Errors of Measurement Questions in English

(A) Given: $A = 2.5 \, m/s$,$\Delta A = 0.5 \, m/s$ and $B = 0.10 \, s$,$\Delta B = 0.01 \, s$.
We need to find the value of $X = AB$.
First,calculate the product: $X = AB = 2.5 \times 0.10 = 0.25 \, m$.
For the propagation of error in multiplication,the relative error is given by: $\frac{\Delta X}{X} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$.
Substituting the values: $\frac{\Delta X}{0.25} = \frac{0.5}{2.5} + \frac{0.01}{0.10}$.
$\frac{\Delta X}{0.25} = 0.2 + 0.1 = 0.3$.
$\Delta X = 0.3 \times 0.25 = 0.075$.
Rounding to the appropriate significant figures,$\Delta X \approx 0.08$.
Thus,$AB = (0.25 \pm 0.08) \, m$.

(C) Given the formula for the time period of a simple pendulum: $T = 2\pi \sqrt{\frac{l}{g}}$.
Squaring both sides,we get $T^2 = 4\pi^2 \frac{l}{g}$.
Rearranging for $g$,we get $g = 4\pi^2 \frac{l}{T^2}$.
Taking the relative error,we use the rule for multiplication and division: $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2 \frac{\Delta T}{T}$.
Given that the fractional error in $l$ is $\frac{\Delta l}{l} = y$ and the fractional error in $T$ is $\frac{\Delta T}{T} = x$.
Substituting these values,we get $\frac{\Delta g}{g} = y + 2x$ or $2x + y$.

(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.
The relative error in density is given by $\frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} + 3 \frac{\Delta l}{l}$.
To find the percentage error,we multiply by $100$:
$\frac{\Delta \rho}{\rho} \times 100 = \left( \frac{\Delta M}{M} \times 100 \right) + 3 \left( \frac{\Delta l}{l} \times 100 \right)$.
Given that the percentage error in mass $\frac{\Delta M}{M} \times 100 = 4\%$ and the percentage error in length $\frac{\Delta l}{l} \times 100 = 3\%$.
Substituting these values:
Percentage error in density $= 4\% + 3(3\%) = 4\% + 9\% = 13\%$.

(C) The least count of the stopwatch is the absolute error in the measurement,given as $\Delta t = 0.2\, s$.
The measured time for $20\, oscillations$ is $t = 25\, s$.
The percentage error in the measurement of time is calculated using the formula:
$\text{Percentage Error} = \left( \frac{\Delta t}{t} \right) \times 100\, \%$
Substituting the given values:
$\text{Percentage Error} = \left( \frac{0.2}{25} \right) \times 100\, \%$
$\text{Percentage Error} = 0.2 \times 4\, \% = 0.8\, \%$

(C) The surface area of a sphere is given by $A = 4\pi r^2$.
Using the rules of error propagation,the relative error in $A$ is given by $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.
Given that the percentage error in the radius is $\frac{\Delta r}{r} \times 100 = 0.3\%$.
Therefore,the percentage error in the surface area is $\frac{\Delta A}{A} \times 100 = 2 \times (0.3\%) = 0.6\%$.
Thus,the Assertion is correct.
The Reason states the formula $\frac{\Delta A}{A} = \frac{4\Delta r}{r}$,which is incorrect because the correct relation is $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.
Hence,the Assertion is correct but the Reason is incorrect.

(A) The kinetic energy $E$ is given by the formula $E = \frac{1}{2}mv^2$.
Taking the relative error,we use the formula for propagation of errors:
$\frac{\Delta E}{E} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v}$.
Given that the percentage error in mass $\frac{\Delta m}{m} \times 100 = 1\%$ and the percentage error in velocity $\frac{\Delta v}{v} \times 100 = 2\%$.
Substituting these values into the error equation:
$\frac{\Delta E}{E} \times 100 = (\frac{\Delta m}{m} \times 100) + 2 \times (\frac{\Delta v}{v} \times 100)$
$\frac{\Delta E}{E} \times 100 = 1\% + 2 \times 2\% = 1\% + 4\% = 5\%$.
Since both the Assertion and the Reason are correct and the Reason correctly explains the propagation of error formula used in the Assertion,the correct option is $A$.

(C) The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{\ell}{g}}$.
Squaring both sides and rearranging for $g$,we get $g = \frac{4\pi^2 \ell}{T^2}$.
The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta T}{T}$.
Given: $\ell = 25.0 \; cm$,so $\Delta \ell = 0.1 \; cm$. The time for $40$ oscillations is $t = 50 \; s$,so the time period $T = \frac{50}{40} = 1.25 \; s$. The resolution of the stopwatch is $\Delta t = 1 \; s$,so $\Delta T = \frac{\Delta t}{40} = \frac{1}{40} \; s$.
Substituting these values: $\frac{\Delta g}{g} = \frac{0.1}{25.0} + 2 \times \frac{1/40}{50/40} = \frac{0.1}{25} + 2 \times \frac{1}{50} = 0.004 + 0.04 = 0.044$.
Therefore,the percentage error is $0.044 \times 100 = 4.4 \%$.

