Errors of Measurement Questions in English

(C) The pressure $p$ on a circular plate is given by the formula $p = \frac{F}{A}$,where $F$ is the force and $A$ is the area of the plate. Since the plate is circular,$A = \pi R^2$,where $R$ is the radius.
Therefore,$p = \frac{F}{\pi R^2}$.
Using the rules of propagation of errors,the relative error in $p$ is given by:
$\frac{\Delta p}{p} = \frac{\Delta F}{F} + 2 \frac{\Delta R}{R}$.
To find the percentage error,we multiply by $100$:
$\frac{\Delta p}{p} \times 100 = \left( \frac{\Delta F}{F} \times 100 \right) + 2 \left( \frac{\Delta R}{R} \times 100 \right)$.
Given that the percentage error in force $\frac{\Delta F}{F} \times 100 = 5 \%$ and the percentage error in radius $\frac{\Delta R}{R} \times 100 = 3 \%$,we substitute these values:
$\text{Percentage error in } p = 5 \% + 2(3 \%) = 5 \% + 6 \% = 11 \%$.
Thus,the percentage error in the measurement of pressure is $11 \%$.

(B) The quantity is given by $X = \frac{L}{T^2}$.
According to the rule of propagation of errors,the maximum relative error in $X$ is given by $\frac{\Delta X}{X} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$.
Given that the percentage error in $L$ is $\frac{\Delta L}{L} \times 100 \% = 3 \%$ and the percentage error in $T$ is $\frac{\Delta T}{T} \times 100 \% = 2 \%$.
Therefore,the maximum percentage error in $\frac{L}{T^2}$ is $\left( \frac{\Delta L}{L} \times 100 \% \right) + 2 \left( \frac{\Delta T}{T} \times 100 \% \right)$.
Substituting the values: $3 \% + 2 \times (2 \%) = 3 \% + 4 \% = 7 \%$.

(A) Let the maximum temperature be $T_1 = 44^{\circ} C \pm 0.5^{\circ} C$ and the minimum temperature be $T_2 = 22^{\circ} C \pm 0.5^{\circ} C$.
The temperature difference is given by $\Delta T = T_1 - T_2$.
The value of the difference is $44^{\circ} C - 22^{\circ} C = 22^{\circ} C$.
According to the rule of error propagation for subtraction,when two quantities are subtracted,their 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 $22^{\circ} C \pm 1^{\circ} C$.

(B) The surface area $S$ of a sphere is given by $S = 4\pi r^2$,where $r$ is the radius.
The relative error in surface area is given by $\frac{\Delta S}{S} = 2 \frac{\Delta r}{r}$.
Given $\frac{\Delta S}{S} \times 100 = 1.2 \%$,we have $2 \frac{\Delta r}{r} \times 100 = 1.2 \%$,which implies $\frac{\Delta r}{r} \times 100 = 0.6 \%$.
The volume $V$ of a sphere is given by $V = \frac{4}{3}\pi r^3$.
The relative error in volume is given by $\frac{\Delta V}{V} = 3 \frac{\Delta r}{r}$.
Substituting the value of $\frac{\Delta r}{r}$,we get $\frac{\Delta V}{V} \times 100 = 3 \times 0.6 \% = 1.8 \%$.

(D) Given,current $I = (10 \pm 0.5) \text{ A}$ and potential difference $V = (100 \pm 6) \text{ V}$.
According to Ohm's law,$R = \frac{V}{I}$.
The relative error in resistance is given by $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.
To find the percentage error,we multiply by $100$:
$\frac{\Delta R}{R} \times 100 = \left( \frac{\Delta V}{V} \times 100 \right) + \left( \frac{\Delta I}{I} \times 100 \right)$.
Substituting the given values:
$\frac{\Delta R}{R} \times 100 = \left( \frac{6}{100} \times 100 \right) + \left( \frac{0.5}{10} \times 100 \right)$.
$\frac{\Delta R}{R} \times 100 = 6\% + 5\% = 11\%$.

