Given below are two statements: One is labeled as Assertion (A) and other is labeler as Reason (R). Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S), if density of the liquid is ρ and radius of the drop is r, then T = K ρ r 3 S 3 2 is dimensionally correct, where K is dimensionless. Reason (R) : Using dimensional analysis we get R.H.S. having different dimension than that of time period. In the light of above statements, choose the correct answer from the options given below.

Both (A) and (R) are true and (R) is the correct explanation of (A)

Both (A) and (R) are true but (R) is not the correct explanation of (A)

Identify the pair of physical quantities which have different dimensions

Specific heat capacity and Latent heat

Wave number and Rydberg’s constant

Stress and Coefficient of elasticity

Velocity ( υ ) and acceleration (a) in two system of units 1 and 2 are related as υ 2 = n m 2 υ 1 a n d a 2 = a 1 m n respectively. Here m and n are constants. The relations for distance and time in two system respectively are:

n 3 m 3 L 1 = L 2 a n d n 2 m T 1 = T 2

L 1 = n 4 m 2 L 2 a n d T 1 = n 2 m T 2

L 1 = n 2 m L 2 a n d T 2 = n 4 m 2 T 2

n 2 m L 1 , = L 2 a n d n 4 m 2 T 1 = T 2

Units and Measurements Class 11 NCERT Solutions – Stepwise Explanation

Q: 2.26 It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s?

2.26 Total time in 100 years = 100 * 365 * 24 * 60 * 60 s Error in 100 years = 0.02 sec Error in 1 sec= 0.02 / (100 * 365 * 24 * 60 * 60) s = 6.34 * 10-12

Q: 2.6 Which of the following is the most precise device for measuring length? (a) A vernier calipers with 20 divisions on the sliding scale (b) A screw gauge of pitch 1 mm and 100 divisions on the circular scale (c) An optical instrument that can measure length to within a wavelength of light ?

2.6 (a) The least count of a vernier calliper = 0.1 mm (b) The least count of the screw gauge = 1/100 mm = 0.01 mm (c) The least count of the optical instrument = wave length of the light = 10-5cm = 10-3 mm = 0.001 mm Hence the answer is (c)

Q: 2.31 The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?

2.31 Speed of light = 3 * 108 m/s, Time taken = 3 billion years = 3 * 109 years = 3 * 365 * 24 * 60 * 60 * 109 s The distance = speed x time = 3 * 108 * 3 * 365 * 24 * 60 * 60 * 109 m = 2.838 * 1025 m

Q: 2.27 Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase: 970 kg/m3. Are the two densities of the same order of magnitude? If so, why?

2.27 The diameter of the sodium atom = 2.5 Å = 2.5 * 10-10 m Radius of the sodium atom = 1.25 * 10-10 m Volume of the sodium atom = (4/3)(π) r3 = (4/3) * (22/7) * (1.25 * 10-10) m3 = 8.18 * 10-30 Mass of 1 mole atom of sodium = 23 g = 23 * 10-3 kg 1 mole of sodium contains 6.023 * 1023 atoms. Hence mass of 1 sodium atom = (23 * 10-3)/ (6.023 * 1023) = 3.818 * 10-26 kg Atomic mass density of sodium = M/V = (3.818 * 10-26)/ (8.18 * 10-30) kg/m3 = 4667.48 kg/m3 The density of sodium in its solid state is 4667.48 kg/m3 but in the crystalline phase is 970 kg/m3. In crystalline phase there are voids in between atoms, hence the density is less.

Updated on Jul 9, 2025 17:17 IST

By Pallavi Pathak, Assistant Manager Content

NCERT Solutions for Class 11 Physics Chapter 1 – Units and Measurements is created by the subject matter experts at Shiksha.com. It provides a comprehensive study material with clear explanations of the basic concepts like SI units, physical quantities, significant figures, measurement errors, and dimensional analysis.
Units and Measurement Class 11 NCERT solutions are according to the latest CBSE syllabus. It helps students to build a strong foundation in Physics. These solutions will help you in improving your concept clarity and problem-solving skills. It helps prepare for the CBSE Board examination and other competitive exams like NEET and JEE Main.

