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Click the links below to teach / view the detailed explanation of Chapter (Units & Measurements)

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Units & Dimensions (Introduction)

RULES FOR USING SYMBOLS OF SI UNITS

ACCURACY, PRECISION AND ERROR ANALYSIS

SIGNIFICANT FIGURES

DIMENSIONS AND DIMENSIONAL ANALYSIS

OBJECTIVE TYPE QUESTIONS

JEE MAINS MCQ’s (LEVEL – 1)

JEE MAINS MCQ’s (LEVEL – 2)

JEE MAINS MCQ’s (LEVEL – 3)

JEE MAINS MCQ’s (LEVEL – 4)

NCERT Solutions for Class 11 Physics Chapter 2 – Free PDF Download

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements is the best study resource you can get to understand the main topics and to score good grades in the examination. These solutions provide appropriate answers to the textbook questions. To get a grip on this chapter, students can make use of the NCERT Solutions for Class 11 Physics available at ANAND CLASSES. Students can also find solutions to exemplary problems, worksheets, questions from previous question papers, numerical problems, MCQs, short answer questions, tips and tricks.

Chapter 2 of NCERT Solutions for Class 11 Physics mainly helps in understanding the fundamentals of units and measurements according to the latest CBSE Syllabus. In our daily lives, most of the activities depend on this and it is very important for the students to learn it effectively. Students can access the Physics NCERT Solutions for Class 11 to comprehend the key concepts present in this chapter.

What is measurement ?

To measure a physical quantity like length, mass and time we require a standard of measurement. This standard of measurement is called the unit of that physical quantity. For example, the unit of length is metres and a standard length of 1 metre has a precise definition. To measure the length of an object is to determine how many times this standard length metre is contained in the length of a room. The comparison of a physical quantity with a standard quantity is called measurement.

Physical Quantities

Those quantities which can describe the laws of physics are called the physical quantity. A physical quantity is one that can be measured. Thus, length, mass, time, pressure, temperature, current and resistance are the physical quantities.

Classification of physical quantities

The physical quantities are classified into

(i) Fundamental quantities or base quantities

(ii) Derived quantities

The physical quantities that are independent of each other are called fundamental quantities. All the other quantities which can be expressed in terms of the fundamental quantities are called the derived quantities.

Units

The reference standard used to measure the physical quantities is called the unit.

Properties of Unit

The unit should be of some suitable size
The unit must be well-defined
The unit should be easily reproducible at all places
The unit must not change with time
The unit should not change with physical conditions like temperature, pressure etc.
The unit must be easily comparable experimentally with similar physical quantities.

Types of Units

(i) Fundamental Units

The units defined for the fundamental quantities are called fundamental units.

(ii) Derived Units

The units of all other physical quantities which are derived from the fundamental units are called the derived units.

System of Units

(1) FPS System: In this system, the unit of length is foot, unit of mass is pound and the unit of time is second.

(2) CGS System: In this system, the units of length, mass and time are centimetre, gram and second, respectively.

(3) MKS System: In this system, the unit of length, mass and time are meters, kilogram and second, respectively.

(4) SI System: This system is widely used in all measurements throughout the world. The system is based on seven basic units and two supplementary units.

Basic Units
Quantity	Unit	Symbol of the unit
Length	metre	m
Mass	kilogram	kg
Time	second	s
Temperature	kelvin	K
Electric current	ampere	A
Number of particles	mole	mol
Luminous intensity	candela	cd
Supplementary Units
Plane angle	radian	rad
Solid angle	Steradian	sr

Definition of Basic and Supplementary Units

Basic Units

1. Metre (m): One metre is the distance travelled by light in the vacuum during a time interval of (1/299792458) seconds.

2. Kilogram (kg): It is the mass of a platinum-iridium cylinder kept at the National Bureau of weights and measurements, Paris.

3. Second (s): The second is the time taken by the light of a specified wavelength emitted by a cesium-133 atom to execute 9192631770 vibrations.

4. Ampere (A): One ampere is that current which when passed through two straight parallel conductors of infinite length and of negligible cross-section kept at a distance of 1 metre apart in the vacuum produces between them a force equal to 2 x 10^-7 newton per metre length.

5. Kelvin (K): It is the fraction 1/273.6 of the thermodynamic temperature of the triple point of water.

6. Candela (cd): A candela is defined as 1/60^thof luminous intensity of 1 square centimetre of a perfect black body maintained at the freezing temperature of platinum (1773 ⁰C).

7. Mole (md): One mole is the amount of substance that contains elementary units equal to the number of atoms in 0.012 kg of carbon-12.

Supplementary Units

1. Radian (rad): The radian is the angle subtended at the centre of the circle by the arc whose length is equal to the radius of the circle.

2. Steradian (Sr): The steradian is the solid angle subtended at the centre of a sphere by a spherical surface of an area equal to the square of its radius.

Dimensional Formula

The dimensional formula of any physical quantity is the formula that tells which of the fundamental units have been used for the measurement of that physical quantity.

How dimensional formula is written for a physical quantity

(1) The formula of the physical quantity must be written. The quantity must be on the left-hand side of the equation.

(2) All the quantities on the right-hand side of the formula must be written in terms of fundamental quantities like mass, length and time.

(3) Replace mass, length and time with M, L and T.

(4) Write the powers of the terms.

Characteristics of Dimensions

(1) Dimensions do not depend on the system of units.

(2) Quantities with similar dimensions can be added or subtracted from each other.

(3) Dimensions can be obtained from the units of the physical quantities and vice versa.

(4) Two different quantities can have the same dimension.

