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Class 11 Physics Chapter 1

Units and Measurements NCERT Solutions, Complete PDF Content and Important Questions

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements PDF Download (2026) + Important Questions Welcome to Class 11 Physics! Stepping into the 11th grade is an exciting milestone, but it also means stepping up your game. Physics is no longer just about reading theory; it's about understanding the language of the universe. And that language starts right here with Class 11 Physics Chapter 1: Units and Measurements . Whether you are aiming to top your CBSE 2026 board exams or have your eyes set on cracking competitive exams like JEE and NEET, this chapter sets the ultimate foundation. You cannot solve complex mechanics or thermodynamics problems later if your units and dimensions are messed up! In this complete guide, we will break down the chapter into simple, easy-to-understand chunks. We've got detailed NCERT Solutions, a handy list of formulas, and extra board-style important questions to get your preparation on track.

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ChapterUnits and Measurements

SubjectPhysics

Class11

BoardCBSE

DifficultyModerate

Exam Weightage~3 to 5 Marks (Crucial for competitive exams)

Learning Objectives

Understand the difference between fundamental (base) and derived physical quantities. Master the 7 base SI units and their standard definitions.
Calculate and express absolute, relative, and percentage errors in measurements.
Apply the rules of significant figures in mathematical operations.
Use dimensional analysis to check the correctness of equations, convert units, and.
Derive relationships between physical quantities.

Key Concepts, Definitions and Formulas

Physical Quantity: Anything that can be measured directly or indirectly and in terms of which laws of physics can be expressed
Fundamental Quantities: Quantities that are independent of each other (Length, Mass, Time, Electric Current, Temperature, Luminous Intensity,
Amount of Substance). Derived Quantities: Quantities derived from fundamental quantities (e.g., Velocity, Force,
Energy). Errors in Measurement: The difference between the true value and the measured value of a quantity
Absolute Error: $\Delta a_i = |a_{mean} - a_i|$
Relative Error: $\frac{\Delta a_{mean}}{a_{mean}}$
Percentage Error: $\frac{\Delta a_{mean}}{a_{mean}} \times 100\%$
Significant Figures: The reliable digits in a measurement plus the first uncertain digit. (e.g., 0.00230 has 3 significant figures)
Dimensional Formula: The expression showing how and which base quantities represent a physical quantity
Velocity: $[M^0 L^1 T^{-1}]$
Force: $[M^1 L^1 T^{-2}]$
Energy/Work: $[M^1 L^2 T^{-2}]$
Principle of Homogeneity: A physical equation is dimensionally correct only if the dimensions of all the terms on both sides of the equation are the same

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Full NCERT Solutions

NCERT Q1.1: Fill in the blanks:(a) The volume of a cube of side 1 cm is equal to ___ $m^3$(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to _ $(mm)^2$(c) A vehicle moving with a speed of 18 km/h covers _ m in 1 s(d) The relative density of lead is 11.3. Its density is _ $g \text{ cm}^{-3}$ or ___ $kg \text{ m}^{-3}$.

Answer: (a) Side of cube, $a = 1 \text{ cm} = 10^{-2} \text{ m}$. Volume $V = a^3 = (10^{-2} \text{ m})^3 = 10^{-6} \text{ m}^3$. Answer: $10^{-6}$ (b) Radius $r = 2.0 \text{ cm} = 20 \text{ mm}$. Height $h = 10.0 \text{ cm} = 100 \text{ mm}$. Total surface area $S = 2\pi r(r + h) = 2 \times 3.14 \times 20 \times (20 + 100) = 125.6 \times 120 = 15072 \text{ mm}^2$.Using significant figures (2 sig figs from the given data), it is $1.5 \times 10^4 \text{ mm}^2$. Answer: $1.5 \times 10^4$ (c) Speed $v = 18 \text{ km/h}$. To convert to m/s, multiply by $5/18$:$v = 18 \times \frac{5}{18} \text{ m/s} = 5 \text{ m/s}$. Distance covered in 1 second is 5 m. Answer: $5$ (d) Relative density = $\frac{\text{Density of substance}}{\text{Density of water}}$. Density of water = $1 \text{ g cm}^{-3}$ or $10^3 \text{ kg m}^{-3}$. Density of lead = $11.3 \times 1 \text{ g cm}^{-3} = 11.3 \text{ g cm}^{-3}$. In kg/m3, density = $11.3 \times 10^3 \text{ kg m}^{-3}$. Answer: $11.3$, $1.13 \times 10^4$