(B) To determine the precision of the clocks,we calculate the range of variation for each clock over the seven days.
For Clock $1$,the readings range from $11:59:08$ to $12:01:50$. The total variation is $162 \; s$.
For Clock $2$,the readings range from $10:14:53$ to $10:15:24$. The total variation is $31 \; s$.
Precision in an experiment depends on the consistency of the clock,not its absolute accuracy (zero error),as a constant zero error can be easily corrected by calibration.
Since Clock $2$ has a much smaller variation $(31 \; s)$ compared to Clock $1$ $(162 \; s)$,it is more precise.
Therefore,Clock $2$ is preferred for precision time interval measurements.

(N/A) The mean period of oscillation of the pendulum is:
$T = \frac{(2.63 + 2.56 + 2.42 + 2.71 + 2.80) \; s}{5} = \frac{13.12}{5} \; s = 2.624 \; s \approx 2.62 \; s$.
The absolute errors in the measurements are:
$|\Delta T_1| = |2.63 - 2.62| = 0.01 \; s$
$|\Delta T_2| = |2.56 - 2.62| = 0.06 \; s$
$|\Delta T_3| = |2.42 - 2.62| = 0.20 \; s$
$|\Delta T_4| = |2.71 - 2.62| = 0.09 \; s$
$|\Delta T_5| = |2.80 - 2.62| = 0.18 \; s$
The mean absolute error is:
$\Delta T_{\text{mean}} = \frac{0.01 + 0.06 + 0.20 + 0.09 + 0.18}{5} \; s = \frac{0.54}{5} \; s = 0.108 \; s \approx 0.11 \; s$.
The relative error is:
$\text{Relative error} = \frac{\Delta T_{\text{mean}}}{T} = \frac{0.11}{2.62} \approx 0.042$.
The percentage error is:
$\text{Percentage error} = \frac{\Delta T_{\text{mean}}}{T} \times 100\% = \frac{0.11}{2.62} \times 100\% \approx 4.2\% \approx 4\%$.

(D) Given temperatures are $t_{1} = 20^{\circ}C \pm 0.5^{\circ}C$ and $t_{2} = 50^{\circ}C \pm 0.5^{\circ}C$.
To find the temperature difference $\Delta t = t_{2} - t_{1}$,we subtract the values:
$\Delta t = (50 - 20)^{\circ}C = 30^{\circ}C$.
When subtracting two quantities,the absolute errors are added.
Therefore,the error in the difference is $\Delta(\Delta t) = \Delta t_{1} + \Delta t_{2} = 0.5^{\circ}C + 0.5^{\circ}C = 1.0^{\circ}C$.
Thus,the temperature difference is $30^{\circ}C \pm 1^{\circ}C$.

(A) Given $V = 100 \; V$ and $\Delta V = 5 \; V$. The percentage error in $V$ is $(\Delta V / V) \times 100 = (5 / 100) \times 100 = 5 \%$.
Given $I = 10 \; A$ and $\Delta I = 0.2 \; A$. The percentage error in $I$ is $(\Delta I / I) \times 100 = (0.2 / 10) \times 100 = 2 \%$.
Since $R = V / I$,the relative error in $R$ is given by $\Delta R / R = \Delta V / V + \Delta I / I$.
Therefore,the percentage error in $R$ is $5 \% + 2 \% = 7 \%$.

(A) For the series combination,the equivalent resistance is $R = R_{1} + R_{2}$.
$R = (100 + 200) \pm (3 + 4) \ \Omega = 300 \pm 7 \ \Omega$.
$(b)$ For the parallel combination,the equivalent resistance $R^{\prime}$ is given by $\frac{1}{R^{\prime}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$.
$R^{\prime} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{100 \times 200}{100 + 200} = \frac{20000}{300} \approx 66.7 \ \Omega$.
To find the error $\Delta R^{\prime}$,we use $\frac{\Delta R^{\prime}}{R^{\prime 2}} = \frac{\Delta R_{1}}{R_{1}^{2}} + \frac{\Delta R_{2}}{R_{2}^{2}}$.
$\Delta R^{\prime} = R^{\prime 2} \left( \frac{\Delta R_{1}}{R_{1}^{2}} + \frac{\Delta R_{2}}{R_{2}^{2}} \right) = (66.7)^{2} \left( \frac{3}{100^{2}} + \frac{4}{200^{2}} \right)$.
$\Delta R^{\prime} = 4448.89 \left( \frac{3}{10000} + \frac{4}{40000} \right) = 4448.89 \left( 0.0003 + 0.0001 \right) = 4448.89 \times 0.0004 \approx 1.8 \ \Omega$.
Thus,the equivalent resistance is $66.7 \pm 1.8 \ \Omega$.