(D) Given: $C = (4.0 \pm 0.2) \mu F$ and $V = (10.0 \pm 0.1) V$.
The charge $Q$ on the capacitor is given by $Q = C \times V$.
Calculating the mean value: $Q = 4.0 \mu F \times 10.0 V = 40 \mu C$.
The relative error in charge is given by $\frac{\Delta Q}{Q} = \frac{\Delta C}{C} + \frac{\Delta V}{V}$.
Substituting the values: $\frac{\Delta Q}{Q} = \frac{0.2}{4.0} + \frac{0.1}{10.0} = 0.05 + 0.01 = 0.06$.
The percentage error in charge is $\frac{\Delta Q}{Q} \times 100 = 0.06 \times 100 = 6 \%$.
Therefore,the charge on the capacitor is $40 \mu C \pm 6 \%$.

(C) Given the relation $S = \frac{a^{1/2} b}{c^3 d^4}$.
To find the maximum relative error in $S$,we use the formula for propagation of errors:
$\frac{\Delta S}{S} = \frac{1}{2} \frac{\Delta a}{a} + \frac{\Delta b}{b} + 3 \frac{\Delta c}{c} + 4 \frac{\Delta d}{d}$.
Substituting the given percentage errors:
$\left( \frac{\Delta S}{S} \times 100 \right) = \frac{1}{2} \times (2 \%) + (1 \%) + 3 \times (1 \%) + 4 \times (1 \%)$.
Calculating the values:
$= 1 \% + 1 \% + 3 \% + 4 \% = 9 \%$.
Therefore,the percentage error in $S$ is $9 \%$.

(B) The given expression is $Z = \frac{A B^{1/2}}{C^2}$.
Using the rules of propagation of errors,the relative error in $Z$ is given by:
$\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{1}{2} \frac{\Delta B}{B} + 2 \frac{\Delta C}{C}$.
Given that the percentage error in $A, B,$ and $C$ is $1\%$ each,we have:
$\frac{\Delta A}{A} \times 100 = 1\%$,$\frac{\Delta B}{B} \times 100 = 1\%$,and $\frac{\Delta C}{C} \times 100 = 1\%$.
Substituting these values into the expression for percentage error in $Z$:
$\frac{\Delta Z}{Z} \times 100 = \left( \frac{\Delta A}{A} \times 100 \right) + \frac{1}{2} \left( \frac{\Delta B}{B} \times 100 \right) + 2 \left( \frac{\Delta C}{C} \times 100 \right)$.
$\frac{\Delta Z}{Z} \times 100 = 1\% + \frac{1}{2}(1\%) + 2(1\%) = 1\% + 0.5\% + 2\% = 3.5\%$.

(D) Given that,number of observations,$n=5$.
The readings are $a_1=2.4 \ m, a_2=2.5 \ m, a_3=2.6 \ m, a_4=2.8 \ m, a_5=3.0 \ m$.
Mean value of observations,$\bar{a} = \frac{a_1+a_2+a_3+a_4+a_5}{5} = \frac{2.4+2.5+2.6+2.8+3.0}{5} = \frac{13.3}{5} = 2.66 \ m$.
Absolute errors in individual observed values are:
$|\Delta a_1| = |2.4 - 2.66| = 0.26 \ m$
$|\Delta a_2| = |2.5 - 2.66| = 0.16 \ m$
$|\Delta a_3| = |2.6 - 2.66| = 0.06 \ m$
$|\Delta a_4| = |2.8 - 2.66| = 0.14 \ m$
$|\Delta a_5| = |3.0 - 2.66| = 0.34 \ m$
Mean absolute error,$\Delta \bar{a} = \frac{0.26+0.16+0.06+0.14+0.34}{5} = \frac{0.96}{5} = 0.192 \ m$.
Relative error = $\frac{\Delta \bar{a}}{\bar{a}} = \frac{0.192}{2.66} \approx 0.072$.

(B) Given: Length $l = 1 \ m$,Breadth $b = 50 \ cm = 0.5 \ m$.
The relative error in length is $\frac{\delta l}{l} = 1 \% = 0.01$ and in breadth is $\frac{\delta b}{b} = 1 \% = 0.01$.
The area of the rectangle is $A = l \times b = 1 \times 0.5 = 0.5 \ m^2$.
The relative error in area is given by $\frac{\delta A}{A} = \frac{\delta l}{l} + \frac{\delta b}{b}$.
Substituting the values: $\frac{\delta A}{A} = 0.01 + 0.01 = 0.02$.
The absolute error in area is $\delta A = A \times 0.02 = 0.5 \times 0.02 = 0.01 \ m^2$.
Therefore,the area of the table including error is $A \pm \delta A = (0.5 \pm 0.01) \ m^2$.