(5) When two dimensions are multiplied or divided it will form the dimension of the third quantity.

Dimensional Analysis

The dimensional formula can be used to

(1) To check the correctness of the equation.

(2) Convert the unit of the physical quantity from one system to another.

(3) Deduce the relation connecting the physical quantities.

Units and Dimensions Of A Few Derived Quantities

Physical Quantity	Unit	Dimensional Formula
Displacement	m	M⁰L¹T⁰
Area	m²	M⁰L²T⁰
Volume	m³	M⁰L³T⁰
Velocity	ms^-1	M⁰L¹T^-1
Acceleration	ms^-2	M⁰L¹T^-2
Density	Kg m^-3	M¹L^-3T⁰
Momentum	Kg ms^-1	M¹L¹T^-1
Work/Energy/Heat	Joule (or) Kg m²/sec²	M¹L²T^-2
Power	Watt (W) (or) Joule/sec	M¹L²T^-3
Angular velocity	rad s^-1	M⁰L⁰T^-1
Angular acceleration	rad s^-2	M⁰L⁰T^-2
Moment of Inertia	Kg m²	M¹L²T⁰
Force	Newton (or) Kg m/sec²	M¹L¹T^-2
Pressure	Newton/m (or) Kg m^-1/sec²	M¹L^-1T^-2
Impulse	Newton sec (or) Kg m/sec	M¹L¹T^-1
Inertia	Kg m²	M¹L²T⁰
Electric Current	Ampere (or) C/sec	QT^-1
Resistance/Impedance	Ohm (or) Kg m²/sec C²	ML²T^-1Q^-2
EMF/Voltage/Potential	Volt (or) Kg m²/sec² C	ML²T^-2Q^-1
Permeability	henry/m (or) Kg m/C²	MLQ^-2
Permittivity	Farad/m (or) sec²C²/Kgm³	T²Q²M^-1L^-3
Frequency	Hertz (or) sec^-1	T^-1
Wavelength	m	L¹

Principle of Homogeneity

According to the principle of homogeneity of dimensions, all the terms in a given physical equation must be the same.

Ex. s = ut + (½) at²

Dimensionally

[L] = [LT^-1.T] + [LT^-2. T²]
[L] = [L] + [L]

Defects of Dimensional Analysis

While deriving the formula the proportionality constant cannot be found.
The equation of a physical quantity that depends on more than three independent physical quantities cannot be deduced.
This method cannot be used if the physical quantity depends on more parameters than the number of fundamental quantities.
The equations containing trigonometric functions and exponential functions cannot be derived

Points to Remember

Those quantities which can describe the laws of physics are called the physical quantity. Example:- length, mass and time
Physical quantities can be classified as fundamental quantities and derived quantities.
The reference standard used to measure the physical quantities is called the unit. Units are classified as fundamental units and derived units.
SI system is the most commonly used system of units
The SI is based on seven basic units and two supplementary units.
The dimensional formula of any physical quantity is the formula that tells which of the fundamental units have been used for the measurement of that physical quantity.
The dimensional formula follows the principle of homogeneity

Solved Examples

(1) The diameter of a cylinder is measured using vernier callipers with no zero error. It is found that the zero of the vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is

a) 5.112 cm

b) 5. 124 cm

c) 5.136 cm

d) 5.148 cm

Answer: b) 5. 124 cm

Solution:

Least Count of a Vernier is given by

L.C = 1 Main Scale Division/Number of divisions on Vernier Scale

L.C = 1M.S.D/n

One main scale division = 0.05 cm

n = 50

L.C = 0.05/50 = 0.001 cm

Diameter of the cylinder = Main Scale Reading + (Least Count x Vernier Scale Reading)

= 5.10 + (24 x 0.001) = 5.124 cm

(2) The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is

a) 0.9%

b) 2.4 %

c) 3.1 %

d) 4.2 %

Answer: c) 3.1 %

Solution:

Given,

Pitch = 0.5 mm

Circular scale division = 50

Main scale division = 2.5 mm

Least Count = Pitch/ Circular scale division= 0.5/50 = 0.01 mm

Circular scale division reading = 20

Relative error = 2%

Screw gauge reading = Main scale reading + (Least Count x Circular scale division reading)

= 2.5 + (0.01 x 20)

= 2.7 mm

Density, ρ = mass/volume

(begin{array}{l}rho =frac{M}{frac{4pi }{3}(frac{D}{2})^{3}}end{array} )

The relative percentage error in density is

(begin{array}{l}frac{Delta rho }{rho }times 100=left ( frac{Delta M}{M}+frac{3Delta D}{D} right )times 100end{array} )

(begin{array}{l}frac{Delta rho }{rho }times 100= left ( 2+3times frac{0.01}{2.7}times 100 right )end{array} )

= 3.1%

(3) The dimensional formula for relative refractive index is

a) [M⁰L¹T^-1]

b) [M⁰L⁰T⁰]

c) [M⁰L¹T¹]

d) [MLT^-1]

Answer: b) [M⁰L⁰T⁰]

Solution:

The relative refractive index is the ratio of the refractive index of the medium to the refractive index of the vacuum. Hence, it is a dimensionless quantity.