(A) Given the expression $Z = \frac{A^{4} B^{1/3}}{C D^{3/2}}$.
According to the rules of propagation of errors,for a quantity $Z = \frac{A^p B^q}{C^r D^s}$,the relative error is given by $\frac{\Delta Z}{Z} = p\frac{\Delta A}{A} + q\frac{\Delta B}{B} + r\frac{\Delta C}{C} + s\frac{\Delta D}{D}$.
Applying this to the given equation,the powers are $p=4$,$q=1/3$,$r=1$,and $s=3/2$.
Therefore,the relative error in $Z$ is $\frac{\Delta Z}{Z} = 4\frac{\Delta A}{A} + \frac{1}{3}\frac{\Delta B}{B} + \frac{\Delta C}{C} + \frac{3}{2}\frac{\Delta D}{D}$.

(D) The formula for the acceleration due to gravity is $g = 4 \pi^2 L / T^2$.
Taking the relative error,we have $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$.
Given $L = 20.0 \; cm$ and $\Delta L = 1 \; mm = 0.1 \; cm$. Thus,$\frac{\Delta L}{L} = \frac{0.1}{20.0} = 0.005$.
For the time period,$T = \frac{t}{n}$,where $t = 90 \; s$ and $n = 100$. The resolution of the watch is $\Delta t = 1 \; s$.
Since $T = t/n$,the relative error is $\frac{\Delta T}{T} = \frac{\Delta t}{t} = \frac{1}{90}$.
Substituting these into the error formula: $\frac{\Delta g}{g} = 0.005 + 2 \left( \frac{1}{90} \right) = 0.005 + 0.0222 = 0.0272$.
The percentage error is $\frac{\Delta g}{g} \times 100 = 0.0272 \times 100 \approx 3 \%$.

(A) Given the relation: $P = \frac{a^3 b^2}{c^{1/2} d}$.
The relative error in $P$ is given by: $\frac{\Delta P}{P} = 3 \frac{\Delta a}{a} + 2 \frac{\Delta b}{b} + \frac{1}{2} \frac{\Delta c}{c} + \frac{\Delta d}{d}$.
To find the percentage error: $\left( \frac{\Delta P}{P} \times 100 \right) \% = \left( 3 \times \frac{\Delta a}{a} \times 100 + 2 \times \frac{\Delta b}{b} \times 100 + \frac{1}{2} \times \frac{\Delta c}{c} \times 100 + \frac{\Delta d}{d} \times 100 \right) \%$.
Substituting the given percentage errors: $\left( \frac{\Delta P}{P} \times 100 \right) \% = (3 \times 1 + 2 \times 3 + 0.5 \times 4 + 2) \% = (3 + 6 + 2 + 2) \% = 13 \%$.
The calculated value of $P$ is $3.763$. Since the percentage error is $13 \%$,which has two significant figures,the result should be rounded off to two significant figures. Thus,$3.763$ rounded to two significant figures is $3.8$.

(N/A)

Mistake	Error
$(1)$ $A$ mistake is caused by carelessness,wrong calculations,or improper methods of measurement.	$(1)$ An error is caused by the imperfect design of the measuring instrument or the observation limitations of individuals.
$(2)$ By taking proper care,a mistake can be eliminated completely.	$(2)$ By taking more observations or using an instrument with a smaller least count,an error can be minimized but not completely eliminated.

(N/A) Errors in the measurement of physical quantities are categorized as follows:
$(a)$ Systematic Error
$(b)$ Random Error
$(a)$ Systematic Error:
Systematic errors are those that tend to be in one direction,either positive or negative. These errors arise from known sources.
$(i)$ Instrumental Error: These arise due to imperfect design or calibration of the measuring instrument,or zero error in the instrument. For example,a thermometer may be inadequately calibrated,reading $104^{\circ}C$ instead of $100^{\circ}C$ at $STP$.
$(ii)$ Imperfection in Experimental Technique or Procedure: These occur due to flaws in the experimental setup. For example,if a thermometer is not placed in proper contact with the body,it will not measure the actual temperature.
$(iii)$ Personal Error: These arise due to individual bias,lack of proper setting of the apparatus,or carelessness. For example,parallax error occurs if the observer's head is not positioned correctly while reading a scale.
Systematic errors can be minimized by improving experimental techniques,selecting better instruments,and removing personal bias.
$(b)$ Random Error:
Errors that occur irregularly and randomly with respect to sign and size are called random errors. These arise due to unpredictable fluctuations in experimental conditions (e.g.,temperature,voltage supply). For example,if the same person repeats an observation,they may get different readings each time. These can be positive or negative and are minimized by taking the arithmetic mean of a large number of observations.

(N/A) The smallest value that can be measured by a measuring instrument is called its least count.
- All readings or measured values are accurate only up to this value.
- The error associated with the resolution of the instrument is called the least count error.
- The least count of a vernier caliper is $0.01 \text{ cm}$,and the least count of a spherometer is $0.001 \text{ cm}$.
- Least count error belongs to the category of random errors but is limited in size.
- It can occur alongside both systematic and random errors.
- The least count of a standard meter scale is $1 \text{ mm}$.
- Least count error can be reduced by using instruments of higher precision and by improving experimental techniques.
- By repeating the observation several times and taking the arithmetic mean of all observations,the mean value becomes very close to the true value of the measured quantity.

