta m_2$.
$\Delta m = 0.1 + 0.1 = 0.2 \, g$.
Thus,the mass of the liquid with possible limits of accuracy is $(7.2 \pm 0.2) \, g$.

(D) Given:
$l = (4.00 \pm 0.01) \; cm \implies \frac{\Delta l}{l} = \frac{0.01}{4.00} = 0.0025$
$r = (0.250 \pm 0.001) \; cm \implies \frac{\Delta r}{r} = \frac{0.001}{0.250} = 0.004$
$m = (6.25 \pm 0.01) \; g \implies \frac{\Delta m}{m} = \frac{0.01}{6.25} = 0.0016$
Density $\rho = \frac{m}{V} = \frac{m}{\pi r^2 l}$
The relative error in density is given by $\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$
Percentage error $= \left( \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l} \right) \times 100\%$
$= (0.0016 + 2 \times 0.004 + 0.0025) \times 100\%$
$= (0.0016 + 0.008 + 0.0025) \times 100\%$
$= 0.0121 \times 100\% = 1.21\%$

(B) The mean value of the refractive index is given by:
$\bar{n} = \frac{1.54 + 1.53 + 1.44 + 1.54 + 1.56 + 1.45}{6} = \frac{9.06}{6} = 1.51$
The absolute errors in each measurement are:
$|\Delta n_1| = |1.51 - 1.54| = 0.03$
$|\Delta n_2| = |1.51 - 1.53| = 0.02$
$|\Delta n_3| = |1.51 - 1.44| = 0.07$
$|\Delta n_4| = |1.51 - 1.54| = 0.03$
$|\Delta n_5| = |1.51 - 1.56| = 0.05$
$|\Delta n_6| = |1.51 - 1.45| = 0.06$
The mean absolute error is the average of these absolute errors:
$\Delta \bar{n} = \frac{0.03 + 0.02 + 0.07 + 0.03 + 0.05 + 0.06}{6}$
$\Delta \bar{n} = \frac{0.26}{6} \approx 0.0433$
Rounding to two decimal places,we get $\Delta \bar{n} = 0.04$.

(A) Given: $L = 10.0 \pm 0.1 \; cm$,$b = 1.00 \pm 0.01 \; cm$,$t = 0.100 \pm 0.001 \; cm$.
Volume $V = L \times b \times t = 10.0 \times 1.00 \times 0.100 = 1.000 \; cm^{3}$.
The relative error in volume is given by $\frac{\Delta V}{V} = \frac{\Delta L}{L} + \frac{\Delta b}{b} + \frac{\Delta t}{t}$.
Substituting the values: $\frac{\Delta V}{V} = \frac{0.1}{10.0} + \frac{0.01}{1.00} + \frac{0.001}{0.100}$.
$\frac{\Delta V}{V} = 0.01 + 0.01 + 0.01 = 0.03$.
Therefore,the absolute error $\Delta V = 0.03 \times V = 0.03 \times 1.000 = 0.03 \; cm^{3}$.

(B) The volume of the block is given by $V = l \times b \times t$.
The relative error in volume is given by $\frac{\Delta V}{V} = \frac{\Delta l}{l} + \frac{\Delta b}{b} + \frac{\Delta t}{t}$.
The percentage error in volume is $\frac{\Delta V}{V} \times 100 = \left( \frac{\Delta l}{l} + \frac{\Delta b}{b} + \frac{\Delta t}{t} \right) \times 100$.
Substituting the given values:
$\text{Percentage error} = \left( \frac{0.01}{15.12} + \frac{0.01}{10.15} + \frac{0.01}{5.28} \right) \times 100$.
Calculating each term:
$\frac{0.01}{15.12} \times 100 \approx 0.0661$
$\frac{0.01}{10.15} \times 100 \approx 0.0985$
$\frac{0.01}{5.28} \times 100 \approx 0.1894$
Summing these values:
$0.0661 + 0.0985 + 0.1894 = 0.3540$.
Rounding to the appropriate significant figures,we get $0.36\%$.