This MCQ module is based on: Errors in Measurement
TOPIC 2 OF 33
Errors in Measurement
🎓 Class 11
Physics
CBSE
Theory
Ch 1 – Units and Measurements
⏱ ~14 min
🌐 Language: [gtranslate]
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Errors in Measurement
Errors in Measurement
No measurement is perfectly accurate. Every measured value differs from the true value by some amount called the error. Understanding errors is crucial because it tells us how much confidence we can place in our results.
Types of Errors
| Error Type | Description | Example |
|---|---|---|
| Systematic Errors | Errors that consistently shift measurements in one direction (always too high or always too low). They have a definite cause. | A zero-error in a vernier caliper; a thermometer that always reads 2°C too high. |
| Random Errors | Errors that fluctuate unpredictably from measurement to measurement due to unknown or uncontrollable factors. | Slight variations in reaction time when using a stopwatch; fluctuations due to small air currents. |
| Least Count Error | The smallest measurement possible with a given instrument. It sets the minimum uncertainty. | A metre ruler with mm markings has a least count of 1 mm; readings are uncertain by ±0.5 mm. |
Systematic errors sub-types:
(i) Instrumental errors — arise from faulty calibration or defects in the instrument.
(ii) Personal errors — arise from the observer's habits (parallax, bias in reading).
(iii) Environmental errors — arise from external conditions (temperature, humidity, vibrations).
(i) Instrumental errors — arise from faulty calibration or defects in the instrument.
(ii) Personal errors — arise from the observer's habits (parallax, bias in reading).
(iii) Environmental errors — arise from external conditions (temperature, humidity, vibrations).
Reducing Errors: Systematic errors can be minimised by using better instruments and improving experimental technique. Random errors are reduced by taking a large number of measurements and computing their mean.
Absolute Error, Relative Error, and Percentage Error
Suppose a physical quantity is measured \(n\) times, yielding values \(a_1, a_2, a_3, \ldots, a_n\). The arithmetic mean (best estimate of the true value) is:
\[a_{\text{mean}} = \bar{a} = \frac{1}{n}\sum_{i=1}^{n} a_i = \frac{a_1 + a_2 + \cdots + a_n}{n}\]
Absolute Error
The absolute error in each measurement is the magnitude of the difference between the individual measurement and the mean:
\[\Delta a_i = |a_i - \bar{a}|\]
Mean Absolute Error
The average of all the absolute errors gives the mean absolute error:
\[\Delta \bar{a} = \frac{1}{n}\sum_{i=1}^{n}|\Delta a_i| = \frac{|\Delta a_1| + |\Delta a_2| + \cdots + |\Delta a_n|}{n}\]
The measurement is then reported as: \(a = \bar{a} \pm \Delta\bar{a}\)
Relative Error and Percentage Error
\[\text{Relative Error} = \frac{\Delta\bar{a}}{\bar{a}}\]
\[\text{Percentage Error} = \frac{\Delta\bar{a}}{\bar{a}} \times 100\%\]
Worked Examples — Error Analysis
Example 1: Complete Error Analysis from Repeated Measurements
The diameter of a wire is measured five times with a screw gauge. The readings are: 3.11 mm, 3.15 mm, 3.14 mm, 3.12 mm, 3.16 mm. Find the mean, absolute errors, mean absolute error, relative error, and percentage error.
Step 1: Calculate the mean
\[\bar{d} = \frac{3.11 + 3.15 + 3.14 + 3.12 + 3.16}{5} = \frac{15.68}{5} = 3.136 \text{ mm}\] Rounding to 2 decimal places: \(\bar{d} = 3.14\) mm.