(4) A thin copper wire of length l metre increases in length by 2% when heated through 10°C. What is the percentage increase in the area when a square copper sheet of length l metre is heated through 10°C?

a) 4%

b) 8%

c) 16%

d) None of these

Answer: a) 4%

Solution:

△l = l αΔT

△l/l = 2/100 = α x 100

α = 2/1000

β = 2α = 4/1000

△A = A βΔT

△A/A = βΔT

= (4/1000) x 10

= 4/100

Percentage increase in area = (4/100) x 100

= 4%

(5) The period of oscillation of a simple pendulum in the experiment is recorded as 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s, respectively. The average absolute error is

a) 0.1 s

b) 0.11 s

c) 0.01 s

d) 1.0 s

Answer: b) 0.11 s

Solution:

Average value = (2.63 + 2.56 + 2.42 + 2.71 + 2.80)/5

= 2.62 sec

Now,

|△T₁| = 2.63 – 2.62 = 0.01

|△T₂| = 2.62- 2.56 = 0.06

|△T₃| = 2.62- 2.42 = 0.20

|△T₄| = 2.71- 2.62 = 0.09

|△T₄| = 2.80- 2.62 = 0.18

Mean absolute error,

(begin{array}{l}Delta T=frac{left | Delta T_{1} right |+left | Delta T_{2} right |+left | Delta T_{3} right |+left | Delta T_{4} right |+left | Delta T_{5} right |}{5}end{array} )

= (0.01 + 0.06 + 0.20 + 0.09 + 0.18)/5

= 0.54/5 = 0.108 = 0.11 sec

Practice Problems

(1) Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of –0.03 mm. While measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is

a) 3.32 mm

b) 3.73 mm

c) 3.67 mm

d) 3.38 mm

(2) To find the distance d over which a signal can be seen clearly in foggy conditions, a railway engineer uses dimensional analysis and assumes that the distance depends on the mass density ρ of the fog, intensity (power/area) S of the light from the signal and its frequency f. The engineer finds that d is proportional to S^1/n . The value of n

a) 2

b) 3

c) 1

d) 4

(3) The energy (E), angular momentum (L) and universal gravitational constant (G) are chosen as fundamental quantities. The dimensions of the universal gravitational constant in the dimensional formula of Planck’s constant (h) is

a) zero

b) -1

c) 5/3

d) 1

(4) The current-voltage relation of the diode is given by I = (e^1000V/T – 1) mA, where the applied V is in volts and the temperature T is in degree Kelvin. If a student makes an error measuring ±0.01 V while measuring the current of 5 mA at 300K, what will be the error in the value of current in mA?

a) 0.2 mA

b) 0.02 mA

c) 0.5 mA

d) 0.05 mA

(5) A student performs an experiment to determine Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ± 0.01 mm. Take, g= 9.8 ms^-2 (exact). The Young’s modulus obtained from the reading is

a) (2 ± 0.3) x 10¹¹N/m²

b) (2 ± 0.2) x 10¹¹N/m²

c) (2 ± 0.1) x 10¹¹N/m²

d) (2 ± 0.05) x 10¹¹N/m²

Frequently Asked Questions

Define S.I unit of Current.

The SI unit of current is Ampere. One ampere is that current which when passed through two straight parallel conductors of infinite length and of negligible cross-section kept at a distance of 1 metre apart in the vacuum produces between them a force equal to 2 x 10-7 newton per metre length.

Name the physical quantities whose units are ohm and Hertz.

The physical quantities whose units are ohm and Hertz are resistance and frequency, respectively.

What are the types of units?

Fundamental units and derived units

A vernier calliper is used to measure the mean diameter of a thin brass rod. Why is a set of 50 measurements more accurate than a set of 5 measurements?

The random errors can be reduced by increasing the number of measurements.

Which of the following is not a unit of time?
(a) Second
(b) Parsec
(c) Year
(d) Light Year

(b) Parsec and (d) Light Year
Solution:
Parsec and Light year are units to measure large distances. For example, the distance between the sun and the earth or other celestial bodies. So they are the units of length and not time.

NCERT Solutions for Class 11 Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements is the best study resource you can get to understand the main topics and to score good grades in the examination. These solutions provide appropriate answers to the textbook questions. Students can also find solutions to exemplary problems, worksheets, questions from previous question papers, numerical problems, MCQs, short answer questions, tips and tricks.

Chapter 2 of NCERT Solutions for Class 11 Physics mainly helps in understanding the fundamentals of units and measurements according to the latest CBSE Syllabus 2022-23. In our daily lives, most of the activities depend on this and it is very important for the students to learn it effectively. From buying milk in the morning to the pounds of bread for the breakfast, from buying sugar for milk to the kilograms of rice needed for lunch, everything depends on the units and measurements.

2.1 Fill in the blanks

(a) The volume of a cube of side 1 cm is equal to …..m³
(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^–1 covers….m in 1 s

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

Answer:

(a) Volume of cube, V = (1 cm)³ = (10^-2 m)³ = 10^-6 m³.

(b) Surface area = curved area + area on top /base = 2πrh + 2πr² = 2πr (h + r)

r = 2 cm = 20 mm

h = 10 cm = 100 mm

Surface area = 2πr (h + r) = 2 x 3.14 x 20 (100 + 20) = 15072 mm²

Hence, answer is 15072 mm²

1 km = 1000 m

1 hr = 60 x 60 = 3600 s

1 km/hr = 1000 m/3600 s = 5/18 m/s

18 km/h = = (18 x 1000)/3600
= 5 m/s

Distance travelled by the vehicle in 1 s = 5 m

(d) The Relative density of lead is 11.3 g cm^-3

=> 11.3 x 10³kg m^-3 [1 kilogram = 10³g, 1 meter = 10²cm]

=> 11.3 x 10³ kg m^-4

2.2 Fill in the blanks by suitable conversion of units

(a) 1 kg m² s^–2 = ….g cm² s^–2

(b) 1 m = ….. ly
(c) 3.0 m s^–2 = …. km h^–2
(d) G = 6.67 × 10^–11 N m² (kg)^–2 = …. (cm)3s^–2 g^–1
Answer:

