Concept Card: Errors in Measurement & Propagation of Errors
1. Classification of Experimental Errors:
Error is the difference between the true value and the measured value of a physical quantity:
- Systematic Errors: Errors that tend to occur consistently in one direction (either consistently positive or consistently negative).
- Instrumental errors: Imperfect calibration or zero error of the measuring instrument.
- Imperfection in technique/procedure: Heat loss in calorimetry, air resistance effects, buoyancy errors.
- Personal errors: Parallax errors and bias in taking readings.
- Remedy: Identified and eliminated by calibration, zero correction, and refined procedures. Cannot be minimized by simple averaging.
- Random Errors: Unpredictable, irregular variations occurring due to random fluctuations in experimental conditions (e.g., temperature changes, line voltage fluctuations, mechanical vibrations).
- Follow a Gaussian normal distribution.
- Remedy: Minimized by repeating the measurement $n$ times. If $n$ observations are taken, the random error is reduced by a factor of $\frac{1}{\sqrt{n}}$.
- Least Count Error: The smallest division on the measuring scale determines the resolution limit and represents the absolute uncertainty of a single observation.
2. Mathematical Definitions of Errors:
For a set of $n$ measurements $a_1, a_2, \dots, a_n$ of a physical quantity:
- Arithmetic Mean (Best Estimate of True Value):
$a_{\text{mean}} = \bar{a} = \frac{1}{n}\sum_{i=1}^{n} a_i$ - Absolute Error ($\Delta a_i$):
$\Delta a_i = |a_{\text{mean}} – a_i|$ (always reported as a positive magnitude) - Mean Absolute Error ($\Delta a_{\text{mean}}$):
$\Delta a_{\text{mean}} = \frac{1}{n}\sum_{i=1}^{n} \Delta a_i$
Final reported measurement: $a = a_{\text{mean}} \pm \Delta a_{\text{mean}}$ - Relative / Fractional Error:
$\text{Relative Error} = \frac{\Delta a_{\text{mean}}}{a_{\text{mean}}}$ - Percentage Error ($\%E$):
$\%E = \left(\frac{\Delta a_{\text{mean}}}{a_{\text{mean}}}\right) \times 100\%$
3. Rules for Combination and Propagation of Errors:
- Sum and Difference: If $Z = A + B$ or $Z = A – B$, absolute errors always add:
$\Delta Z = \Delta A + \Delta B$
Maximum relative error: $\frac{\Delta Z}{Z} = \frac{\Delta A + \Delta B}{A \pm B}$ - Product and Quotient: If $Z = A \cdot B$ or $Z = \frac{A}{B}$, relative errors add:
$\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$
$\%Z = \%A + \%B$ - Generalized Power Rule: If $Z = \frac{A^p B^q}{C^r}$:
$\frac{\Delta Z}{Z} = p\left(\frac{\Delta A}{A}\right) + q\left(\frac{\Delta B}{B}\right) + r\left(\frac{\Delta C}{C}\right)$
$\%Z = p(\%A) + q(\%B) + r(\%C)$ - Calculus Method for Non-Linear Equations: If $Z = f(x, y)$:
$\Delta Z_{\text{max}} = \left|\frac{\partial f}{\partial x}\right|\Delta x + \left|\frac{\partial f}{\partial y}\right|\Delta y$
Solved Examples
Example 1 (Mean Absolute Error & Percentage Error Calculation):
The time period of oscillation of a simple pendulum is recorded in 5 successive measurements as: $2.63\text{ s}$, $2.56\text{ s}$, $2.42\text{ s}$, $2.71\text{ s}$, and $2.80\text{ s}$. Determine the mean period, mean absolute error, relative error, and percentage error.