(A) The absolute error in the measurement is $\Delta x = 0.05 \ m$.
The measured length is $x = 5 \ m$.
The relative error is given by the ratio of absolute error to the measured value: $\frac{\Delta x}{x} = \frac{0.05}{5} = 0.01$.
The percentage error is calculated by multiplying the relative error by $100$: $\text{Percentage Error} = \frac{\Delta x}{x} \times 100 \% = 0.01 \times 100 \% = 1 \%$.
Therefore,the correct option is $A$.

(C) The mean mass of the ball is calculated as:
$\bar{M} = \frac{2.61 + 2.58 + 2.40 + 2.73 + 2.80}{5} = \frac{13.12}{5} = 2.624 \,g \approx 2.62 \,g$.
The absolute errors in each measurement are:
$|\Delta M_1| = |2.62 - 2.61| = 0.01 \,g$
$|\Delta M_2| = |2.62 - 2.58| = 0.04 \,g$
$|\Delta M_3| = |2.62 - 2.40| = 0.22 \,g$
$|\Delta M_4| = |2.62 - 2.73| = 0.11 \,g$
$|\Delta M_5| = |2.62 - 2.80| = 0.18 \,g$
The mean absolute error is the average of these absolute errors:
$\Delta \bar{M} = \frac{0.01 + 0.04 + 0.22 + 0.11 + 0.18}{5} = \frac{0.56}{5} = 0.112 \,g \approx 0.11 \,g$.

(C) Given,current passing through the conductor,$I = (5 \pm 0.2) \text{ A}$,where $\Delta I = 0.2 \text{ A}$.
Voltage developed,$V = (60 \pm 6) \text{ V}$,where $\Delta V = 6 \text{ V}$.
By Ohm's law,$V = RI$,so $R = V/I$.
The relative error in resistance $R$ is given by the formula: $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.
Substituting the given values: $\frac{\Delta R}{R} = \frac{6}{60} + \frac{0.2}{5}$.
Calculating the terms: $\frac{\Delta R}{R} = 0.1 + 0.04 = 0.14$.
To find the percentage error,multiply by $100$: $\frac{\Delta R}{R} \times 100 = 0.14 \times 100 = 14\%$.

(C) Density $\rho = \frac{m}{V} = \frac{m}{l^3}$.
Relative error in density is given by $\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3 \frac{\Delta l}{l}$.
Given values: $m = 20 \text{ kg}$,$\Delta m = 10 \text{ g} = 0.01 \text{ kg}$,$l = 100 \text{ cm}$,$\Delta l = 1 \text{ mm} = 0.1 \text{ cm}$.
Substituting these values:
$\frac{\Delta \rho}{\rho} = \frac{0.01}{20} + 3 \times \frac{0.1}{100}$.
$\frac{\Delta \rho}{\rho} = 0.0005 + 3 \times 0.001 = 0.0005 + 0.003 = 0.0035$.
Thus,the relative error is $3.5 \times 10^{-3}$.

(B) Given:
Percentage error in length,$\frac{\Delta L}{L} \times 100 = 3 \%$
Percentage error in force,$\frac{\Delta F}{F} \times 100 = 4 \%$
We know that pressure $P$ is defined as force per unit area:
$P = \frac{F}{A}$
Since the plate is square,the area $A = L^2$.
Therefore,$P = \frac{F}{L^2}$.
The relative error in pressure is given by:
$\frac{\Delta P}{P} = \frac{\Delta F}{F} + 2 \frac{\Delta L}{L}$
To find the percentage error,multiply by $100$:
$\frac{\Delta P}{P} \times 100 = \left( \frac{\Delta F}{F} \times 100 \right) + 2 \left( \frac{\Delta L}{L} \times 100 \right)$
Substituting the given values:
$\frac{\Delta P}{P} \times 100 = 4 \% + 2(3 \%) = 4 \% + 6 \% = 10 \%$
Thus,the permissible error in the calculation of pressure is $10 \%$.

(C) 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: $l = 1.01 \ m$,$\Delta l = 0.01 \ m$,$T_{total} = 63 \ s$,$\Delta T_{total} = 3 \ s$.
The time period $T = \frac{T_{total}}{30} = \frac{63}{30} = 2.1 \ s$.
The error in time period $\Delta T = \frac{\Delta T_{total}}{30} = \frac{3}{30} = 0.1 \ s$.
Substituting these values into the error formula:
$\frac{\Delta g}{g} = \frac{0.01}{1.01} + 2 \times \frac{0.1}{2.1} \approx 0.0099 + 0.0952 \approx 0.1051$.
Percentage error = $0.1051 \times 100 \% \approx 10.5 \%$.
Rounding to the nearest provided option,the percentage error is $10 \%$.