Step 2: Calculate absolute errors
\[\begin{aligned} |\Delta d_1| &= |3.11 - 3.14| = 0.03 \text{ mm}\\ |\Delta d_2| &= |3.15 - 3.14| = 0.01 \text{ mm}\\ |\Delta d_3| &= |3.14 - 3.14| = 0.00 \text{ mm}\\ |\Delta d_4| &= |3.12 - 3.14| = 0.02 \text{ mm}\\ |\Delta d_5| &= |3.16 - 3.14| = 0.02 \text{ mm} \end{aligned}\] Step 3: Mean absolute error
\[\Delta\bar{d} = \frac{0.03 + 0.01 + 0.00 + 0.02 + 0.02}{5} = \frac{0.08}{5} = 0.016 \approx 0.02 \text{ mm}\] Step 4: Relative error
\[\frac{\Delta\bar{d}}{\bar{d}} = \frac{0.02}{3.14} = 0.0064\] Step 5: Percentage error
\[\frac{\Delta\bar{d}}{\bar{d}} \times 100\% = 0.0064 \times 100\% = \boxed{0.64\%}\] Result: The diameter is \(d = 3.14 \pm 0.02\) mm with a percentage error of 0.64%.
\[\bar{d} = \frac{3.11 + 3.15 + 3.14 + 3.12 + 3.16}{5} = \frac{15.68}{5} = 3.136 \text{ mm}\] Rounding to 2 decimal places: \(\bar{d} = 3.14\) mm.
Step 2: Calculate absolute errors
\[\begin{aligned} |\Delta d_1| &= |3.11 - 3.14| = 0.03 \text{ mm}\\ |\Delta d_2| &= |3.15 - 3.14| = 0.01 \text{ mm}\\ |\Delta d_3| &= |3.14 - 3.14| = 0.00 \text{ mm}\\ |\Delta d_4| &= |3.12 - 3.14| = 0.02 \text{ mm}\\ |\Delta d_5| &= |3.16 - 3.14| = 0.02 \text{ mm} \end{aligned}\] Step 3: Mean absolute error
\[\Delta\bar{d} = \frac{0.03 + 0.01 + 0.00 + 0.02 + 0.02}{5} = \frac{0.08}{5} = 0.016 \approx 0.02 \text{ mm}\] Step 4: Relative error
\[\frac{\Delta\bar{d}}{\bar{d}} = \frac{0.02}{3.14} = 0.0064\] Step 5: Percentage error
\[\frac{\Delta\bar{d}}{\bar{d}} \times 100\% = 0.0064 \times 100\% = \boxed{0.64\%}\] Result: The diameter is \(d = 3.14 \pm 0.02\) mm with a percentage error of 0.64%.
Combination of Errors
When we compute a derived quantity from measured values, how do the individual errors combine? The rules depend on whether the quantities are added, multiplied, or raised to powers.
Error in Sum or Difference
If \(Z = A + B\) or \(Z = A - B\), then the maximum absolute error in \(Z\) is:
\[\Delta Z = \Delta A + \Delta B\]
Important: When two quantities are added or subtracted, their absolute errors add up, regardless of whether the operation is addition or subtraction.
Error in Product or Quotient
If \(Z = A \times B\) or \(Z = \frac{A}{B}\), then the maximum relative error in \(Z\) is:
\[\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}\]
Error in a Quantity Raised to a Power
If \(Z = A^n\), then:
\[\frac{\Delta Z}{Z} = n\,\frac{\Delta A}{A}\]
General Rule for Combined Errors
For a general formula \(Z = \dfrac{A^p \cdot B^q}{C^r}\), the maximum fractional (relative) error is:
\[\frac{\Delta Z}{Z} = p\,\frac{\Delta A}{A} + q\,\frac{\Delta B}{B} + r\,\frac{\Delta C}{C}\]
Example 2: Error in Period Measurement
A student measures the total time for 20 oscillations of a pendulum as \(t = 39.6 \pm 0.2\) s. What is the percentage error in the time period \(T\)?
Solution:
The time period \(T = \frac{t}{n}\), where \(n = 20\) (exact number, no error).
\[T = \frac{39.6}{20} = 1.98 \text{ s}\] \[\Delta T = \frac{\Delta t}{n} = \frac{0.2}{20} = 0.01 \text{ s}\] \[\text{Percentage error in } T = \frac{\Delta T}{T} \times 100\% = \frac{0.01}{1.98} \times 100\% = \boxed{0.51\%}\]
The time period \(T = \frac{t}{n}\), where \(n = 20\) (exact number, no error).
\[T = \frac{39.6}{20} = 1.98 \text{ s}\] \[\Delta T = \frac{\Delta t}{n} = \frac{0.2}{20} = 0.01 \text{ s}\] \[\text{Percentage error in } T = \frac{\Delta T}{T} \times 100\% = \frac{0.01}{1.98} \times 100\% = \boxed{0.51\%}\]
Example 3: Error in Density (Product/Quotient of Powers)
The density of a material is calculated using \(\rho = \frac{M}{\pi r^2 L}\). If the percentage errors in measuring \(M\), \(r\), and \(L\) are 1%, 0.5%, and 2% respectively, find the maximum percentage error in \(\rho\).