(a) 1 kg m² s^–2 = ….g cm² s^–2

1 kg m² s^-2= 1kg x 1m² x 1s ^-2

We know that,

1kg = 10³

1m = 100cm = 10²cm

When the values are put together, we get:

1kg x 1m² x 1s^-2= 10³g x (10²cm)² x 1s^-2 = 10³g x 10⁴ cm² x 1s^-2 = 10⁷ gcm²s^-2

=>1kg m² s^-2 = 10⁷ gcm²s^-2

(b) 1 m = ….. ly

Using the formula,

Distance = speed x time

Speed of light = 3 x 10⁸ m/s

Time = 1 yr = 365 days = 365 x 24 hr = 365 x 24 x 60 x 60 sec

Put these values in the formula mentioned above, we get:

One light year distance = (3 x 10⁸ m/s) x (365 x 24 x 60 x 60) = 9.46×10¹⁵m

9.46 x 10¹⁵ m = 1ly

So that, 1m = 1/9.46 x 10¹⁵ly

=> 1.06 x 10^-16ly

=>1 meter = 1.06 x 10^-16ly

1 km = 1000m so that 1m = 1/1000 km

3.0 m s^-2 = 3.0 (1/1000 km) (1/3600 hour) ^-2 = 3.0 x 10^-3 km x ((1/3600)^-2h^-2)

= 3 x 10^-3km x (3600)² hr^-2= 3.88 x 10⁴ km h^-2

=> 3.0 m s^-2 = 3.88 x 10⁴ km h^-2

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

G = 6.67 x 10^-11 N m² (kg)^-2

We know that,

1N = 1kg m s^-2

1 kg = 10³ g

1m = 100cm= 10² cm

Put the values together, we get:

=> 6.67 x 10^-11 Nm² kg^-2 = 6.67 x 10^-11 x (1kg m s ^-2) (1m²) (1kg^-2)

Solve the following and cancelling out the units, we get:

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

Put the above values together to convert kg to g and m to cm

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

=> 6.67 x 10^-8cm³ s^-2g ^-1

=>G = 6.67 x 10^-11Nm²(kg)^-2 = 6.67 x 10^-8 (cm)³ s^-2 g ^-1

2.3 A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J =1 kg m²s^–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 of 4.2 α^–1 β^–2 γ² in terms of the new units.

Answer

1 calorie = 4.2 J = 4.2 kg m²s^–2

The standard formula for the conversion is

frac{Given , unit}{new , unit} = left ( frac{M_{1}}{M_{2}} right )^{x}left ( frac{L_{_{1}}}{L_{2}} right )^{y} left ( frac{T_{1}}{T_{2}} right )^{z}

n e w u n i t G i v e n u n i t = (M ^{2} M ^{1})^{x} (L ^{2} L ^{1})^{y} (T ^{2} T ^{1})^{z}

Dimensional formula for energy = left [ M^{1}L^{2}T^{-2} right ]

[M^{1} L^{2} T^{-2}]

Here, x = 1, y = 2 and z =- 2M₁ = 1 kg, L₁ = 1m, T₁ = 1sand M₂ = α kg, L₂ = β m, T₂ = γ s frac{Calorie}{new , unit} = 4.2left ( frac{1}{alpha } right )^{1}left ( frac{1}{beta } right )^{2} left ( frac{1}{gamma } right )^{-2}

n e w u n i t C a l o r i e = 4.2 (α 1)^{1} (β 1)^{2} (γ 1)^{-2}

Calorie = 4.2 α^–1 β^–2 γ²2.4 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.(a) Atoms are small objectAnswer:(a) In comparison with a soccer ball, atoms are very small(b) When compared with a bicycle, jet plane travels at high speed.(c) When compared with the mass of a cricket ball, the mass of Jupiter is very large.(d) As compared with the air inside a lunch box, the air inside the room has a large number of molecules.(e) A proton is massive when compared with an electron.(f) Like comparing the speed of a bicycle and a jet plane, the speed of light is more than the speed of sound.2.5 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:

Distance between them = Speed of light x Time taken by light to cover the distance

Speed of light = 1 unit

Time taken = 8 x 60 + 20 = 480 + 20 = 500s

The distance between Sun and Earth = 1 x 500 = 500 units.

2.6 Which of the following is the most precise device for measuring length :
(a) a vernier callipers 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?

Answer:

(a) Least count = 1-

frac{9}{10} $10 9$ =

frac{1}{10} $10 1$ = 0.01cm

(b) Least count =

frac{pitch}{number of divisions} $n u m b e r o f d i v i s i o n s p i t c h$

frac{1}{10000} $10000 1$ = 0.001 cm

= 0.00001 cm

We can come to the conclusion that the optical instrument is the most precise device used to measure length.

2.7. A student measures the thickness of a human hair by looking at it through a
microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of the hair?

Answer

Magnification of the microscope = 100
Average width of the hair in the field of view of the microscope = 3.5 mm

Actual thickness of hair =3.5 mm/100 = 0.035 mm

2. 8. Answer the following :
(a)You are given a thread and a metre scale. How will you estimate the diameter of
the thread?
(b)A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do
you think it is possible to increase the accuracy of the screw gauge arbitrarily by
increasing the number of divisions on the circular scale?
(c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why
is a set of 100 measurements of the diameter expected to yield a more reliable
estimate than a set of 5 measurements only?