(C) Given the formula $X = \frac{A^2 B}{C^{1/3} D^3}$.
Using the principle of propagation of errors,the relative error in $X$ is given by:
$\frac{\Delta X}{X} = 2 \left( \frac{\Delta A}{A} \right) + \left( \frac{\Delta B}{B} \right) + \frac{1}{3} \left( \frac{\Delta C}{C} \right) + 3 \left( \frac{\Delta D}{D} \right)$.
Now,calculate the contribution of each term to the percentage error:
Contribution from $A = 2 \times 2 \% = 4 \%$.
Contribution from $B = 1 \times 2 \% = 2 \%$.
Contribution from $C = \frac{1}{3} \times 4 \% = 1.33 \%$.
Contribution from $D = 3 \times 5 \% = 15 \%$.
Comparing these values,the minimum contribution to the percentage error in $X$ is from $C$.

(D) Let the mass of the scale be $M = 100 \ g$ acting at its center of mass $(50 \ cm)$. Let the mass of each plate be $m = 200 \ g$ at $0 \ cm$ and $100 \ cm$. The pivot is at $x = 45 \ cm$.
Taking moments about the pivot at $45 \ cm$:
Clockwise moments: $M_{veg} \times (100 - 45) + m \times (100 - 45) + M \times (50 - 45) = 300 \times (45 - 0) + m \times (45 - 0)$
$M_{veg} \times 55 + 200 \times 55 + 100 \times 5 = 300 \times 45 + 200 \times 45$
$55 M_{veg} + 11000 + 500 = 13500 + 9000$
$55 M_{veg} + 11500 = 22500$
$55 M_{veg} = 11000$
$M_{veg} = 200 \ g$.
The actual weight is $300 \ g$ and the measured weight is $200 \ g$. The error is $|300 - 200| = 100 \ g$.

(D) Step $1$: Calculate the mean value of the observations.
Mean value $\bar{x} = \frac{2.62 + 2.68 + 2.58 + 2.57 + 2.54 + 2.55}{6} = \frac{15.54}{6} = 2.59 \ mPa \cdot s$.
Step $2$: Calculate the absolute errors for each observation $|\Delta x_i| = |x_i - \bar{x}|$.
$|\Delta x_1| = |2.62 - 2.59| = 0.03$
$|\Delta x_2| = |2.68 - 2.59| = 0.09$
$|\Delta x_3| = |2.58 - 2.59| = 0.01$
$|\Delta x_4| = |2.57 - 2.59| = 0.02$
$|\Delta x_5| = |2.54 - 2.59| = 0.05$
$|\Delta x_6| = |2.55 - 2.59| = 0.04$
Step $3$: Calculate the mean absolute error.
Mean absolute error $\Delta \bar{x} = \frac{0.03 + 0.09 + 0.01 + 0.02 + 0.05 + 0.04}{6} = \frac{0.24}{6} = 0.04 \ mPa \cdot s$.

(C) Pressure $P$ is defined as force $F$ per unit area $A$,so $P = F/A$.
For a square plate of side $s$,the area $A = s^2$.
Therefore,$P = F/s^2$.
The relative error in pressure is given by $\frac{\Delta P}{P} = \frac{\Delta F}{F} + 2 \frac{\Delta s}{s}$.
Given,$\frac{\Delta F}{F} \times 100 = 3 \%$ and $\frac{\Delta s}{s} \times 100 = 1 \%$.
Substituting these values,the percentage error in pressure is $\frac{\Delta P}{P} \times 100 = 3 \% + 2(1 \%) = 3 \% + 2 \% = 5 \%$.
Thus,the correct option is $C$.