Solution:
\[\rho = \frac{M}{\pi r^2 L}\] Taking the general error formula: \(\frac{\Delta\rho}{\rho} = \frac{\Delta M}{M} + 2\frac{\Delta r}{r} + \frac{\Delta L}{L}\)
(Note: \(\pi\) is exact, so no error from it. The power of \(r\) is 2, so its error is multiplied by 2.)
\[\frac{\Delta\rho}{\rho} \times 100\% = 1\% + 2(0.5\%) + 2\% = 1\% + 1\% + 2\% = \boxed{4\%}\]
\[\rho = \frac{M}{\pi r^2 L}\] Taking the general error formula: \(\frac{\Delta\rho}{\rho} = \frac{\Delta M}{M} + 2\frac{\Delta r}{r} + \frac{\Delta L}{L}\)
(Note: \(\pi\) is exact, so no error from it. The power of \(r\) is 2, so its error is multiplied by 2.)
\[\frac{\Delta\rho}{\rho} \times 100\% = 1\% + 2(0.5\%) + 2\% = 1\% + 1\% + 2\% = \boxed{4\%}\]
Example 4: Error in a Complex Expression
A physical quantity \(X\) is related to other measurable quantities by: \(X = \frac{a^2 b^3}{c\sqrt{d}}\). If the percentage errors in \(a, b, c, d\) are 1%, 2%, 3%, and 4% respectively, find the percentage error in \(X\).
Solution:
Rewriting: \(X = a^2 \cdot b^3 \cdot c^{-1} \cdot d^{-1/2}\)
Applying the general formula:
\[\frac{\Delta X}{X} \times 100\% = 2\left(\frac{\Delta a}{a}\right) + 3\left(\frac{\Delta b}{b}\right) + 1\left(\frac{\Delta c}{c}\right) + \frac{1}{2}\left(\frac{\Delta d}{d}\right)\] \[= 2(1\%) + 3(2\%) + 1(3\%) + \frac{1}{2}(4\%)\] \[= 2\% + 6\% + 3\% + 2\% = \boxed{13\%}\]
Rewriting: \(X = a^2 \cdot b^3 \cdot c^{-1} \cdot d^{-1/2}\)
Applying the general formula:
\[\frac{\Delta X}{X} \times 100\% = 2\left(\frac{\Delta a}{a}\right) + 3\left(\frac{\Delta b}{b}\right) + 1\left(\frac{\Delta c}{c}\right) + \frac{1}{2}\left(\frac{\Delta d}{d}\right)\] \[= 2(1\%) + 3(2\%) + 1(3\%) + \frac{1}{2}(4\%)\] \[= 2\% + 6\% + 3\% + 2\% = \boxed{13\%}\]
Example 5: Error in Resistance
In an experiment, the resistance \(R\) is determined from \(V = IR\), i.e. \(R = V/I\). If \(V = 100 \pm 5\) V and \(I = 10 \pm 0.2\) A, find the percentage error in \(R\) and express the result.
Solution:
\[R = \frac{V}{I} = \frac{100}{10} = 10\;\Omega\] Percentage error in \(V\): \(\frac{5}{100}\times 100\% = 5\%\)
Percentage error in \(I\): \(\frac{0.2}{10}\times 100\% = 2\%\)
For a quotient: \(\frac{\Delta R}{R}\times 100\% = 5\% + 2\% = 7\%\)
\(\Delta R = 7\% \text{ of } 10 = 0.7\;\Omega\)
\[\boxed{R = 10 \pm 0.7\;\Omega \quad\text{(percentage error = 7\%)}}\]
\[R = \frac{V}{I} = \frac{100}{10} = 10\;\Omega\] Percentage error in \(V\): \(\frac{5}{100}\times 100\% = 5\%\)
Percentage error in \(I\): \(\frac{0.2}{10}\times 100\% = 2\%\)
For a quotient: \(\frac{\Delta R}{R}\times 100\% = 5\% + 2\% = 7\%\)
\(\Delta R = 7\% \text{ of } 10 = 0.7\;\Omega\)
\[\boxed{R = 10 \pm 0.7\;\Omega \quad\text{(percentage error = 7\%)}}\]
Accuracy vs Precision
Accuracy and precision are two distinct concepts that are often confused:
- Accuracy refers to how close a measured value is to the true value.