Answer

(a) The thread should be wrapped around a pencil a number of times so as to form a coil having its turns touching each other closely. Measure the length of this coil with a metre scale. If L be the length of the coil and n be the number of turns of the coil then the diameter of the thread is given by the relation

Diameter = L/n.
(b) Least count of the screw gauge = Pitch/number of divisions on the circular scale

So, theoretically when the number of divisions on the circular scale is increased the least count of the screw gauge will decrease. Hence, the accuracy of the screw gauge will increase. However, this is only a theoretical idea. Practically, there will be many other difficulties when the number of turns is increased.

(c) The probability of making random errors can be reduced to a larger extent in 100 observations than in the case of 5 observations.

2.9 . The photograph of a house occupies an area of 1.75 cm²on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m². What is the linear magnification of the projector-screen arrangement?

Answer

Arial Magnification = Area of the image/Area of the object

= 1.55/1.75 x 10⁴

= 8.857x 10³

Linear Magnification = √Arial magnification

= √8.857x 10³

= 94. 1

2.10 State the number of significant figures in the following :
(a) 0.007 m²
(b) 2.64 × 10²⁴ kg
(c) 0.2370 g cm^–3
(d) 6.320 J
(e) 6.032 N m^–2
(f) 0.0006032 m²

Answer:

(a) 0.007 m²

The given value is 0.007 m².

Only one significant digit. It is 7

(b) 2.64 × 10²⁴ kg

Answer:

The value is 2.64 × 10²⁴ kg

For the determination of significant values, the power of 10 is irrelevant. The digits 2, 6, and 4 are significant figures. The number of significant digits is 3.

(c) 0.2370 g cm^–3

Answer:

The value is 0.2370 g cm^–3

For the given value with decimals, all the numbers 2, 3, 7, and 0 are significant. The 0 before the decimal point is not significant

(d) All the numbers are significant. The number of significant figures here is 4.

(e) 6, 0, 3, 2 are significant figures. Therefore, the number of significant figures is 4.

(f) 6, 0, 3, 2 are significant figures. The number of significant figures is 4.

2. 11. The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

Answer

Area of the rectangular sheet = length x breadth

= 4.234 x 1.005 = 4.255 m²= 4.3 m²

Volume of the rectangular sheet = length x breadth x thickness = 4.234 x 1.005 x 2.01 x 10^-2 = 8.55 x 10^-2 m³.

2.12 The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is

(a) the total mass of the box,

(b) the difference in the masses of the pieces to correct significant figures?

Answer:

The mass of the box = 2.30 kg

and the mass of the first gold piece = 20.15 g

The mass of the second gold piece = 20.17 g

The total mass = 2.300 + 0.2015 + 0.2017 = 2.7032 kg

Since 1 is the least number of decimal places, the total mass = 2.7 kg.

The mass difference = 20.17 – 20.15 = 0.02 g

Since 2 is the least number of decimal places, the total mass = 0.02 g.

2.13 A physical quantity P is related to four observables a, b, c and d as follows:

P =

frac{a^{3}b^{2}}{sqrt{c}d} $c d a ^{3} b ^{2}$

The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result?

Answer:

frac{a^{3}b^{2}}{sqrt{c}d} $c d a ^{3} b ^{2}$

frac{Delta P}{P} $P Δ P$ =

frac{3Delta a}{a} $a 3Δ a$ +

frac{2Delta b}{b} $b 2Δ b$ +

frac{1}{2} $2 1$

frac{Delta c}{c} $c Δ c$ +

frac{Delta d}{d} $d Δ d$

(

frac{Delta P}{P} $P Δ P$ x 100 ) % = ( 3 x

frac{Delta a}{a} $a Δ a$ x 100 + 2 x

frac{Delta b}{b} $b Δ b$ x 100 +

frac{1}{2} $2 1$

frac{Delta c}{c} $c Δ c$ x 100 +

frac{Delta d}{d} $d Δ d$ x 100 ) %

= 3 x 1 + 2 x 3 +

frac{1}{2} $2 1$ x 4 + 2

= 3 + 6 + 2 + 2 = 13 %

P = 4.235

Delta P $Δ P$ = 13 % of P

frac{13P}{100} $100 13 P$

frac{13times 4.235}{100} $100 13 \times 4.235$

= 0.55

The error lies in the first decimal point, so the value of p = 4.3

2.14 A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion:

(a) y = a sin (

frac{2pi t}{T} $T 2 π t$ )

(b) y = a sin vt

(c) y =

frac{a}{T} $T a$ sin

frac{t}{a} $a t$

(d) y =

asqrt{2} $a 2$ ( sin

frac{2pi t}{T} $T 2 π t$ + cos

frac{2pi t}{T} $T 2 π t$ )

Answer:

(a) y = a sin

frac{2pi t}{T} $T 2 π t$

Dimension of y = M⁰ L¹T⁰

The dimension of a = M⁰ L¹ T⁰

Dimension of sin

frac{2pi t}{T} $T 2 π t$ = M⁰ L⁰ T⁰

Since the dimensions on both sides are equal, the formula is dimensionally correct.

(b) It is dimensionally incorrect, as the dimensions on both sides are not equal.

(d) y =

asqrt{2} $a 2$ ( sin

frac{2pi t}{T} $T 2 π t$ + cos

frac{2pi t}{T} $T 2 π t$ )

Dimension of y = M⁰ L¹ T⁰

The dimension of a = M⁰ L¹ T⁰

Dimension of

frac{t}{T} $T t$ = M⁰ L⁰ T⁰

The formula is dimensionally correct.