NCERT Q1.2: Fill in the blanks by suitable conversion of units:(a) $1 \text{ kg m}^2 \text{ s}^{-2}$ = ___ $g \text{ cm}^2 \text{ s}^{-2}$(b) 1 m = _ ly(c) $3.0 \text{ m s}^{-2}$ = _ $\text{km h}^{-2}$(d) $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$ = ___ $(\text{cm})^3 \text{ s}^{-2} \text{ g}^{-1}$.

Answer: (a) $1 \text{ kg} = 10^3 \text{ g}$ and $1 \text{ m} = 10^2 \text{ cm}$.$1 \text{ kg m}^2 \text{ s}^{-2} = (10^3 \text{ g}) \times (10^2 \text{ cm})^2 \times \text{s}^{-2} = 10^3 \times 10^4 \text{ g cm}^2 \text{ s}^{-2} = 10^7 \text{ g cm}^2 \text{ s}^{-2}$. Answer: $10^7$ (b) 1 Light year (ly) = distance traveled by light in 1 year = $3 \times 10^8 \text{ m/s} \times (365 \times 24 \times 60 \times 60 \text{ s}) = 9.46 \times 10^{15} \text{ m}$. Therefore, $1 \text{ m} = \frac{1}{9.46 \times 10^{15}} \text{ ly} \approx 1.057 \times 10^{-16} \text{ ly}$. Answer: $1.057 \times 10^{-16}$ (c) $3.0 \text{ m s}^{-2} = 3.0 \times (10^{-3} \text{ km}) / (\frac{1}{3600} \text{ h})^2$.$= 3.0 \times 10^{-3} \times (3600)^2 \text{ km h}^{-2} = 3.0 \times 10^{-3} \times 12960000 = 3.88 \times 10^4 \text{ km h}^{-2}$. Answer: $3.9 \times 10^4$ (d) Convert $N$ to base units: $1 \text{ N} = 1 \text{ kg m s}^{-2}$. So, $G = 6.67 \times 10^{-11} (\text{kg m s}^{-2})(\text{m}^2)(\text{kg}^{-2}) = 6.67 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}$. Now, $1 \text{ m} = 10^2 \text{ cm}$ and $1 \text{ kg} = 10^3 \text{ g}$.$G = 6.67 \times 10^{-11} \times (10^2 \text{ cm})^3 \times (10^3 \text{ g})^{-1} \times \text{s}^{-2}$$= 6.67 \times 10^{-11} \times 10^6 \times 10^{-3} \text{ cm}^3 \text{ g}^{-1} \text{ s}^{-2} = 6.67 \times 10^{-8} \text{ cm}^3 \text{ s}^{-2} \text{ g}^{-1}$. Answer: $6.67 \times 10^{-8}$

NCERT Q1.3: A calorie is a unit of heat or energy and it equals about 4.2 J where $1 \text{ J} = 1 \text{ kg m}^2 \text{ s}^{-2}$. Suppose we employ a system of units in which the unit of mass equals $\alpha \text{ kg}$, the unit of length equals $\beta \text{ m}$, the unit of time is $\gamma \text{ s}$. Show that a calorie has a magnitude $4.2 \alpha^{-1} \beta^{-2} \gamma^2$ in terms of the new units.