Solution:
1. Mean time period:
$T_{\text{mean}} = \frac{2.63 + 2.56 + 2.42 + 2.71 + 2.80}{5} = \frac{13.12}{5} = 2.624\text{ s} \approx 2.62\text{ s}$
2. Absolute errors in each reading:
$|\Delta T_1| = |2.62 – 2.63| = 0.01\text{ s}$
$|\Delta T_2| = |2.62 – 2.56| = 0.06\text{ s}$
$|\Delta T_3| = |2.62 – 2.42| = 0.20\text{ s}$
$|\Delta T_4| = |2.62 – 2.71| = 0.09\text{ s}$
$|\Delta T_5| = |2.62 – 2.80| = 0.18\text{ s}$
3. Mean absolute error:
$\Delta T_{\text{mean}} = \frac{0.01 + 0.06 + 0.20 + 0.09 + 0.18}{5} = \frac{0.54}{5} = 0.108\text{ s} \approx 0.11\text{ s}$
Reported period: $T = (2.62 \pm 0.11)\text{ s}$.
4. Relative error and percentage error:
$\text{Relative Error} = \frac{0.11}{2.62} = 0.04198 \approx 0.042$
$\text{Percentage Error} = 0.04198 \times 100\% = 4.2\%$.
Example 2 (Errors in Parallel Combination of Resistors):
Two resistors $R_1 = (100 \pm 3)\,\Omega$ and $R_2 = (200 \pm 4)\,\Omega$ are connected in parallel. Determine the equivalent resistance and the maximum percentage error in the equivalent resistance.
Solution:
1. Equivalent resistance $R_p$:
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \implies R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{100 \times 200}{100 + 200} = \frac{20000}{300} = 66.7\,\Omega$
2. Error formula using differentiation:
Differentiating $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$:
$-\frac{\Delta R_p}{R_p^2} = -\frac{\Delta R_1}{R_1^2} – \frac{\Delta R_2}{R_2^2} \implies \frac{\Delta R_p}{R_p} = R_p \left[\frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}\right]$
Substitute the numerical values:
$\frac{\Delta R_p}{R_p} = 66.7 \times \left[\frac{3}{(100)^2} + \frac{4}{(200)^2}\right] = 66.7 \times \left[\frac{3}{10000} + \frac{4}{40000}\right] = 66.7 \times \left[3 \times 10^{-4} + 1 \times 10^{-4}\right]$
$\frac{\Delta R_p}{R_p} = 66.7 \times 4 \times 10^{-4} = 0.02668$
3. Maximum percentage error:
$\%R_p = 0.02668 \times 100\% \approx 2.7\%$.
Absolute error: $\Delta R_p = 0.02668 \times 66.7\,\Omega \approx 1.8\,\Omega$.
Reported value: $R_p = (66.7 \pm 1.8)\,\Omega$.
Example 3 (Density Calculation with Propagation of Errors):
A rectangular metal block has mass $m = (50.0 \pm 0.5)\text{ g}$, length $l = (10.0 \pm 0.1)\text{ cm}$, breadth $b = (5.0 \pm 0.1)\text{ cm}$, and thickness $h = (2.0 \pm 0.1)\text{ cm}$. Find the maximum percentage error in the density of the metal block.
Solution:
Density is given by $\rho = \frac{m}{V} = \frac{m}{l \cdot b \cdot h}$.
Using the product and quotient error rule:
$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta l}{l} + \frac{\Delta b}{b} + \frac{\Delta h}{h}$
Calculate individual fractional errors:
$\frac{\Delta m}{m} = \frac{0.5}{50.0} = 0.01 = 1.0\%$
$\frac{\Delta l}{l} = \frac{0.1}{10.0} = 0.01 = 1.0\%$
$\frac{\Delta b}{b} = \frac{0.1}{5.0} = 0.02 = 2.0\%$
$\frac{\Delta h}{h} = \frac{0.1}{2.0} = 0.05 = 5.0\%$
Total percentage error:
$\%\rho = 1.0\% + 1.0\% + 2.0\% + 5.0\% = 9.0\%$.
Example 4 (Determination of $g$ using Simple Pendulum):
In an experiment to determine acceleration due to gravity $g$ using a simple pendulum, $T = 2\pi \sqrt{\frac{L}{g}}$. The measured length is $L = (100.0 \pm 0.1)\text{ cm}$, and the time for $20$ oscillations is measured as $t = (40.0 \pm 0.2)\text{ s}$ using a stopwatch. Find the percentage error in the measured value of $g$.