2.15 A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m_o of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :

m =

frac{m_{0}}{sqrt{1 – nu ^{2}}} $1- ν ^{2} m ^{0}$

Guess where to put the missing c.

Answer:

The relation given is

frac{m_{0}}{sqrt{1 – nu ^{2}}} $1- ν ^{2} m ^{0}$

We can get,

frac{m_{0}}{m} $m m ^{0}$ =

sqrt{1-nu ^{2}} $1 - ν^{2}$

frac{m_{0}}{m} $m m ^{0}$ is dimensionless. Therefore, the right hand side should also be dimensionless.

To satisfy this,

sqrt{1-nu ^{2}} $1 - ν^{2}$ should become

sqrt{1-frac{nu ^{2}}{c^{2}}} $1 - c ^{2} ν ^{2}$ .

Thus, m =

m_{0}sqrt{1-frac{nu ^{2}}{c^{2}}} $m_{0} 1 - c ^{2} ν ^{2}$ .

2.16 The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10^–10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m³ of a mole of hydrogen atoms?

Answer:

hydrogen atom radius = 0.5 A = 0.5 x 10^-10m

Volume =

frac{4}{3}pi r^{3} $3 4 π r^{3}$

frac{4}{3} $3 4$ x

frac{22}{7} $7 22$ x (0.5 x 10^-10)³

= 0.524 x 10^-30 m³

1 hydrogen mole contains 6.023 x 10²³ hydrogen atoms.

Volume of 1 mole of hydrogen atom = 6.023 x 10²³ x 0.524 x 10^-30

= 3.16 x 10^-7 m³.

2.17 One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large?

Answer:

Radius = 0.5 A = 0.5 x 10^-10 m

Volume =

frac{4}{3}pi r^{3} $3 4 π r^{3}$

frac{4}{3} $3 4$ x

frac{22}{7} $7 22$ x ( 0.5 x 10^-10)³

= 0.524 x 10^-30 m³

1 hydrogen mole contains 6.023 x 10²³ hydrogen atoms.

Volume of 1 mole of hydrogen atom = 6.023 x 10²³ x 0.524 x 10^-30

= 3.16 x 10^-7 m³

V_m = 22.4 L = 22.4 x 10^-3 m³

frac{V_{m}}{V_{a}} $V ^{a} V ^{m}$ =

frac{22.4times 10^{-3}}{3.16 times 10^{-7}} $3.16 \times 10 ^{-7} 22.4 \times 10 ^{-3}$ = 7.1 x 10⁴

The molar volume is 7.1 x 10⁴times more than the atomic volume. Hence, the inter-atomic separation in hydrogen gas is larger than the size of the hydrogen atom.

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 (hilltops, 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).

Answer:

An imaginary line which joins the object and the observer’s eye is called the line of sight. When we observe the nearby objects, they move fast in the opposite direction as the line of sight changes constantly. Whereas, the distant objects seem to be stationary as the line of sight does not change rapidly.

2.19. The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit ≈ 3 × 10¹¹m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1” (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1” (second of arc) from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres?

Answer

Diameter of Earth’s orbit = 3 × 10¹¹ m

Radius of Earth’s orbit r = 1.5 × 10¹¹ m

Let the distance parallax angle be θ=1″ (s) = 4.847 × 10^–6 rad.

Let the distance of the star be D.

Parsec is defined as the distance at which the average radius of the Earth’s orbit subtends an angle of 1″

Therefore, D = 1.5 × 10¹¹ /4.847 × 10^–6= 0.309 x 10¹⁷

Hence 1 parsec ≈ 3.09 × 10¹⁶ m.

2. 20. The nearest star to our solar system is 4.29 light-years away. How much is this
distance in terms of parsecs? How much parallax would this star (named Alpha
Centauri) show when viewed from two locations of the Earth six months apart in its
orbit around the Sun?

Answer

1 light year is the distance travelled by light in a year

1 light year = 3 x 108 x 365 x 24 x 60 x 60 = 9.46 x 10¹⁵ m

Therefore, distance travelled by light in 4.29 light years = 4.29 x 9.46 x 10¹⁵ = 4.058 x 10¹⁶ m

Parsec is also a unit of distance
1 parsec = 3.08 x 10¹⁶ m
Therefore, the distance travelled by light in parsec is given as

4.29 light years =4.508 x 10¹⁶/3.80 x 10¹⁶ = 1.318 parsec = 1.32 parsec.

Using the relation,
θ = d / D
here,
d is the diameter of Earths orbit, d = 3 × 10¹¹ m
D is the distance of the star from the earth, D = 405868.32 × 10¹¹ m
∴ θ = 3 × 10¹¹/ 405868.32 × 10¹¹ = 7.39 × 10^-6 rad
But the angle covered in 1 sec = 4.85 × 10^–6 rad
∴ 7.39 × 10^-6 rad = 7.39 × 10^-6/ 4.85 × 10^-6= 1.52″

2.21 Precise measurements of physical quantities are a need for science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.

Answer:

Precise measurement is essential for the development of science. The ultra-short laser pulse is used for measurement of time intervals. X-ray spectroscopy is used to find the interatomic separation. To measure the mass of atoms, the mass spectrometer is developed.

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: a mass of the Sun = 2.0 × 10³⁰ kg, radius of the Sun = 7.0 × 10⁸ m.