(D) The formula for the time period is $T = 2 \pi \sqrt{\frac{m}{k}}$.
Squaring both sides,we get $T^2 = 4 \pi^2 \frac{m}{k}$,which implies $k = \frac{4 \pi^2 m}{T^2}$.
The relative error in $k$ is given by $\frac{\Delta k}{k} = \frac{\Delta m}{m} + 2 \frac{\Delta T}{T}$.
Given: $m = 10 \ g$,$\Delta m = 10 \ mg = 0.01 \ g$.
Time for $50$ oscillations is $t = 60 \ s$,with resolution $\Delta t = 2 \ s$.
The time period $T = \frac{t}{50} = \frac{60}{50} = 1.2 \ s$.
The error in time period is $\Delta T = \frac{\Delta t}{50} = \frac{2}{50} = 0.04 \ s$.
Substituting these values: $\frac{\Delta k}{k} = \frac{0.01}{10} + 2 \times \frac{0.04}{1.2} = 0.001 + 2 \times \frac{0.04}{1.2} = 0.001 + 0.0666... = 0.06766...$
Percentage error = $0.06766 \times 100 \% \approx 6.76 \%$.

(C) Step $1$: Calculate the mean length of the rod:
$\ell_{\text{mean}} = \frac{20.00 + 19.75 + 17.01 + 18.25}{4} = \frac{75.01}{4} = 18.7525 \ cm \approx 18.75 \ cm$.
Step $2$: Calculate the absolute errors for each measurement:
$|\Delta \ell_1| = |20.00 - 18.75| = 1.25 \ cm$
$|\Delta \ell_2| = |19.75 - 18.75| = 1.00 \ cm$
$|\Delta \ell_3| = |17.01 - 18.75| = 1.74 \ cm$
$|\Delta \ell_4| = |18.25 - 18.75| = 0.50 \ cm$
Step $3$: Calculate the mean absolute error:
$\Delta \ell_{\text{mean}} = \frac{1.25 + 1.00 + 1.74 + 0.50}{4} = \frac{4.49}{4} = 1.1225 \ cm \approx 1.12 \ cm$.
Step $4$: Calculate the relative error:
$\text{Relative error} = \frac{\Delta \ell_{\text{mean}}}{\ell_{\text{mean}}} = \frac{1.12}{18.75} \approx 0.06$.

(D) The resistance $R$ is calculated using Ohm's law as $R = V/I$.
Given $V = 10 \text{ V}$ and $I = 5 \text{ A}$,so $R = 10 / 5 = 2 \text{ } \Omega$.
The relative error in resistance is given by the formula $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.
Given the least counts: $\Delta V = 500 \text{ mV} = 0.5 \text{ V}$ and $\Delta I = 200 \text{ mA} = 0.2 \text{ A}$.
Substituting these values into the error formula:
$\Delta R = R \times (\frac{\Delta V}{V} + \frac{\Delta I}{I})$
$\Delta R = 2 \times (\frac{0.5}{10} + \frac{0.2}{5})$
$\Delta R = 2 \times (0.05 + 0.04)$
$\Delta R = 2 \times (0.09) = 0.18 \text{ } \Omega$.

(C) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
Since the diameter $D = 2r$,we can express the volume in terms of the diameter as $V = \frac{4}{3} \pi (\frac{D}{2})^3 = \frac{\pi}{6} D^3$.
Taking the logarithmic differentiation,we get $\frac{\Delta V}{V} = 3 \frac{\Delta D}{D}$.
Given that the percentage error in the diameter measurement is $\frac{\Delta D}{D} \times 100 = 2\%$.
Therefore,the percentage error in the volume is $\frac{\Delta V}{V} \times 100 = 3 \times (\frac{\Delta D}{D} \times 100) = 3 \times 2\% = 6\%$.

(B) The density $\rho$ of a cylinder is given by the formula $\rho = \frac{m}{V} = \frac{m}{\pi (d/2)^2 l} = \frac{4m}{\pi d^2 l}$.
To find the relative error,we use the formula $\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta l}{l}$.
Substituting the given values:
$\frac{\Delta \rho}{\rho} = \frac{0.02}{97.42} + 2 \times \left( \frac{0.02}{20.20} \right) + \frac{0.05}{8.35}$.
Calculating each term:
$\frac{0.02}{97.42} \approx 0.000205$,
$2 \times \left( \frac{0.02}{20.20} \right) \approx 0.001980$,
$\frac{0.05}{8.35} \approx 0.005988$.
Adding these values: $\frac{\Delta \rho}{\rho} \approx 0.000205 + 0.001980 + 0.005988 = 0.008173$.
The percentage error is $\frac{\Delta \rho}{\rho} \times 100 \% \approx 0.008173 \times 100 \% = 0.8173 \% \approx 0.82 \%$.