Solution:
From $T = 2\pi \sqrt{\frac{L}{g}} \implies g = 4\pi^2 \frac{L}{T^2}$.
Since $T = \frac{t}{n}$ where $n = 20$ is an exact count, $\frac{\Delta T}{T} = \frac{\Delta t}{t}$.
Thus, the fractional error formula is:
$\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\left(\frac{\Delta t}{t}\right)$
Substitute the values:
$\frac{\Delta L}{L} = \frac{0.1\text{ cm}}{100.0\text{ cm}} = 0.001 = 0.1\%$
$\frac{\Delta t}{t} = \frac{0.2\text{ s}}{40.0\text{ s}} = 0.005 = 0.5\%$
$\frac{\Delta g}{g} = 0.1\% + 2(0.5\%) = 0.1\% + 1.0\% = 1.1\%$.
The maximum percentage error in $g$ is $1.1\%$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A physical quantity $P$ is determined from the relation $P = \frac{A^3 B^2}{\sqrt{C} \cdot D}$. The percentage errors in measuring $A, B, C$, and $D$ are $1\%$, $2\%$, $4\%$, and $3\%$ respectively. The maximum percentage error in $P$ is:
(A) $12\%$
(B) $14\%$
(C) $10\%$
(D) $8\%$
Problem 2 (JEE Main – Single Correct):
A student measures the diameter of a wire using a screw gauge with least count $0.001\text{ cm}$ and lists measurements as $0.050\text{ cm}$, $0.052\text{ cm}$, $0.051\text{ cm}$, and $0.051\text{ cm}$. The relative error in the diameter is approximately:
(A) $0.01$
(B) $0.02$
(C) $0.05$
(D) $0.005$
Problem 3 (JEE Main – Single Correct):
The resistance of a wire is given by $R = \frac{V}{I}$. If $V = (50 \pm 2)\text{ V}$ and $I = (5 \pm 0.2)\text{ A}$, the percentage error in $R$ is:
(A) $4\%$
(B) $8\%$
(C) $6\%$
(D) $2\%$
Problem 4 (JEE Main – Single Correct):
The error in the measurement of the radius of a solid sphere is $0.8\%$. The maximum percentage error in the calculated volume of the sphere is:
(A) $0.8\%$
(B) $1.6\%$
(C) $2.4\%$
(D) $3.2\%$
Problem 5 (JEE Main – Single Correct):
Two temperatures measured by a thermometer are $T_1 = (20.0 \pm 0.4)^\circ\text{C}$ and $T_2 = (45.0 \pm 0.3)^\circ\text{C}$. The temperature difference $\Delta T = T_2 – T_1$ along with its absolute error should be stated as:
(A) $(25.0 \pm 0.1)^\circ\text{C}$
(B) $(25.0 \pm 0.7)^\circ\text{C}$
(C) $(25.0 \pm 0.35)^\circ\text{C}$
(D) $(25.0 \pm 0.5)^\circ\text{C}$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements is/are TRUE regarding errors in physical measurements?
(A) Systematic errors cannot be minimized by simply taking a large number of readings.
(B) Random errors are caused by unknown and unpredictable causes.
(C) The fractional error is a dimensionless quantity.
(D) If $Z = A – B$, the relative error in $Z$ is always smaller than the relative error in $A$.
Problem 7 (JEE Advanced – One or More Correct):
A physical quantity $Q$ is given by $Q = \frac{A – B}{A + B}$, where $A$ and $B$ are independently measured positive quantities with absolute errors $\Delta A$ and $\Delta B$. Which of the following expressions represent the maximum fractional error $\frac{\Delta Q}{Q}$?
(A) $\frac{\Delta A + \Delta B}{A – B} + \frac{\Delta A + \Delta B}{A + B}$
(B) $\frac{2(B \Delta A + A \Delta B)}{A^2 – B^2}$
(C) $\frac{\Delta A}{A} + \frac{\Delta B}{B}$
(D) $\frac{\Delta A – \Delta B}{A + B}$
Problem 8 (JEE Advanced – One or More Correct):
The heat generated in a resistor is given by $H = I^2 R t$. The percentage errors in the measurement of electric current $I$, resistance $R$, and time $t$ are $2\%$, $1\%$, and $1\%$ respectively. Which of the following statements is/are correct?