- Precision refers to how close repeated measurements are to each other.
A helpful analogy is target shooting. Imagine four different scenarios:
Activity — Investigating Accuracy and Precision
L4 Analyse
Predict first: If you use two different stopwatches to measure the period of a pendulum 10 times each, which source of error (systematic or random) would cause a consistent offset in the mean value?
- Set up a simple pendulum of known length (say, 1 m).
- Calculate the expected period using \(T = 2\pi\sqrt{l/g}\) (theoretical value).
- Measure the time for 20 oscillations using a digital stopwatch. Repeat 5 times.
- Compute the mean, absolute errors, and percentage error.
- Compare your mean value with the theoretical value to assess accuracy.
- Look at how close your 5 readings are to each other to assess precision.
Analysis:
For \(l = 1\) m: \(T_{\text{theory}} = 2\pi\sqrt{1/9.8} \approx 2.007\) s.
If your mean is close to 2.01 s, your measurement is accurate.
If all 5 readings are within 0.02 s of each other, your measurement is precise.
If the mean is, say, 2.10 s (consistently off), you likely have a systematic error — perhaps in measuring the length, or in the reaction time being biased.
For \(l = 1\) m: \(T_{\text{theory}} = 2\pi\sqrt{1/9.8} \approx 2.007\) s.
If your mean is close to 2.01 s, your measurement is accurate.
If all 5 readings are within 0.02 s of each other, your measurement is precise.
If the mean is, say, 2.10 s (consistently off), you likely have a systematic error — perhaps in measuring the length, or in the reaction time being biased.
Interactive: Error Calculator L3 Apply
Enter up to 5 repeated measurements and compute the error analysis automatically:
Competency-Based Questions
In a physics laboratory, a student measures the radius of a metal sphere using a vernier caliper (least count 0.01 cm). Five readings are taken: 2.15 cm, 2.13 cm, 2.14 cm, 2.16 cm, 2.12 cm. The student then needs to calculate the volume of the sphere and determine the uncertainty in the result.
Q1. L1 Remember Define absolute error and relative error.
Answer: Absolute error is the magnitude of the difference between a measured value and the mean value: \(|\Delta a_i| = |a_i - \bar{a}|\). Relative error is the ratio of mean absolute error to the mean value: \(\frac{\Delta\bar{a}}{\bar{a}}\). It is a dimensionless quantity that indicates the fractional uncertainty in the measurement.
Q2. L3 Apply Using the five readings given above, calculate the mean radius, mean absolute error, and percentage error in the radius measurement. (3 marks)
Answer:
Mean: \(\bar{r} = \frac{2.15+2.13+2.14+2.16+2.12}{5} = \frac{10.70}{5} = 2.14\) cm
Absolute errors: |0.01|, |0.01|, |0.00|, |0.02|, |0.02| cm
Mean absolute error: \(\Delta\bar{r} = \frac{0.01+0.01+0.00+0.02+0.02}{5} = \frac{0.06}{5} = 0.012 \approx 0.01\) cm
Percentage error: \(\frac{0.01}{2.14}\times 100\% = 0.47\%\)
Mean: \(\bar{r} = \frac{2.15+2.13+2.14+2.16+2.12}{5} = \frac{10.70}{5} = 2.14\) cm
Absolute errors: |0.01|, |0.01|, |0.00|, |0.02|, |0.02| cm
Mean absolute error: \(\Delta\bar{r} = \frac{0.01+0.01+0.00+0.02+0.02}{5} = \frac{0.06}{5} = 0.012 \approx 0.01\) cm
Percentage error: \(\frac{0.01}{2.14}\times 100\% = 0.47\%\)
Q3. L3 Apply The volume of a sphere is \(V = \frac{4}{3}\pi r^3\). What is the percentage error in the volume? (2 marks)
Answer:
Since \(V \propto r^3\), the percentage error in \(V\) is:
\[\frac{\Delta V}{V}\times 100\% = 3 \times \frac{\Delta r}{r}\times 100\% = 3 \times 0.47\% = \boxed{1.41\%}\] The power of \(r\) is 3, so the error is magnified threefold.