(N/A) The estimation of error refers to the process of determining the maximum possible uncertainty or error in an experimental measurement and expressing it alongside the final result. This provides an indication of the precision and reliability of the measurement.
Error is estimated in three primary ways:
$(1)$ Absolute error: The magnitude of the difference between the individual measured value and the true value (or mean value) of the quantity.
$(2)$ Relative error and Fractional error: The ratio of the mean absolute error to the mean value of the measured quantity.
$(3)$ Percentage error: The relative error expressed as a percentage by multiplying it by $100$.

$(a)$ Absolute Error:
The magnitude of the difference between the individual measurement and the true value of the quantity is called the absolute error of the measurement. It is denoted by $|\Delta a|$. In the absence of any other method,we consider the arithmetic mean as the true value.
Consider a physical quantity '$a$'. Let its measurements be $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$. The average value is:
$a_{\text{mean}} = \frac{a_{1}+a_{2}+a_{3}+\ldots+a_{n}}{n} = \frac{1}{n} \sum_{i=1}^{n} a_{i}$
The absolute error in each measurement is:
$\Delta a_{1} = a_{1} - a_{\text{mean}}, \Delta a_{2} = a_{2} - a_{\text{mean}}, \ldots, \Delta a_{n} = a_{n} - a_{\text{mean}}$.
The mean absolute error is:
$(\Delta a)_{\text{mean}} = \frac{|\Delta a_{1}| + |\Delta a_{2}| + \ldots + |\Delta a_{n}|}{n} = \frac{1}{n} \sum_{i=1}^{n} |\Delta a_{i}|$.
$(b)$ Relative Error:
The ratio of the mean absolute error to the mean value of the quantity is called the relative error.
Relative Error $= \frac{(\Delta a)_{\text{mean}}}{a_{\text{mean}}}$.
$(c)$ Percentage Error:
When the relative error is expressed in percentage,it is called the percentage error.
Percentage Error $= \frac{(\Delta a)_{\text{mean}}}{a_{\text{mean}}} \times 100\%$.

(N/A) When an experiment involves several measurements,the errors in all individual measurements combine to affect the final result.
For example,in the measurement of density $(d = m/V)$,the errors in the measurement of mass $(m)$ and volume $(V)$ will combine to produce an error in the calculated density.
Errors combine according to the following mathematical operations:
$1$. Addition: If $z = a + b$,then the absolute error $\Delta z = \Delta a + \Delta b$.
$2$. Subtraction: If $z = a - b$,then the absolute error $\Delta z = \Delta a + \Delta b$.
$3$. Multiplication: If $z = ab$,then the relative error $\frac{\Delta z}{z} = \frac{\Delta a}{a} + \frac{\Delta b}{b}$.
$4$. Division: If $z = a/b$,then the relative error $\frac{\Delta z}{z} = \frac{\Delta a}{a} + \frac{\Delta b}{b}$.
$5$. Power: If $z = a^n$,then the relative error $\frac{\Delta z}{z} = n \frac{\Delta a}{a}$.

(N/A) Let two physical quantities $A$ and $B$ have measured values $A \pm \Delta A$ and $B \pm \Delta B$,respectively.
$(i)$ For addition:
Let $Z$ be the quantity obtained by the addition of $A$ and $B$.
$Z = A + B$
Let the error in $Z$ be $\Delta Z$.
$Z \pm \Delta Z = (A \pm \Delta A) + (B \pm \Delta B)$
$Z \pm \Delta Z = (A + B) \pm (\Delta A + \Delta B)$
Since $Z = A + B$,we have $\pm \Delta Z = \pm \Delta A \pm \Delta B$.
For the maximum absolute error,$\Delta Z = \Delta A + \Delta B$.
$(ii)$ For subtraction:
Let the difference of $A$ and $B$ be $Z$.
$Z = A - B$ (where $A > B$)
$Z \pm \Delta Z = (A \pm \Delta A) - (B \pm \Delta B)$
$Z \pm \Delta Z = (A - B) \pm \Delta A \mp \Delta B$
Since $Z = A - B$,we have $\pm \Delta Z = \pm \Delta A \mp \Delta B$.
The possible values for $\Delta Z$ are $\Delta A - \Delta B$ or $-\Delta A + \Delta B$ or $\Delta A + \Delta B$ or $-\Delta A - \Delta B$.
For the maximum absolute error,$\Delta Z = \Delta A + \Delta B$.