(A) The maximum percentage error in $H$ is $6\%$.
(B) The measurement of current $I$ contributes the maximum percentage uncertainty to $H$.
(C) If the percentage error in $I$ is reduced to $1\%$, the error in $H$ becomes $4\%$.
(D) The percentage error in $H$ is independent of the value of resistance $R$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
The velocity of a transverse wave in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension in the string and $\mu$ is the linear mass density. If the percentage error in tension is $3\%$ and the percentage error in linear mass density is $1\%$, the maximum percentage error in the wave speed is $x\%$. Find the value of $10 \times x$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
The focal length of a thin convex lens is determined using the lens formula $\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$. The object distance is measured as $u = -(20 \pm 0.2)\text{ cm}$ and the real image distance is measured as $v = +(30 \pm 0.3)\text{ cm}$. The percentage error in the focal length $f$ is $p\%$. Find the value of $10 \times p$.
Solutions & Explanations
Answer Key Summary:
1. (A) | 2. (A) | 3. (B) | 4. (C) | 5. (B) | 6. (A, B, C) | 7. (A, B) | 8. (A, B, C, D) | 9. 20 | 10. 10
Solution 1:
$P = \frac{A^3 B^2}{C^{1/2} D}$.
$\frac{\Delta P}{P} = 3\left(\frac{\Delta A}{A}\right) + 2\left(\frac{\Delta B}{B}\right) + \frac{1}{2}\left(\frac{\Delta C}{C}\right) + \left(\frac{\Delta D}{D}\right)$.
$\%P = 3(1\%) + 2(2\%) + \frac{1}{2}(4\%) + 3\% = 3\% + 4\% + 2\% + 3\% = 12\%$.
Correct Answer: (A)
Solution 2:
Mean diameter $d_{\text{mean}} = \frac{0.050 + 0.052 + 0.051 + 0.051}{4} = \frac{0.204}{4} = 0.051\text{ cm}$.
Absolute errors:
$|\Delta d_1| = |0.051 – 0.050| = 0.001\text{ cm}$
$|\Delta d_2| = |0.051 – 0.052| = 0.001\text{ cm}$
$|\Delta d_3| = |0.051 – 0.051| = 0.000\text{ cm}$
$|\Delta d_4| = |0.051 – 0.051| = 0.000\text{ cm}$
Mean absolute error $\Delta d_{\text{mean}} = \frac{0.001 + 0.001 + 0 + 0}{4} = 0.0005\text{ cm}$.
$\text{Relative error} = \frac{0.0005}{0.051} \approx 0.0098 \approx 0.01$.
Correct Answer: (A)
Solution 3:
$R = \frac{V}{I} \implies \frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$.
$\frac{\Delta V}{V} = \frac{2}{50} = 4\%$
$\frac{\Delta I}{I} = \frac{0.2}{5} = 4\%$
$\%R = 4\% + 4\% = 8\%$.
Correct Answer: (B)
Solution 4:
Volume of sphere $V = \frac{4}{3}\pi r^3 \implies \frac{\Delta V}{V} = 3\left(\frac{\Delta r}{r}\right)$.
$\%V = 3 \times 0.8\% = 2.4\%$.
Correct Answer: (C)
Solution 5:
For subtraction $\Delta T = T_2 – T_1 = 45.0 – 20.0 = 25.0^\circ\text{C}$.
Absolute errors in difference always add:
$\Delta(\Delta T) = \Delta T_1 + \Delta T_2 = 0.4 + 0.3 = 0.7^\circ\text{C}$.
Reported value: $(25.0 \pm 0.7)^\circ\text{C}$.
Correct Answer: (B)
Solution 6:
– (A) True: Systematic errors are unidirectional and constant; averaging does not eliminate them.
– (B) True: Random errors fluctuate symmetrically and follow Gaussian distribution.
– (C) True: Fractional error is a ratio of identical quantities $\frac{\Delta x}{x}$, hence dimensionless.