Answer:

Mass = 2 x 10³⁰ kg

Radius = 7 x 10⁸ m

Volume V =

frac{4}{3}pi r^{3} $3 4 π r^{3}$

frac{4}{3} $3 4$ x

frac{22}{7} $7 22$ x (7 x 10⁸)³

frac{88}{21} $21 88$ x 512 x 10²⁴m³ = 2145.52 x 10²⁴ m³

Density =

frac{Mass}{Volume} $V o l u m e M a s s$ =

frac{3times 10^{30}}{2145.52times 10^{24}} $2145.52 \times 10 ^{24} 3 \times 10 ^{30}$ = 1.39 x 10³ kg/m⁵.

The density is in the range of solids and liquids. Its density is due to the high gravitational attraction on the outer layer by the inner layer of the sun.

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

Answer:

Distance of the planet Jupiter from Earth, D= 824.7 million kilometres = 824.7 x 10⁶ km

Angular diameter θ = 35.72 “= 35.72 x 4.85 x 10^-6 rad
= 173.242 x 10^-6 rad
Diameter of Jupiter d = θ x D= 173.241 x 10^-6x 824.7 x 10⁶ km
=142871 = 1.43 x 10⁵ km

2.25. A man walking briskly in rain with speed v must slant his umbrella forward making an angle θ with the vertical. A student derives the following relation between θ and v: tan θ = v and checks that the relation has a correct limit: as v →0, θ →0, as
expected. (We are assuming there is no strong wind and that the rain falls vertically
for a stationary man). Do you think this relation can be correct? If not, guess the
correct relation.

Answer

According to the principle of homogeneity of dimensional equations,
Dimensions of L.H.S = Dimensions of R.H.S

In relation v = tan θ, tan θ is a trigonometric function and it is dimensionless. The dimension of v is [L¹ T^-1]. Therefore, this relation is incorrect.
To make the relation correct, the L.H.S must be divided by the velocity of rain, u.

Therefore, the relation becomes
v/u= tan θ

This relation is correct dimensionally

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?

Answer

Total time = 100 years = 100 x 365 x 24 x 60 x 60 s

Error in 100 years = 0.02 s
Error in 1 second=0.02/100 x 365 x 24 x 60 x 60
=6.34 x 10^-12 s
Accuracy of the standard cesium clock in measuring a time-interval of 1 s is 10^-12 s

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 m^–3. Are the two densities of the same order of magnitude? If so, why?

Answer

The diameter of sodium= 2.5 A = 2.5 x 10^-10 m

Therefore, the radius is 1.25 x 10^-10 m

Volume of sodium atom, V= (4/3)πr³

= (4/3) x (22/7) x (1.25 x 10^-10) ³= 8.177 x 10^-30 m³

Mass of one mole atom of sodium = 23 g = 23 x 10^-3 kg

1 mole of sodium contains 6.023 x 10²³ atoms

Therefore, the mass of one sodium atom, M= 23 x 10^-3/6.023 x 10²³= 3.818 x 10^-26 kg

Atomic mass density of sodium, ρ= M/V
=3.818 x 10^-26/8.177 x 10^-30

= 0.46692 x 10⁴= 4669.2 kg m^-3
The density of sodium in its solid state is 4669.2 kg m^-3but in the crystalline phase, density is 970 kg m^-3. Hence, both are in a different order. In solid-phase, atoms are tightly packed but in the crystalline phase, atoms arrange a sequence which contains void. So, density in solid-phase is greater than in the crystalline phase.

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 = r₀A^1/3
where r is the radius of the nucleus, A its mass number, and r₀is 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 the sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

Answer:

Radius of the nucleus

r = r₀ A^1/3

r_o = 1.2 f = 1.2 x 10^-15 m

Considering the nucleus is spherical. Volume of nucleus
= 4/3 πr³ = 4/3 π [r₀ A^1/3]³ = 4/3 πr₀³A

Mass of nucleus = mA

m is the average mass of the nucleon

A is the number of nucleons

Nuclear mass density = Mass of nucleus/Volume of nucleus
= mA/(4/3πr³) = 3mA/4πr³= 3mA/4πr₀³A

= 3m/4πr₀³

Using m = 1.66 x 10^-27 kg and r_o = 1.2 f = 1.2 x 10^-15 m in the above equation

= 3 x 1.66 x 10^-27 /4 x 3.14 x ( 1.2 x 10^-15)³= 4.98 x 10^-27/21. 703 x 10^-45= 2.29 x 10¹⁷ kg/m³

So, the nuclear mass density is much larger than atomic mass density for a sodium atom we got in 2.27.

2.29. A LASER is a source of very intense, monochromatic, and the 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 ?

Answer

Time taken for the laser beam to return to Earth after reflection by the Moon’s surface = 2.56 s

The speed of laser light ,c = 3 x 10⁸m/s.

Let d be the distance of Moon from the Earth,

The time taken by laser signal to reach the Moon, t = 2d/c

Therefore, d = tc/2 = (2.56 x 3 x 10⁸)/2 = 3.84 x 10⁸ m

2. 30. A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects underwater. In a submarine equipped with a SONAR, the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in water = 1450 m s^–1).

Answer:

Speed of sound in water,v = 1450 m s^–1

Time between generation and the reception of the echo after reflection, 2t= 77.0 s

Time taken for the sound waves to reach the submarine, t = 77.0/2 = 38. 5 s

Then v = d/t

Distance of enemy submarine, d = tv

Therefore, d=vt=(1450 x 38. 5) =55825 m=55.8 x 10³ m or 55.8 km.

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?

Answer

Time taken by light from the quasar to reach the observer, t = 3.0 billion years = 3.0 x 10⁹ years = 3.0 x 10⁹x 365 x 24 x 60 x 60 s

= 94608000 x 10⁹ s

= 9.46 x 10¹⁶m

Speed of light = 3 x 10⁸ m/s
Distance of quasar from Earth = 3.0 x 10⁸ x 9.46 x 10¹⁶m
= 28.38 x 10²⁴ m

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.

