Units and Measurements — Class 11 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Physics chapter "Units and Measurements" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Units and Measurements" — 8 important questions with detailed answers for CBSE board e…
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NCERT Solutions for Class 11 Physics — Units and Measurements. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Units and Measurements — All 8 Questions Solved
Q1. Define the SI unit of length and explain how the metre is defined.
The SI unit of length is the metre (m).
Definition: The metre is defined as the length of the path travelled by light in vacuum during a time interval of 1/299,792,458 of a second.
This definition was adopted at the 17th General Conference on Weights and Measures (CGPM) in 1983. It is based on the speed of light, which is a fundamental constant of nature (c = 299,792,458 m/s).
Advantages of this definition:
- It relates the metre to a fundamental physical constant
- It is reproducible in any well-equipped laboratory
- It has extremely high precision
- It does not depend on a physical artifact
Q2. A student measures the thickness of a single sheet of paper using a screw gauge. If the main scale reading is 0 mm and the circular scale reading is 45 divisions (least count = 0.01 mm), calculate the thickness of the paper.
Given:
Main scale reading (MSR) = 0 mm
Circular scale reading (CSR) = 45 divisions
Least count (LC) = 0.01 mm
Formula:
Reading = MSR + (CSR × LC)
Step 1: Identify the main scale reading
MSR = 0 mm
Step 2: Calculate the circular scale contribution
CSR contribution = 45 × 0.01 mm = 0.45 mm
Step 3: Add to get total reading
Thickness = 0 + 0.45 = 0.45 mm
Answer: The thickness of the paper is 0.45 mm or 4.5 × 10⁻⁴ m
Q3. Express the following in scientific notation with appropriate significant figures: (a) 0.000456 kg (b) 45000 m (c) 0.0050 A
Step 1: Identify significant figures and convert to scientific notation
(a) 0.000456 kg
- Significant figures: 3 (the zeros before 4 are not significant)
- Scientific notation: 4.56 × 10⁻⁴ kg
(b) 45000 m
- Significant figures: 2 (assuming only 4 and 5 are significant; trailing zeros without decimal point are ambiguous)
- Scientific notation: 4.5 × 10⁴ m
- If all zeros are significant: 4.5000 × 10⁴ m
(c) 0.0050 A
- Significant figures: 2 (the two zeros before 5 are not significant; the zero after 5 is significant)
- Scientific notation: 5.0 × 10⁻³ A
Answers:
(a) 4.56 × 10⁻⁴ kg (3 sf)
(b) 4.5 × 10⁴ m (2 sf)
(c) 5.0 × 10⁻³ A (2 sf)
Q4. The radius of a circular disc is measured as 5.21 ± 0.05 cm. Calculate the area of the disc with proper uncertainty.
Given:
Radius r = 5.21 ± 0.05 cm
Formula:
Area A = πr²
Step 1: Calculate central value of area
A = π × (5.21)²
A = π × 27.1441
A = 85.26 cm²
Step 2: Calculate relative error in radius
Δr/r = 0.05/5.21 = 0.0096 or 0.96%
Step 3: Calculate relative error in area
For A = πr², we have ΔA/A = 2(Δr/r)
ΔA/A = 2 × 0.0096 = 0.0192 or 1.92%
Step 4: Calculate absolute error in area
ΔA = A × (ΔA/A) = 85.26 × 0.0192 = 1.64 cm²
Step 5: Round to appropriate significant figures
ΔA ≈ 1.6 cm² or 2 cm² (rounded to 1 sf)
Final Answer: A = (85 ± 2) cm² or (85.3 ± 1.6) cm²
Q5. What is the difference between accuracy and precision? Give an example of each.