(N/A) Suppose two physical quantities $A$ and $B$ have measured values $A \pm \Delta A$ and $B \pm \Delta B$ respectively,where $\Delta A$ and $\Delta B$ are their absolute errors.
For a product: Let $Z = AB$. Then the measured value is $Z \pm \Delta Z = (A \pm \Delta A)(B \pm \Delta B) = AB \pm A \Delta B \pm B \Delta A \pm \Delta A \Delta B$.
Dividing by $Z = AB$,we get $1 \pm \frac{\Delta Z}{Z} = 1 \pm \frac{\Delta A}{A} \pm \frac{\Delta B}{B} \pm \frac{\Delta A}{A} \cdot \frac{\Delta B}{B}$.
Since $\frac{\Delta A}{A}$ and $\frac{\Delta B}{B}$ are very small,their product is neglected. For maximum relative error,$\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$.
For a quotient: Let $Z = \frac{A}{B}$. Then $Z \pm \Delta Z = \frac{A \pm \Delta A}{B \pm \Delta B} = \frac{A(1 \pm \Delta A/A)}{B(1 \pm \Delta B/B)} = Z(1 \pm \Delta A/A)(1 \pm \Delta B/B)^{-1}$.
Using binomial expansion $(1 \pm x)^n \approx 1 \pm nx$ for small $x$,we get $1 \pm \frac{\Delta Z}{Z} \approx (1 \pm \frac{\Delta A}{A})(1 \mp \frac{\Delta B}{B}) \approx 1 \pm \frac{\Delta A}{A} \mp \frac{\Delta B}{B}$.
Thus,the maximum relative error is $\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$.

(N/A) $1$. Error in measurement: An error is the difference between the measured value and the true value of a physical quantity. It is an inherent uncertainty in any measurement due to limitations of instruments,environmental factors,or the observer's technique. Errors are classified into systematic and random errors and cannot be completely eliminated,only minimized.
$2$. Mistake in measurement: $A$ mistake (or blunder) is an error caused by human carelessness,such as misreading a scale,incorrect recording of data,or using the wrong formula. Unlike errors,mistakes are avoidable and can be eliminated by being careful and verifying the procedure.

(N/A) The least count is defined as the smallest value that can be measured accurately by a measuring instrument. For example,the least count of a standard meter scale is $1 \ mm$ or $0.1 \ cm$.
Least count error is a type of systematic error associated with the resolution of the instrument. It occurs because the instrument cannot measure values smaller than its least count. This error can be minimized by using instruments with higher precision (smaller least count) and by improving experimental techniques.

(N/A) Relative error is defined as the ratio of the mean absolute error to the mean value of the quantity measured.
Mathematically,it is expressed as: $\text{Relative Error} = \frac{\Delta a_{\text{mean}}}{a_{\text{mean}}}$.
Fractional error is another term used for relative error. It represents the uncertainty in a measurement relative to the size of the measurement itself. When the relative error is expressed in percentage,it is called percentage error,given by: $\text{Percentage Error} = \frac{\Delta a_{\text{mean}}}{a_{\text{mean}}} \times 100\%$.

(A) Absolute error is defined as the difference between the measured value and the true value of a physical quantity. Since it represents a physical quantity,it has the same units as the measured quantity.
Relative error is defined as the ratio of the mean absolute error to the mean value of the quantity. Since it is a ratio of two quantities with the same units,the units cancel out,making it a dimensionless quantity.
Fractional error is essentially the same as relative error (often expressed as a percentage when multiplied by $100$),and it is also dimensionless.
Therefore,absolute error has a unit,whereas relative and fractional errors have no units.

(B) No,error cannot be completely eliminated.
Measurement is always subject to some degree of uncertainty due to limitations of the measuring instrument,environmental conditions,and human limitations.
While we can minimize errors through careful calibration,better techniques,and statistical averaging,it is impossible to achieve a perfectly accurate measurement with zero error.

(N/A) When two physical quantities $A$ and $B$ having absolute errors $\Delta A$ and $\Delta B$ respectively are added or subtracted,the result is $Z = A \pm B$.
The absolute error in the result $Z$,denoted by $\Delta Z$,is given by the sum of the absolute errors of the individual quantities.
That is,$\Delta Z = \Delta A + \Delta B$.
This means that when two quantities are added or subtracted,the absolute errors in the individual quantities are added to get the absolute error in the final result.

(N/A) When two physical quantities $A$ and $B$ are multiplied or divided to obtain a result $Z$ (where $Z = AB$ or $Z = A/B$),the relative error in the result $Z$ is equal to the sum of the relative errors in the individual quantities $A$ and $B$.
Mathematically,if $Z = AB$ or $Z = A/B$,then the maximum fractional error or relative error is given by:
$\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$
Here,$\Delta A$ and $\Delta B$ are the absolute errors in $A$ and $B$ respectively,and $\Delta Z$ is the absolute error in the result $Z$.

(A) The error in measurement is defined as the difference between the measured value and the true value of a physical quantity.
Mathematically,it is expressed as: $\text{Error} = \text{Measured Value} - \text{True Value}$.
This discrepancy arises due to various factors such as instrumental limitations,environmental conditions,or human observation errors.