Answer: Dimensional formula for energy (calorie/Joule) is $[M^1 L^2 T^{-2}]$. Given system (SI): $n_1 = 4.2$, $M_1 = 1 \text{ kg}$, $L_1 = 1 \text{ m}$, $T_1 = 1 \text{ s}$. New system: $n_2 = ?$, $M_2 = \alpha \text{ kg}$, $L_2 = \beta \text{ m}$, $T_2 = \gamma \text{ s}$. Using the conversion formula: $n_2 = n_1 \left[\frac{M_1}{M_2}\right]^a \left[\frac{L_1}{L_2}\right]^b \left[\frac{T_1}{T_2}\right]^c$ Here $a=1, b=2, c=-2$.$n_2 = 4.2 \left[\frac{1 \text{ kg}}{\alpha \text{ kg}}\right]^1 \left[\frac{1 \text{ m}}{\beta \text{ m}}\right]^2 \left[\frac{1 \text{ s}}{\gamma \text{ s}}\right]^{-2}$$n_2 = 4.2 (\alpha^{-1}) (\beta^{-2}) (\gamma^2)$$n_2 = 4.2 \alpha^{-1} \beta^{-2} \gamma^2$. Hence Proved.

NCERT Q1.4: Explain this statement clearly: "To call a dimensional quantity 'large' or 'small' is meaningless without specifying a standard for comparison". In view of this, reframe the following statements wherever necessary:(a) atoms are very small objects(b) a jet plane moves with great speed(c) the mass of Jupiter is very large

Answer: A physical quantity cannot be judged as large or small in isolation. For instance, a length of 1 meter is small compared to the distance between stars but very large compared to the size of a bacterial cell. We must state the reference point. Reframed statements:(a) Atoms are very small objects compared to a tennis ball. (b) A jet plane moves with great speed compared to a high-speed train. (c) The mass of Jupiter is very large compared to the mass of the Earth.

NCERT Q1.5: A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

Answer: Given: Speed of light, $c = 1 \text{ new unit / s}$. Time taken, $t = 8 \text{ min } 20 \text{ s} = (8 \times 60) + 20 = 480 + 20 = 500 \text{ s}$. Distance = Speed $\times$ Time Distance = $1 \text{ new unit / s} \times 500 \text{ s} = 500 \text{ new units}$. Answer: 500 new units.

Extra Important Questions for 2026 Exams

Important Q1: Which of the following is NOT a fundamental SI unit? A) Meter B) Ampere C) Newton D) Kelvin

Answer: C) Newton. (It is a derived unit of force).

Important Q2: The number of significant figures in 0.005020 is: A) 3 B) 4 C) 6 D) 7

Answer: B)

Important Q3: If the percentage error in the measurement of the radius of a sphere is 2%, what is the percentage error in the measurement of its volume? A) 2% B) 4% C) 6% D) 8%

Answer: C) 6%. (Volume $V = \frac{4}{3}\pi r^3$. Percentage error = $3 \times (\text{error in } r) = 3 \times 2\% = 6\%$). Difficulty: Moderate

Important Q4: State the Principle of Homogeneity of dimensions.

Answer: It states that a physical equation is correct only if the dimensions of all the terms separated by '+', '-', or '=' signs are exactly the same.

Important Q5: Name any two physical quantities that have the same dimensional formula.

Answer: Work and Energy both have the dimensional formula $[M^1 L^2 T^{-2}]$. (Another example: Impulse and Momentum).

Important Q6: Round off the number 3.745 to three significant figures.

Answer: The digit to be dropped is

Important Q7: Write the dimensional formula of the universal gravitational constant (G).

Answer: From $F = G \frac{m_1 m_2}{r^2}$, we get $G = \frac{F r^2}{m_1 m_2}$. Dimensions = $\frac{[M L T^{-2}] [L^2]}{[M^2]} = [M^{-1} L^3 T^{-2}]$. Difficulty: Moderate

Important Q8: Deduce the formula for the time period (T) of a simple pendulum using dimensional analysis, assuming it depends on the mass (m) of the bob, length (l) of the pendulum, and acceleration due to gravity (g).

Answer: Let $T \propto m^a l^b g^c$. Therefore, $T = k m^a l^b g^c$ (where k is a dimensionless constant). Substituting dimensions on both sides:$[M^0 L^0 T^1] = [M]^a [L]^b [L T^{-2}]^c = [M^a L^{b+c} T^{-2c}]$ Equating powers of M, L, and T:$a = 0$$b + c = 0$$-2c = 1 \implies c = -1/2$ So, $b = 1/2$. Substituting values: $T = k m^0 l^{1/2} g^{-1/2} = k \sqrt{\frac{l}{g}}$.