Answer

From examples 2.3 and 2.4 we get the following data

Distance of the Moon from Earth = 3.84 x 10⁸ m

Distance of the Sun from Earth = 1.496 x 10¹¹ m

Sun’s diameter = 1.39 x 10⁹ m

Sun’s angular diameter,θ = 1920″ = 1920 x 4.85 x 10^-6 rad = 9.31 x 10^-3 rad [1″ = 4.85 x 10^-6 rad]

During a total solar eclipse, the disc of the moon completely covers the disc of the sun, so the angular diameter of both the sun and the moon must be equal.

Therefore, Angular diameter of the moon, θ = 9.31 x 10^-3 rad
The earth-moon distance, S = 3.8452 x 10⁸ m

Therefore, the diameter of the moon, D = θ x S
= 9.31 x 10^-3x 3.8452 x 10⁸ m = 35.796 x 10⁵ m

Frequently Asked Questions on NCERT Solutions for Class 11 Physics Chapter 2

Why should the students download the NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements PDF?

The NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements PDF contains diagrams and answers for all the questions present in the textbook. Each and every question is answered by keeping in mind the understanding abilities of students. The solutions created strictly adhere to the latest CBSE Syllabus 2022-23 and exam pattern to help students face the exams without fear. It also improves their time management skills which are important from the exam point of view.

Why are NCERT Solutions for Class 11 Physics Chapter 2 beneficial for the students?

The benefits of using the NCERT Solutions for Class 11 Physics Chapter 2 are –
1. Completely solved answers for all the questions present in the NCERT textbook are available in PDF format.
2. Simple and easy to understand language is used to make learning fun for the students.
3. Subject matter experts prepare the solutions after conducting vast research on each concept.
4. The solutions not only help students with their exam preparation but also for various competitive exams like JEE, NEET, etc.
5. PDF format of solutions are available in chapter wise and exercise wise format to help students learn the concepts with ease.

What are SI Units according to NCERT Solutions for Class 11 Physics Chapter 2?

The International System Of Units (SI) is the metric system that is used universally as a standard for measurements. SI units play a vital role in scientific and technological research and development.

MCQs on Class 11 Chapter 2 Units and Measurements

Check the multiple-choice questions for the 11th Class Physics units and measurements chapter. Each MCQ will have four options here, out of which only one is correct. Students have to pick the correct option and check the answer provided here.

1. The symbol to represent “Amount of Substance” is ________

Answer: (d) mol

Explanation: The symbol to represent Amount of Substance is mol.

2. Which among the following is the Supplementary Unit——–

Mass
Time
Solid angle
Luminosity

Answer: (c) Solid angle

Explanation: Supplementary units are plane angles and solid angles. Other units mentioned are base units.

3. What is the unit of solid angle?

second
Steradian
kilogram
candela

Answer: (b) Steradian

Explanation: Steradian is the unit of solid angle.

4. AU is the unit of ________

Astronomy Unit
Astronomical unit
Astrological Unit
Archaeological Unit

Answer: (b) Astronomical unit

Explanation: Astronomical unit is the average distance of the Sun from the Earth. It is represented by the symbol AU.

5. Dimensions of kinetic energy is the same as that of ________

Acceleration
Velocity
Work
Force

Answer: (c) Work

Explanation: Dimensions of kinetic energy and work are the same.

6. Farad is the unit of ________

Luminosity
Wavelength
Permittivity
Inertia

Answer: (c) Permittivity

Explanation: Permittivity is the unit of Farad.

7. Electron volt is a unit of

Luminosity
Frequency
Force
Energy

Answer: (d) Energy

Explanation: One of the units of energy is electron volt.

8. Joule second is the unit of

Force
Angular momentum
Energy
Power

Answer: (b) Angular momentum

Explanation: Angular momentum is also known as rotational momentum. The total angular momentum of a closed system remains constant.

9. The smallest value which is measured using an instrument is known as ________

Absolute count
Least count
Round off value
Minimum count

Answer: (b) Least count

10. Which is the system of unit

SMS system
MKP system
FPS System
CJS System

Answer: (c) FPS System

Explanation: FPS System is one of the systems of units. Foot, Pound, and Second (FPS system).

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Table of contents

NCERT Solutions for Class 11 Physics Chapter 2 – Free PDF Download

What is measurement ?

Physical Quantities

Units

System of Units

Definition of Basic and Supplementary Units

Dimensional Formula

Dimensional Analysis

Units and Dimensions Of A Few Derived Quantities

Principle of Homogeneity

Defects of Dimensional Analysis

Points to Remember

Solved Examples

Practice Problems

Frequently Asked Questions

Define S.I unit of Current.

Name the physical quantities whose units are ohm and Hertz.

What are the types of units?

A vernier calliper is used to measure the mean diameter of a thin brass rod. Why is a set of 50 measurements more accurate than a set of 5 measurements?

Which of the following is not a unit of time? (a) Second (b) Parsec (c) Year (d) Light Year

NCERT Solutions for Class 11 Physics Chapter 2

Why should the students download the NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements PDF?

Why are NCERT Solutions for Class 11 Physics Chapter 2 beneficial for the students?

What are SI Units according to NCERT Solutions for Class 11 Physics Chapter 2?

MCQs on Class 11 Chapter 2 Units and Measurements

Which of the following is not a unit of time?
(a) Second
(b) Parsec
(c) Year
(d) Light Year