Table of content

Class 11 Physics Chapter 1 Units and Measurements: Key Topics, Weightage
Chapter 1 Units and Measurements Important Formulas & Concepts
NCERT Solutions for Class XI Physics Units and Measurement PDF Download
Units and Measurements Solutions

Class 11 Physics Chapter 1 Units and Measurements: Key Topics, Weightage

Class 11 Physics Chapter 1 Units and Measurements is an extremely important chapter for the CBSE exam. Questions from this chapter also come in other prominent entrance exams, including NEET and JEE Main. Before preparing any chapter, always check the topics covered in the chapter, as it makes your learning a planned one.

These are the topics covered in the Physics Class 11 Chapter 1 Exercise Solutions:

Exercise	Topics Covered
1.1	Introduction
1.2	The International System of Units
1.3	Significant Figures
1.4	Dimensions of Physical Quantities
1.5	Dimensional Formulae and Dimensional Equations
1.6	Applications of Dimensional Analysis

Units and Measurements Weightage in NEET, JEE Main Exams

Exam	Number of Questions	Weightage
NEET	1 question	2%
JEE Main	1-2 questions	2-3%

Chapter 1 Units and Measurements Important Formulas & Concepts

Following are the important concepts and formulas of the Units and Measurements class 11 physics NCERT solutions:

Physical Quantities include Fundamental Quantities (that can be broken down, such as time, length, and mass) and Derived Quantities created out of the Fundamental Quantities.

Fundamental (Base) Units, such as kilogram, meter, and second and derived units, which are a combination of base units. Eg-

Quantity	Unit	Symbol
Length	Meter	m
Mass	Kilogram	kg
Time	Second	s
Temperature	Kelvin	K
Electric Current	Ampere	A
Luminous Intensity	Candela	cd
Amount of Substance	Mole	mol

Dimensions

Mass → [M], Length → [L], Time → [T], etc.

Dimensional Formula Examples

Quantity	Dimensional Formula
Velocity	[M⁰ L¹ T⁻¹]
Acceleration	[M⁰ L¹ T⁻²]
Force	[M¹ L¹ T⁻²]
Work/Energy	[M¹ L² T⁻²]
Power	[M¹ L² T⁻³]
Pressure	[M¹ L⁻¹ T⁻²]

Dimensional Analysis to convert units, derive new relations, and check the correctness of equations.

Significant Figures - The rules are that zeros between significant digits are significant, all non-zero digits are significant, trailing zeros after the decimal point are significant, and leading zeros are not significant.

Error in Measurement

NCERT Solutions for Class XI Physics Units and Measurement PDF Download

Units and Measurement is an important chapter for the students because it forms the foundation of various other concepts. We are providing the solutions for all the NCERT questions given in this chapter for the students. Class 11 Physics Chapter 1 Exercise Solutions PDF can be downloaded from below to get step-by-step solutions created by Shiksha's expert. After downloading the PDF, the students can access the solutions from anywhere and at any time for quick revision.
Get access to the chapter-wise NCERT notes PDF, important topics, solved examples, and weightage information here - Explore Now.

Download Here: NCERT Solution for Class XI Physics Units and Measurements Chapter

Units and Measurements Solutions

Following are the solutions of the NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements:

Q.2.1 Fill in the blanks

Ans.2.1

(a) The volume of a cube of side 1 ${c m}^{3}$ is equal to $10^{- 6} m^{3}$

The volume of the cube = 1 cm × 1 cm × 1 cm = 1 cm³

1 cm3 = 1 × 10^-6 m³

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to1.25 × 10⁴(mm)²

The surface area of a solid cylinder = 2ԯr(r+)h, where r = 2.0 cm = 20 mm, h = 10 cm = 100 mmSurface area = 2 × 22/7 × 20 ×(100 +20) ${m m}^{2}$ = 1.51 × $10^{4}$ ${m m}^{2}$

(c) A vehicle moving with a speed of 18 km h^–1 covers 5 m in 1 s

Vehicle speed = 18 km/h = 18000/3600 m/s = 5 m/s

(d) The relative density of lead is 11.3. Its density is 11.3 g cm^–3 or 11.3 × 10³....kg m^–3.