Since \(V \propto r^3\), the percentage error in \(V\) is:
\[\frac{\Delta V}{V}\times 100\% = 3 \times \frac{\Delta r}{r}\times 100\% = 3 \times 0.47\% = \boxed{1.41\%}\] The power of \(r\) is 3, so the error is magnified threefold.
Q4. L4 Analyse A student finds that when using Instrument A, the measurements are 5.10, 5.11, 5.09, 5.10, 5.10 g. With Instrument B: 5.00, 5.20, 4.90, 5.15, 4.95 g. Both instruments are measuring the same object whose true mass is 5.10 g. Compare the accuracy and precision of the two instruments. (3 marks)
Answer:
Instrument A: Mean = 5.10 g (equals true value, so high accuracy). Readings vary by at most 0.01 g (high precision).
Instrument B: Mean = \(\frac{5.00+5.20+4.90+5.15+4.95}{5} = 5.04\) g (differs from true value by 0.06 g, so lower accuracy). Readings vary by up to 0.30 g (low precision).
Conclusion: Instrument A is both more accurate and more precise than Instrument B.
Instrument A: Mean = 5.10 g (equals true value, so high accuracy). Readings vary by at most 0.01 g (high precision).
Instrument B: Mean = \(\frac{5.00+5.20+4.90+5.15+4.95}{5} = 5.04\) g (differs from true value by 0.06 g, so lower accuracy). Readings vary by up to 0.30 g (low precision).
Conclusion: Instrument A is both more accurate and more precise than Instrument B.
Q5. L5 Evaluate A student claims: "Since systematic errors always push measurements in one direction, taking more readings and averaging will eliminate systematic errors." Is this claim correct? Justify your answer. (3 marks)
Answer: The claim is incorrect. Averaging many readings reduces random errors (which fluctuate about the mean) but does not eliminate systematic errors. Since systematic errors consistently shift all readings in the same direction, the mean will also be shifted by the same amount. To reduce systematic errors, one must recalibrate the instrument, change the experimental technique, or apply a correction factor.
Assertion-Reason Questions
Assertion (A): When two physical quantities are multiplied, the percentage error in the result is the sum of the individual percentage errors.
Reason (R): Errors always add up to give the maximum possible error in the result.
Answer: A. For a product \(Z = AB\), \(\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}\), so percentage errors add. This is because we consider the maximum possible error, which occurs when individual errors compound in the same direction.
Assertion (A): The error in a quantity raised to a power is multiplied by that power.
Reason (R): If \(Z = A^n\), then \(\frac{\Delta Z}{Z} = n\frac{\Delta A}{A}\).
Answer: A. Both are true. The mathematical derivation (using logarithmic differentiation) shows that \(\frac{\Delta Z}{Z} = n\frac{\Delta A}{A}\), which means the relative error gets multiplied by the power. The Reason is the mathematical statement of the Assertion.
Assertion (A): A measurement can be precise without being accurate.
Reason (R): Precision depends on random errors while accuracy depends on systematic errors.
Answer: A. A systematic error (like a miscalibrated instrument) can cause all measurements to be consistently shifted from the true value — the readings will be close to each other (precise) but far from the true value (inaccurate). The Reason correctly explains why this happens.
Frequently Asked Questions - Errors in Measurement
What is the main concept covered in Errors in Measurement?
In NCERT Class 11 Physics Chapter 1 (Units and Measurements), "Errors in Measurement" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Errors in Measurement useful in real-life applications?
Real-life applications of Errors in Measurement from NCERT Class 11 Physics Chapter 1 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Errors in Measurement?
Key formulas in Errors in Measurement (NCERT Class 11 Physics Chapter 1 Units and Measurements) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 1?
NCERT Class 11 Physics Chapter 1 (Units and Measurements) is structured so each part builds on the previous one. Errors in Measurement connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Errors in Measurement?
CBSE board questions from Errors in Measurement typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Errors in Measurement lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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