(N/A) $(1)$ The length and breadth of a thin rectangular plate are $l = 16.2 \text{ cm}$ and $b = 10.1 \text{ cm}$. The least count of the meter scale is $0.1 \text{ cm}$,hence the absolute error in measurement is $0.1 \text{ cm}$.
$l = (16.2 \pm 0.1) \text{ cm}$
$b = (10.1 \pm 0.1) \text{ cm}$
Percentage error in measurement of length:
$\frac{0.1}{16.2} \times 100 \approx 0.6 \%$
$l = (16.2 \pm 0.6 \%) \text{ cm}$
Percentage error in measurement of breadth:
$\frac{0.1}{10.1} \times 100 \approx 1 \%$
$b = (10.1 \pm 1 \%) \text{ cm}$
Area of the rectangular plate:
$A = l \times b = 16.2 \times 10.1 = 163.62 \text{ cm}^2$
Percentage error in $A$:
$\frac{\Delta A}{A} \times 100 = \frac{\Delta l}{l} \times 100 + \frac{\Delta b}{b} \times 100 = 0.6 \% + 1 \% = 1.6 \%$
$\Delta A = \frac{1.6 \times 163.62}{100} \approx 2.6 \text{ cm}^2$
Area: $A = (163.62 \pm 2.6) \text{ cm}^2$
Since the minimum significant digits are $3$,the area should be represented as $A \approx (164 \pm 3) \text{ cm}^2$.
$(2)$ If a set of experimental data is specified to $n$ significant figures,a result obtained by combining the data will generally be valid to $n$ significant figures. However,if data is subtracted,the number of significant figures can be reduced. For example,$12.9 \text{ g} - 7.06 \text{ g} = 5.84 \text{ g}$. In subtraction,we consider the decimal places. Since $12.9$ has one decimal place,the result should be rounded to $5.8 \text{ g}$.

(N/A) Let us consider measuring the lengths of two different objects using the same Vernier calipers,yielding values of $2.20 \pm 0.01 \text{ cm}$ and $8.05 \pm 0.01 \text{ cm}$.
Here,the absolute error in each measurement is the same $(0.01 \text{ cm})$.
The percentage error in the first measurement is $\frac{0.01}{2.20} \times 100 \% = 0.45 \%$.
The percentage error in the second measurement is $\frac{0.01}{8.05} \times 100 \% = 0.12 \%$.
Thus,even though the absolute errors are identical,the percentage error is smaller for the larger measurement and larger for the smaller measurement. Therefore,it can be concluded that the smaller the percentage error,the more accurate the measurement is.

(N/A) The least count of an instrument is the smallest measurement that can be accurately measured by it.
When we use an instrument with a smaller least count,the absolute error in the measurement is reduced.
Since the relative error (or percentage error) is directly proportional to the absolute error,a smaller least count leads to a smaller percentage error.
Therefore,using an instrument with the smallest possible least count increases the precision and accuracy of the measurement.

(N/A) The error in a measurement represents the uncertainty in the numerical value of a physical quantity. Because the measured value can be either slightly greater than or slightly less than the true value,the error is expressed with both positive $(+)$ and negative $(-)$ signs to indicate the range of uncertainty.

(B) If a physical quantity is raised to a power $n$ (i.e.,$x^n$),the relative error in the result is given by $n \times (\Delta x / x)$.
As the exponent $n$ increases,the relative error in the measurement is magnified by a factor of $n$.
Therefore,even a small error in the initial measurement of $x$ can lead to a large error in the final result.
To ensure the final result remains within acceptable limits of accuracy,the initial measurement of quantities with large exponents must be performed with very high precision.

(A) Given the relation $f = x^2$.
To find the relative error,we differentiate both sides with respect to $x$ or use the power rule for errors.
Taking the natural logarithm on both sides: $\ln(f) = 2 \ln(x)$.
Differentiating both sides: $\frac{df}{f} = 2 \frac{dx}{x}$.
For small errors,we replace the differentials with absolute errors: $\frac{\Delta f}{f} = 2 \frac{\Delta x}{x}$.

(A) Given mass $m = 225 \, g$ and absolute error $\Delta m = 0.05 \, g$.
The formula for percentage error is given by $\frac{\Delta m}{m} \times 100 \%$.
Substituting the values: $\text{Percentage error} = \frac{0.05}{225} \times 100 \%$.
$\text{Percentage error} = \frac{5}{225} \%$.
$\text{Percentage error} = \frac{1}{45} \% \approx 0.022 \%$.

(B) Given values are $\theta_1 = 25.5 \pm 0.1 \, ^\circ C$ and $\theta_2 = 35.3 \pm 0.1 \, ^\circ C$.
To find the difference $Z = \theta_1 - \theta_2$,we subtract the mean values: $Z = 25.5 - 35.3 = -9.8 \, ^\circ C$.
When subtracting two quantities,the absolute errors are added: $\Delta Z = \Delta \theta_1 + \Delta \theta_2 = 0.1 + 0.1 = 0.2 \, ^\circ C$.
Therefore,the final result is $\theta_1 - \theta_2 = (-9.8 \pm 0.2) \, ^\circ C$.