Relative density or Specific gravity = Density of the subject / density of water

Density of water = 1 gm/cc= 10^-3 / 10^-6

Q.2.2 Fill in the blanks by suitable conversion of units

Ans.2.2

(a) 1 kg m² s^–2 = 10⁷g cm² s^–2

kg m²s^-2= 10³ × 10⁴= 10⁷

(b) 1 m = 10^-16..... ly

1 m = 1.057× 10^-16 light year

(c) 3.0 m s^–2 = .... km h^–2

= (3/10³) × (3.6 x 10³)² = 3.9 × 10⁴

(d) G = 6.67 × 10^–11 N m² (kg) ^–2 = 6.67 × 10^-8 (cm) ³ s^–2 g^–1.

1N =1 kg = 1000 gm = 10³ gm

G = 6.67 × 10^-11 Nm²(kg)^-2 = 6.67 × 10^-11 x (1 kg m/s²) × ( 1 m²) × ( 1kg ^-2)

= 6.67 × 10^-11 × ( 1 kg^-1 × 1m³ × 1s^-2)

= 6.67 × 10^-11 × (103g)^-1 × (102 cm)³× (1s^-2)

= 6.67 × 10^-8 cm³s^-2g^-1

Q.2.3 A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J = 1 kg m² s^–2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 α ^–1 β ^–2 γ ² in terms of the new units.

Ans.2.3 1

Calorie = 4.2 J = 4.2 kgm²s^-2

Standard formula for conversion

Given unit / New unit = (M1/M2)^x(L1/L2)^y(T1/T2)^z

Formula for energy = M¹L²T^-2

Here x = 1, y = 2, z = -2

In this problem

M1 = 1 kg, L1 = 1 m, T1 = 1 s and

M2 = α kg, L2 = β m, T2 = γ s

So 1 Calorie = 4.2 (1/α)¹(1/β)²(1/γ)^-2

= 4.2α^-1β^-2γ²

Q.2.4 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

(d) The air inside this room contains a large number of molecules

(e) A proton is much more massive than an electron

(f) The speed of sound is much smaller than the speed of light

Ans.2.4

(a) Atoms are very small compared to a cricket ball

(b) Compared to a car on a busy city road, a jet plane moves with great speed

(c) When compared to the average mass of a human being, mass of Jupiter is very large.

(d) When compared to the air inside a football, the air inside this room contains a large number of molecules

(e) A proton has higher mass compared to that of an electron

(f) The speed of sound is much smaller when calling your friend from your balcony than signaling him with a torch light

Commonly asked questions

2.26 It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s?

2.6 Which of the following is the most precise device for measuring length?

(a) A vernier calipers with 20 divisions on the sliding scale

(b) A screw gauge of pitch 1 mm and 100 divisions on the circular scale

(c) An optical instrument that can measure length to within a wavelength of light ?

2.31 The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?

2.27 Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase: 970 kg/m³. Are the two densities of the same order of magnitude? If so, why?

2.28 The unit of length convenient on the nuclear scale is a fermi : 1 f = 10^–15 m. Nuclear sizes obey roughly the following empirical relation :

r = r₀ A^1/3

where r is the radius of the nucleus, A its mass number, and r₀ is a constant equal to about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

2.24 When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72" of arc. Calculate the diameter of Jupiter.

2.29 A LASER is a source of very intense, monochromatic, and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth?

Q.2.1 Fill in the blanks

2.18 Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

2.22 Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):

(a) The total mass of rain-bearing clouds over India during the Monsoon

(b) The mass of an elephant

(c) The wind speed during a storm

(d) The number of strands of hair on your head

(e) The number of air molecules in your classroom

2.32 It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

2.23 The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data: mass of the Sun = 2.0 × 10³⁰ kg, radius of the Sun = 7.0 × 10⁸ m.