Accuracy and precision are related but different concepts in measurement:
Accuracy:
- It refers to how close a measured value is to the true or accepted value
- It is a measure of correctness
- It depends on the skill of the experimenter and the quality of the apparatus
- Example: If the true length of a rod is 10.00 cm and your measurement gives 10.02 cm, your measurement is accurate (within 0.02 cm of true value)
Precision:
- It refers to how close repeated measurements are to each other (reproducibility)
- It is a measure of consistency
- It depends on the least count and resolution of the measuring instrument
- Example: If you measure the same rod four times and get 10.12 cm, 10.13 cm, 10.12 cm, and 10.13 cm, your measurements are precise (all within 0.01 cm of each other) even if the true value is 10.00 cm
Key Difference:
- A measurement can be precise but not accurate (high consistency, but far from true value)
- A measurement can be accurate but not precise (close to true value, but with scattered readings)
- Ideal measurements are both accurate and precise
Q6. Using Vernier callipers with least count 0.01 cm, the main scale reading is 2.3 cm and Vernier scale reading is 7. Calculate the measurement and state the number of significant figures.
Given:
Main scale reading (MSR) = 2.3 cm
Vernier scale reading (VSR) = 7
Least count (LC) = 0.01 cm
Formula:
Total reading = MSR + (VSR × LC)
Step 1: Calculate Vernier contribution
Vernier contribution = 7 × 0.01 = 0.07 cm
Step 2: Add to main scale reading
Total reading = 2.3 + 0.07 = 2.37 cm
Step 3: Identify significant figures
- The main scale reading 2.3 cm has 2 significant figures
- The Vernier reading adds precision to the third decimal place
- Total significant figures: 3 (2, 3, and 7)
Answer: The measurement is 2.37 cm with 3 significant figures
Q7. What is meant by fundamental units and derived units? Give two examples of each and show how one derived unit is derived from fundamental units.
Fundamental Units:
Fundamental units are basic units that are defined independently and are not derived from any other unit. They are the foundation of any system of units.
Examples of fundamental units in SI system:
1. Metre (m) - unit of length
2. Kilogram (kg) - unit of mass
3. Second (s) - unit of time
4. Ampere (A) - unit of electric current
5. Kelvin (K) - unit of temperature
6. Mole (mol) - unit of amount of substance
7. Candela (cd) - unit of luminous intensity
Derived Units:
Derived units are obtained by combining fundamental units through multiplication and division. They are expressed in terms of fundamental units.
Examples of derived units:
1. Newton (N) - unit of force
2. Joule (J) - unit of energy
3. Watt (W) - unit of power
4. Pascal (Pa) - unit of pressure
5. Hertz (Hz) - unit of frequency
Derivation Example - Newton:
Force = mass × acceleration
F = m × a
[F] = [M] × [LT⁻²] = MLT⁻²
Therefore, 1 N = 1 kg⋅m⋅s⁻² = 1 kg⋅m/s²
This shows how the derived unit Newton is expressed in terms of fundamental units (kg, m, s)
Q8. Two measurements are A = (10 ± 1) m and B = (5 ± 1) m. Calculate (A + B) and (A × B) with their uncertainties.
Given:
A = (10 ± 1) m
B = (5 ± 1) m
Part 1: Calculate (A + B)
Step 1: Add central values
A + B = 10 + 5 = 15 m
Step 2: For addition, add absolute uncertainties
Δ(A + B) = ΔA + ΔB = 1 + 1 = 2 m
Answer: A + B = (15 ± 2) m
Part 2: Calculate (A × B)
Step 1: Multiply central values
A × B = 10 × 5 = 50 m²
Step 2: Calculate relative uncertainties
ΔA/A = 1/10 = 0.1 or 10%
ΔB/B = 1/5 = 0.2 or 20%
Step 3: For multiplication, add relative uncertainties
Δ(A × B)/(A × B) = ΔA/A + ΔB/B = 0.1 + 0.2 = 0.3 or 30%
Step 4: Calculate absolute uncertainty
Δ(A × B) = (A × B) × 0.3 = 50 × 0.3 = 15 m²
Answer: A × B = (50 ± 15) m²
Units and Measurements — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which of the following is NOT a fundamental quantity in the SI system of units?
- A. Mass
- B. Length
- C. Time
- D. Volume
Answer: D. Volume — Volume is derived from length (length × width × height), while mass, length, and time are fundamental quantities in the SI system.