(N/A) Given,$t_1 = 39.6\, s$,$t_2 = 39.9\, s$,and $t_3 = 39.5\, s$.
The least count of the measuring instrument is $0.1\, s$ (as measurements have only one significant figure after the decimal point).
Precision in the measurement $=$ Least count of the instrument $= 0.1\, s$.
The average time for $20$ oscillations is given by:
$t_{avg} = \frac{t_1 + t_2 + t_3}{3} = \frac{39.6 + 39.9 + 39.5}{3} = \frac{119.0}{3} = 39.666... \approx 39.7\, s$.
Absolute errors in the measurements are:
$\Delta t_1 = |t_{avg} - t_1| = |39.7 - 39.6| = 0.1\, s$
$\Delta t_2 = |t_{avg} - t_2| = |39.7 - 39.9| = 0.2\, s$
$\Delta t_3 = |t_{avg} - t_3| = |39.7 - 39.5| = 0.2\, s$
Average absolute error (Accuracy) $= \frac{\Delta t_1 + \Delta t_2 + \Delta t_3}{3} = \frac{0.1 + 0.2 + 0.2}{3} = \frac{0.5}{3} \approx 0.17\, s$.
Rounding off to one decimal place,the accuracy of the measurement is $\pm 0.2\, s$.

(D) Given the physical quantity $X = a^2 b^3 c^{\frac{5}{2}} d^{-2}$.
The maximum percentage error in $X$ is given by the formula:
$\frac{\Delta X}{X} \times 100 = \left[ 2 \left( \frac{\Delta a}{a} \times 100 \right) + 3 \left( \frac{\Delta b}{b} \times 100 \right) + \frac{5}{2} \left( \frac{\Delta c}{c} \times 100 \right) + 2 \left( \frac{\Delta d}{d} \times 100 \right) \right]$
Substituting the given percentage errors:
$= [2(1\%) + 3(2\%) + 2.5(3\%) + 2(4\%)]$
$= [2 + 6 + 7.5 + 8] \% = 23.5 \%$
Thus,the percentage error in $X$ is $23.5 \%$.
To round off the value $2.763$,we look at the relative error,which is $23.5 \% = 0.235$. Since the error is in the first decimal place,we round off the result to two significant figures. Therefore,$2.763$ rounded to two significant figures is $2.8$.

(D) The density $\rho$ of a sphere is given by $\rho = \frac{M}{V} = \frac{M}{\frac{4}{3} \pi (D/2)^3} = \frac{6M}{\pi D^3}$.
Taking the natural logarithm on both sides: $\ln \rho = \ln(6/\pi) + \ln M - 3 \ln D$.
Differentiating both sides to find the relative error: $\frac{d\rho}{\rho} = \frac{dM}{M} - 3 \frac{dD}{D}$.
For the maximum percentage error,we add the absolute values of the relative errors: $\left( \frac{d\rho}{\rho} \times 100 \right)_{\text{max}} = \left( \frac{dM}{M} \times 100 \right) + 3 \left( \frac{dD}{D} \times 100 \right)$.
Given $\frac{dM}{M} \times 100 = 6.0 \%$ and $\frac{dD}{D} \times 100 = 1.5 \%$.
Substituting these values: $\text{Max error} = 6.0 + 3(1.5) = 6.0 + 4.5 = 10.5 \%$.
We are given the error as $\left(\frac{x}{100}\right) \% = 10.5 \%$.
Therefore,$\frac{x}{100} = 10.5 \implies x = 1050$.

(A) The arithmetic mean of the given values is taken as the true value.
$t_{\text{mean}} = \frac{1.25 + 1.24 + 1.27 + 1.21 + 1.28}{5} = \frac{6.25}{5} = 1.25 \; s$.
The absolute errors in each measurement are:
$|\Delta t_1| = |1.25 - 1.25| = 0 \; s$
$|\Delta t_2| = |1.24 - 1.25| = 0.01 \; s$
$|\Delta t_3| = |1.27 - 1.25| = 0.02 \; s$
$|\Delta t_4| = |1.21 - 1.25| = 0.04 \; s$
$|\Delta t_5| = |1.28 - 1.25| = 0.03 \; s$
The mean absolute error is:
$\Delta t_{\text{mean}} = \frac{0 + 0.01 + 0.02 + 0.04 + 0.03}{5} = \frac{0.10}{5} = 0.02 \; s$.
The percentage relative error is:
$\text{Percentage error} = \frac{\Delta t_{\text{mean}}}{t_{\text{mean}}} \times 100 = \frac{0.02}{1.25} \times 100 = 1.6 \%$.

(A) The mean value $\bar{x}$ is calculated as:
$\bar{x} = \frac{80.0 + 80.5 + 81.0 + 81.5 + 82.0}{5} = \frac{405}{5} = 81.0$
The absolute error for each observation is $|x_i - \bar{x}|$. The relative error for each is $\frac{|x_i - \bar{x}|}{x_i}$.

Observation $(x_i)$	Absolute Error $|x_i - \bar{x}|$	Relative Error $\frac{|x_i - \bar{x}|}{x_i}$
$80.0$	$1.0$	$0.0125$
$80.5$	$0.5$	$0.00621$
$81.0$	$0.0$	$0.0000$
$81.5$	$0.5$	$0.00613$
$82.0$	$1.0$	$0.01219$

Sum of relative errors $= 0.0125 + 0.00621 + 0 + 0.00613 + 0.01219 = 0.03703$
Mean relative error $= \frac{0.03703}{5} = 0.007406$
Mean percentage error $= 0.007406 \times 100 \% \approx 0.74 \%$.