– (D) False: Since denominator $A – B$ can be very small, relative error $\frac{\Delta A + \Delta B}{A – B}$ can be significantly larger.
Correct Answer: (A, B, C)
Solution 7:
Taking natural logarithm of $Q = \frac{A – B}{A + B}$:
$\ln Q = \ln(A – B) – \ln(A + B)$
Differentiating:
$\frac{dQ}{Q} = \frac{dA – dB}{A – B} – \frac{dA + dB}{A + B}$
For maximum error, add the absolute contributions:
$\frac{\Delta Q}{Q} = \frac{\Delta A + \Delta B}{A – B} + \frac{\Delta A + \Delta B}{A + B}$ → (A) is correct.
Combining the terms over a common denominator:
$\frac{\Delta Q}{Q} = \frac{(\Delta A + \Delta B)(A + B) + (\Delta A + \Delta B)(A – B)}{A^2 – B^2} = \frac{2 A \Delta A + 2 A \Delta B \dots}{A^2 – B^2}$
Alternatively, differentiating directly:
$dQ = \frac{(A + B)(dA – dB) – (A – B)(dA + dB)}{(A + B)^2} = \frac{2 B \, dA – 2 A \, dB}{(A + B)^2}$
Dividing by $Q = \frac{A – B}{A + B}$:
$\frac{dQ}{Q} = \frac{2(B \, dA – A \, dB)}{(A – B)(A + B)} \implies \frac{\Delta Q_{\text{max}}}{Q} = \frac{2(B \Delta A + A \Delta B)}{A^2 – B^2}$ → (B) is also correct.
Correct Answer: (A, B)
Solution 8:
$H = I^2 R t \implies \frac{\Delta H}{H} = 2\left(\frac{\Delta I}{I}\right) + \frac{\Delta R}{R} + \frac{\Delta t}{t}$.
$\%H = 2(2\%) + 1\% + 1\% = 4\% + 1\% + 1\% = 6\%$.
– Current has power $2$, contributing $4\%$ out of $6\%$ total (maximum contributor).
– If $\%I = 1\%$, $\%H = 2(1\%) + 1\% + 1\% = 4\%$.
– Percentage error depends only on relative errors, not the absolute magnitude of $R$.
All options are correct.
Correct Answer: (A, B, C, D)
Solution 9:
$v = T^{1/2} \mu^{-1/2}$.
$\frac{\Delta v}{v} = \frac{1}{2}\left(\frac{\Delta T}{T}\right) + \frac{1}{2}\left(\frac{\Delta \mu}{\mu}\right)$.
$\%v = \frac{1}{2}(3\%) + \frac{1}{2}(1\%) = 1.5\% + 0.5\% = 2.0\%$.
So $x = 2.0 \implies 10 \times x = 10 \times 2 = 20$.
Correct Answer: 20
Solution 10:
For lens formula $\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$:
With $u = -20\text{ cm}$ and $v = +30\text{ cm}$:
$\frac{1}{f} = \frac{1}{30} – \left(-\frac{1}{20}\right) = \frac{1}{30} + \frac{1}{20} = \frac{2 + 3}{60} = \frac{5}{60} = \frac{1}{12} \implies f = 12\text{ cm}$.
Differentiating $\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$:
$-\frac{df}{f^2} = -\frac{dv}{v^2} + \frac{du}{u^2} \implies \frac{\Delta f}{f} = f \left[\frac{\Delta v}{v^2} + \frac{\Delta u}{u^2}\right]$
Substitute $|u| = 20$, $\Delta u = 0.2$, $v = 30$, $\Delta v = 0.3$, $f = 12$:
$\frac{\Delta f}{f} = 12 \left[\frac{0.3}{900} + \frac{0.2}{400}\right] = 12 \left[\frac{1}{3000} + \frac{1}{2000}\right] = 12 \left[\frac{2 + 3}{6000}\right] = 12 \times \frac{5}{6000} = \frac{60}{6000} = 0.01 = 1.0\%$.
Thus, percentage error $p = 1.0\% \implies 10 \times p = 10 \times 1.0 = 10$.
Correct Answer: 10