MCQ 2. The SI unit of luminous intensity is:
- A. Ampere
- B. Candela
- C. Mole
- D. Kelvin
Answer: B. Candela — Candela (cd) is the SI unit for luminous intensity, which measures the power emitted by a light source in a particular direction.
MCQ 3. One femtometer is equivalent to:
- A. 10⁻¹² m
- B. 10⁻¹⁵ m
- C. 10⁻⁹ m
- D. 10⁻⁶ m
Answer: B. 10⁻¹⁵ m — The prefix 'femto' denotes a factor of 10⁻¹⁵. Therefore, one femtometer equals 10⁻¹⁵ meters.
MCQ 4. Which of the following is a unit of pressure?
- A. Joule
- B. Newton
- C. Pascal
- D. Watt
Answer: C. Pascal — Pascal (Pa) is the SI unit of pressure, defined as one Newton per square meter (N/m²). Joule is energy, Newton is force, Watt is power.
MCQ 5. Light-year is a unit of:
- A. Time
- B. Distance
- C. Intensity of light
- D. Speed
Answer: B. Distance — A light-year is the distance light travels in one year. It is a unit used to measure astronomical distances.
MCQ 6. The least count of a vernier calliper is 0.01 cm. This value represents:
- A. The total length of the main scale
- B. The maximum error possible in a measurement
- C. The smallest reading that can be measured accurately
- D. The number of divisions on the vernier scale
Answer: C. The smallest reading that can be measured accurately — The least count of an instrument is the smallest measurement that can be made accurately with that instrument.
MCQ 7. How many significant figures are there in the number 0.0050?
- A. 1
- B. 2
- C. 3
- D. 4
Answer: B. 2 — Leading zeros (0.00) are not significant. Trailing zeros are significant only if the number contains a decimal point. So, 5 and the trailing 0 are significant.
MCQ 8. The dimensional formula for force is:
- A. [MLT⁻¹]
- B. [MLT⁻²]
- C. [ML²T⁻²]
- D. [ML²T⁻¹]
Answer: B. [MLT⁻²] — Force = Mass × Acceleration. Mass has dimension [M], and acceleration has dimension [LT⁻²]. Therefore, force has dimension [MLT⁻²].
MCQ 9. Which of the following pairs has different dimensions?
- A. Work and Energy
- B. Power and Force
- C. Pressure and Stress
- D. Impulse and Momentum
Answer: B. Power and Force — Work and Energy ([ML²T⁻²]) have the same dimensions. Pressure and Stress ([ML⁻¹T⁻²]) have the same dimensions. Impulse ([MLT⁻¹]) and Momentum ([MLT⁻¹]) have the same dimensions. Power ([ML²T⁻³]) and Force ([MLT⁻²]) have different dimensions.
MCQ 10. The range of measurements possible with a metre scale is usually:
- A. 0 to 1 cm
- B. 0 to 10 cm
- C. 0 to 100 cm
- D. 0 to 1000 cm
Answer: C. 0 to 100 cm — A standard metre scale measures up to 1 metre, which is equivalent to 100 centimetres.
MCQ 11. If a student measures the length of a rod as 2.45 cm using a vernier calliper with a least count of 0.01 cm, and the actual length is 2.48 cm, what type of error is most likely present?
- A. Random error
- B. Zero error
- C. Personal error
- D. Instrumental error
Answer: B. Zero error — If the measured value is consistently lower than the actual value (or higher), it often indicates a systematic error like a zero error, where the instrument's zero mark does not coincide with the main scale zero.
MCQ 12. The time period (T) of a simple pendulum is given by T = 2π√(L/g). If the percentage error in the measurement of length (L) is 2% and in gravity (g) is 3%, what is the maximum percentage error in the measurement of the time period?
- A. 1%
- B. 2.5%
- C. 3.5%
- D. 5%
Answer: B. 2.5% — T = 2πL^(1/2)g^(-1/2). The relative error is ΔT/T = (1/2)(ΔL/L) + (1/2)(Δg/g). So, percentage error = (1/2)(2%) + (1/2)(3%) = 1% + 1.5% = 2.5%.