(D) The formula for resistance is $R = \frac{V}{I}$.
For division,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,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 $V = 50 \; V$,$\Delta V = 2 \; V$ and $I = 20 \; A$,$\Delta I = 0.2 \; A$.
Percentage error in $R = \left( \frac{2}{50} \times 100 \right) + \left( \frac{0.2}{20} \times 100 \right)$.
Percentage error in $R = 4\% + 1\% = 5\%$.
Thus,$x = 5$.

(B) The formula for the acceleration due to gravity is $g = \frac{4 \pi^2 \ell}{T^2}$.
Taking the relative error,we have $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta T}{T}$.
Given $\ell = 10 \ cm$ and $\Delta \ell = 1 \ mm = 0.1 \ cm$.
For $200$ oscillations,the total time $t = 100 \ s$ with resolution $\Delta t = 1 \ s$. The time period $T = \frac{t}{200} = \frac{100}{200} = 0.5 \ s$.
The error in time period is $\Delta T = \frac{\Delta t}{200} = \frac{1}{200} \ s$.
Substituting these values: $\frac{\Delta g}{g} = \frac{0.1}{10} + 2 \left( \frac{1/200}{0.5} \right) = 0.01 + 2 \left( \frac{1}{100} \right) = 0.01 + 0.02 = 0.03$.
Percentage error is $\frac{\Delta g}{g} \times 100 = 0.03 \times 100 = 3 \%$.

(A) The resistance $R$ is given by $R = \frac{\rho \ell}{A} = \frac{V}{I}$.
Thus,the resistivity $\rho$ is given by $\rho = \frac{AV}{I\ell} = \frac{\pi d^2 V}{4I\ell}$,where $A = \frac{\pi d^2}{4}$.
The relative error in resistivity is given by $\frac{\Delta \rho}{\rho} = 2\frac{\Delta d}{d} + \frac{\Delta V}{V} + \frac{\Delta I}{I} + \frac{\Delta \ell}{\ell}$.
Given values: $V = 5.0\, V, \Delta V = 0.1\, V, \ell = 10.0\, cm, \Delta \ell = 0.1\, cm, d = 5.00\, mm, \Delta d = 0.01\, mm, I = 2.00\, A, \Delta I = 0.01\, A$.
Substituting these values:
$\frac{\Delta \rho}{\rho} = 2\left(\frac{0.01}{5.00}\right) + \frac{0.1}{5.0} + \frac{0.01}{2.00} + \frac{0.1}{10.0}$
$\frac{\Delta \rho}{\rho} = 0.004 + 0.02 + 0.005 + 0.01 = 0.039$.
The percentage error is $\frac{\Delta \rho}{\rho} \times 100 = 0.039 \times 100 = 3.9\%$.

(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 obtain the relative error formula: $\frac{\Delta V}{V} = 3 \frac{\Delta r}{r}$.
Given $r = 7.50 \, cm$ and $\Delta r = 0.85 \, cm$.
The percentage error in volume is given by $\frac{\Delta V}{V} \times 100 = 3 \times (\frac{\Delta r}{r}) \times 100$.
Substituting the values: $\frac{\Delta V}{V} \times 100 = 3 \times (\frac{0.85}{7.50}) \times 100$.
$\frac{\Delta V}{V} \times 100 = 3 \times 0.11333 \times 100 = 34\%$.
Thus,the value of $x$ is $34$.

(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 $g = \frac{4 \pi^2 L}{T^2}$.
The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$.
Given values are $L = 1.0 \text{ m}$,$\Delta L = 1 \text{ mm} = 0.001 \text{ m}$,$T = 1.95 \text{ s}$,and $\Delta T = 0.01 \text{ s}$.
Substituting these values: $\frac{\Delta g}{g} = \frac{0.001}{1.0} + 2 \times \frac{0.01}{1.95}$.
$\frac{\Delta g}{g} = 0.001 + 0.010256 = 0.011256$.
To find the percentage error,multiply by $100$: $0.011256 \times 100 \approx 1.13 \%$.

(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$,which implies $\ell = \frac{T^2 g}{4\pi^2}$.
The energy $E$ of a simple pendulum for small oscillations is given by $E = \frac{1}{2} m g \ell \theta^2$,where $\theta$ is the amplitude.
Substituting the expression for $\ell$,we get $E = \frac{1}{2} m g \left( \frac{T^2 g}{4\pi^2} \right) \theta^2 = \frac{m g^2 T^2 \theta^2}{8\pi^2}$.
Assuming mass $m$ and amplitude $\theta$ are constant,the relative error in $E$ is given by $\frac{\Delta E}{E} = 2 \frac{\Delta g}{g} + 2 \frac{\Delta T}{T}$.
Given $\frac{\Delta g}{g} = 4\%$ and $\frac{\Delta T}{T} = 3\%$,we have $\frac{\Delta E}{E} = 2(4\%) + 2(3\%) = 8\% + 6\% = 14\%$.
Thus,the accuracy to which $E$ is known is $14\%$.