Important Q9: Explain the different types of systematic errors in measurement.

Answer: Systematic errors tend to be in one direction (either positive or negative). Types include:

Important Q11: Calculate the mean value of the thickness.

Answer: Mean value $a_m = \frac{2.04 + 2.06 + 2.08 + 2.06}{4} = \frac{8.24}{4} = 2.06 \text{ mm}$.

Important Q12: Calculate the absolute error in each measurement.

Answer: Absolute error $\Delta a_i = |a_m - a_i|$$\Delta a_1 = |2.06 - 2.04| = 0.02 \text{ mm}$$\Delta a_2 = |2.06 - 2.06| = 0.00 \text{ mm}$$\Delta a_3 = |2.06 - 2.08| = 0.02 \text{ mm}$$\Delta a_4 = |2.06 - 2.06| = 0.00 \text{ mm}$ Difficulty: Moderate

Important Q13: What is the mean absolute error and percentage error in this experiment?

Answer: Mean absolute error $\Delta a_{mean} = \frac{0.02 + 0.00 + 0.02 + 0.00}{4} = \frac{0.04}{4} = 0.01 \text{ mm}$. Percentage Error = $(\frac{\Delta a_{mean}}{a_m}) \times 100\% = (\frac{0.01}{2.06}) \times 100\% \approx 0.485\%$. Directions: For questions 14 and 15, two statements are given-one labeled Assertion (A) and the other labeled Reason (R). Select the correct answer from the codes (A, B, C, D) given below: A) Both A and R are true and R is the correct explanation of A. B) Both A and R are true but R is NOT the correct explanation of A. C) A is true but R is false. D) A is false and R is true.

Important Q14: Assertion (A): A dimensionally correct equation may not be the exact actual physical equation. Reason (R): Dimensional analysis does not give information about dimensionless constants.

Answer: A) Both A and R are true and R is the correct explanation of A. (For example, $s = ut + \frac{1}{2}at^2$ and $s = ut + at^2$ are both dimensionally correct, but only the first is physically correct because dimensions can't verify the constant $1/2$). Difficulty: Moderate

Important Q3: Reason (R): Leading zeros are never significant.

Answer: D) A is false and R is true. (The significant figure in 0.005 is only 1. Leading zeros before the first non-zero digit merely indicate the position of the decimal point). Difficulty: Moderate

FAQ Section

Is Class 11 Physics Chapter 1 important for JEE/NEET 2026?

Absolutely. While direct questions on units might seem easy, dimensional analysis and error measurement are heavily tested in JEE and NEET. Plus, this concept integrates into almost every other mechanics problem.

Where can I download the NCERT PDF for Class 11 Physics?

You can download the official and updated NCERT PDFs directly from the official NCERT website (ncert.nic.in) under the 'Publications' section.

What are the most important topics in Units and Measurements?

Focus heavily on Dimensional Analysis (checking equation correctness and deriving relations) and Error Analysis (calculating percentage error in combined quantities).

How do I find a dimensional formula easily?

Always break the physical quantity down into its most basic mathematical formula. For example, to find the dimension of Pressure, use $P = \text{Force}/\text{Area}$. Substitute the dimensions of Force and Area to get the final result.

Are there any deleted topics in Chapter 1 for 2026?

Yes, under the rationalized syllabus, topics relating to the estimation of very small distances (like the oleic acid experiment) have been minimized or removed. Always cross-check with the latest CBSE curriculum document

More Class 11 Physics Chapters

Ch 1: Units and Measurements Ch 2: Motion in a Straight Line Ch 3: Motion in a Plane Ch 4: Laws of Motion Ch 5: Work, Energy and Power Ch 6: System of Particles and Rotational Motion Ch 7: Gravitation Ch 8: Mechanical Properties of Solids Ch 9: Mechanical Properties of Fluids Ch 10: Thermal Properties of Matter Ch 11: Thermodynamics Ch 12: Kinetic Theory Ch 13: Oscillations Ch 14: Waves