MCQ 13. The sum of 2.3 cm, 12.12 cm, and 0.123 cm, expressed with the correct number of significant figures, is:
- A. 14.543 cm
- B. 14.54 cm
- C. 14.5 cm
- D. 14.0 cm
Answer: C. 14.5 cm — In addition/subtraction, the result should have the same number of decimal places as the number with the fewest decimal places. 2.3 has one decimal place, 12.12 has two, and 0.123 has three. The sum is 14.543 cm. Rounding to one decimal place gives 14.5 cm.
MCQ 14. The dimension of the ratio of angular momentum to linear momentum is:
- A. [M]
- B. [L]
- C. [T]
- D. [LT⁻¹]
Answer: B. [L] — Angular momentum (J) = [ML²T⁻¹]. Linear momentum (p) = [MLT⁻¹]. The ratio J/p = [ML²T⁻¹] / [MLT⁻¹] = [L].
MCQ 15. Which of the following physical quantities has the same dimensions as Planck's constant (h)?
- A. Energy
- B. Linear momentum
- C. Angular momentum
- D. Power
Answer: C. Angular momentum — Planck's constant has dimensions of energy × time, which is [ML²T⁻²] × [T] = [ML²T⁻¹]. Angular momentum also has dimensions [ML²T⁻¹].
MCQ 16. If 'x' is distance, 'v' is velocity, and 'a' is acceleration, which of the following expressions is dimensionally correct?
- A. x = vt + (1/2)at³
- B. x = vt + at
- C. x = vt² + (1/2)at
- D. x = vt + (1/2)at²
Answer: D. x = vt + (1/2)at² — For dimensional consistency, each term in an equation must have the same dimensions. [x] = [L], [vt] = [LT⁻¹][T] = [L], and [(1/2)at²] = [LT⁻²][T²] = [L]. Thus, x = vt + (1/2)at² is dimensionally correct.
MCQ 17. The unit of permittivity of free space (ε₀) is:
- A. Farad/metre
- B. Farad·metre
- C. Volt/metre
- D. Coulomb·metre
Answer: A. Farad/metre — From Coulomb's law, F = (1/4πε₀) * (q₁q₂/r²), so ε₀ has units of C²N⁻¹m⁻². Alternatively, using the capacitance of a parallel plate capacitor C = ε₀A/d, the unit of ε₀ is Farad/metre.
MCQ 18. A body weighs 20.0 N in air, and its volume is 10.0 cm³. What is the density of the body? (Take g = 10 m/s²) Express your answer with correct significant figures.
- A. 2 g/cm³
- B. 2.0 g/cm³
- C. 2.00 g/cm³
- D. 0.5 g/cm³
Answer: C. 2.00 g/cm³ — Mass m = Weight/g = 20.0 N / 10 m/s² = 2.00 kg. Volume = 10.0 cm³ = 10.0 × 10⁻⁶ m³. Density = mass/volume = 2.00 kg / (10.0 × 10⁻⁶ m³) = 2.00 × 10⁵ kg/m³ = 2.00 g/cm³. Both 20.0 and 10.0 have three significant figures, so the result should also have three.
MCQ 19. Which of the following is a limitation of dimensional analysis?
- A. It can only be used for deriving simple relations
- B. It cannot determine dimensionless constants
- C. It cannot check the correctness of equations
- D. It cannot convert units from one system to another
Answer: B. It cannot determine dimensionless constants — Dimensional analysis cannot determine the numerical value of dimensionless constants (like 2π, 1/2) in an equation. It also cannot differentiate between physical quantities having the same dimensions.
MCQ 20. The maximum percentage error in the measurement of resistance R = V/I, where V = (100 ± 5) V and I = (10 ± 0.2) A, is:
- A. 5%
- B. 7%
- C. 2%
- D. 0.7%
Answer: B. 7% — For R = V/I, the maximum percentage error is (ΔR/R) × 100% = (ΔV/V × 100%) + (ΔI/I × 100%). Percentage error in V = (5/100) × 100% = 5%. Percentage error in I = (0.2/10) × 100% = 2%. Total percentage error = 5% + 2% = 7%.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.