physics class 11 chapter 2 assignment 2.3

NCERT Solutions for Class 11 Physics Chapter 2 Free PDF Download

Ncert solutions for class 11 physics chapter 2 – units and measurement.

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CBSE Class 11 Physics Chapter 15 Units and Measurement NCERT Solutions

Unit refers to the international standards that are accepted worldwide. While on the other hand, measurement refers to the use of the unit for measuring the quantity. In this NCERT Solutions for Class 11 Physics Chapter 2, we will discuss all the topic of the chapter in brief and in an easily understandable language so that students can get what the content of the chapter.

Subtopics covered under NCERT Solutions for Class 11 Physics Chapter 15

2.1 introduction.

This topic discusses what is unit and measurement which we have discussed above.

2.2 The International System of Units

This topic defines that earlier scientist of different countries uses the different unit for measurement. But with time and international usage, they started using SI (Standard Unit).

2.3 Measurement of Length

This topic overview the different unit of measuring the length that is used worldwide.

2.3.1 Measurement of Large Distances- This topic defines the use of parallax method for measuring long distances. Besides, the topic provides various examples of using large distances.

2.3.2 Estimation of very Small Distances: Size of a Molecule- This topic defines the use of the small unit by which we can measure the distance of the molecule.

2.3.3 Range of Lengths- this topic defines the use of such unit that helps to measure the lengths of objects that are spread in this wide universe.

2.4 Measurement of Mass

The weight of any object is its mass and this topic discusses the various unit of measurement of mass.

2.4.1 Range of Masses- It refers to all those objects that are spread over the universe that has a fixed mass and can be measured using these units.

2.5 Measurement of Time

This topic defines the atomic standard of time that is used for measuring the atomic clock and cesium clock.

2.6 Accuracy, Precision of Instruments and Errors in Measurement

This topic firstly defines what are errors and the mistakes we do which causes the errors.

Systematic Errors
Instrumental errors
Imperfection in experimental technique or procedure
Personal errors
Random Errors
Least Count Errors

2.6.1 Absolute Error, Relative Error, and Percentage Error- This topic defines all the three errors and how they affect the results.

2.6.2 Combination of Errors- This topic defines the error that we do while performing several measurements

An error of a sum or a difference
The error of a product or a quotient
Error in case of a measured quantity raised to a power

2.7 Significant Figures

This refers to the first uncertain digit plus the reliable digit. Moreover, this topic several other points related to significant figures.

2.7.1 Rules of Arithmetic Operations with Significant Figures- The topic define the rules related to the arithmetic operations.

2.7.2 Rounding off the Uncertain Digits- This topic defines the rule of rounding off of unclear digits.

2.7.3 Rules of Determining the Uncertainty in the Results of Arithmetic Calculations- This topic defines the rules that help to govern the doubt of the results of mathematics calculations.

2.8 Dimensions of Physical Quantities

This topic defines the scope of physical quantities.

2.9 Dimensional Formulae and Dimensional Equations

This topic describes the dime national equations and formula that we use to equitize a physical quantity.

2.10 Dimensional Analysis and its Applications

This topic defines the various applications by which we can analyze the physical quantities of a dimension.

2.10.1 Checking the Dimensional Consistency of Equations- This topic checks the consistency of dimensional equations.

2.10.2 Deducing Relation among the physical Quantities- This topic defines how we can use the relation of the physical quantities for reasoning.

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NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement is the best study resource you can get to understand the topics and to score good grades in your class 11 final examination. This solution provides appropriate answers to the questions provided in the textbook. Along with the textbook question, this solution has exemplary problems, worksheets, questions from previous question papers, numerical problems, MCQs, short answer questions, tips and tricks.

Download NCERT Solutions Class 11 Physics Chapter 2 PDF:- Download Here

ncert solutions april9 class 11 physics chapter 2 units and measurements 01

Chapter 2 Units and Measurement Class 11 Physics NCERT Solutions

2.1 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 –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 .

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

(b) Surface area = curved area + area on top /base = 2πrh + 2πr 2 = 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 mm 2

Hence, answer is 15072 mm 2

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 10 3 kg m -3 [1 kilogram = 10 3 g, 1 meter = 10 2 cm]

=> 11.3 x 10 3 kg m -4

2.2 Fill in the blanks by suitable conversion of units

(a) 1 kg m 2 s –2 = ….g cm 2 s –2

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

1 kg m 2 s -2 = 1kg x 1m 2 x 1s -2

We know that,

1m = 100cm = 10 2 cm

When the values are put together, we get:

1kg x 1m 2 x 1s -2 = 10 3 g x (10 2 cm) 2 x 1s -2 = 10 3 g x 10 4 cm 2 x 1s -2 = 10 7 gcm 2 s -2

=> 1kg m 2 s -2 = 10 7 gcm 2 s -2

(b) 1 m = ….. ly

Using the formula,

Distance = speed x time

Speed of light = 3 x 10 8 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 10 8 m/s) x (365 x 24 x 60 x 60) = 9.46×10 15 m

9.46 x 10 15 m = 1ly

So that, 1m = 1/9.46 x 10 15 ly

=> 1.06 x 10 -16 ly

=> 1 meter = 1.06 x 10 -16 ly

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) -2 h -2 )

= 3 x 10 -3 km x (3600) 2 hr -2 = 3.88 x 10 4 km h -2

=> 3.0 m s -2 = 3.88 x 10 4 km h -2

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

G = 6.67 x 10 -11 N m 2 (kg) -2

1N = 1kg m s -2

1 kg = 10 3 g

1m = 100cm= 10 2 cm

Put the values together, we get:

=> 6.67 x 10 -11 Nm 2 kg -2 = 6.67 x 10 -11 x (1kg m s -2 ) (1m 2 ) (1kg -2 )

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

=> 6.67 x 10 -11 x (1kg -1 x 1m 3 x 1s -2 )

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

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

=> 6.67 x 10 -8 cm 3 s -2 g -1

=> G = 6.67 x 10 -11 Nm 2 (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 m 2 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 of 4.2 α –1 β –2 γ 2 in terms of the new units.

1 calorie = 4.2 J = 4.2 kg m 2 s –2

The standard formula for the conversion is

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

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

M 1 = 1 kg, L 1 = 1m, T 1 = 1s

and M 2 = α kg, L 2 = β m, T 2 = γ s

Calorie = 4.2 α –1 β –2 γ 2

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.

(a) Atoms are small object

(a) In comparison with a soccer ball, atoms are very small

(b) When compared with a bicycle, jet plane travels at high speed.

(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 ?

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?

(a) Least count = 1- 9 10 frac{9}{10} 1 0 9 = 1 10 frac{1}{10} 1 0 1 = 0.01cm

(b) Least count = p i t c h n u m b e r o f d i v i s i o n s 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

= 1 10000 frac{1}{10000} 1 0 0 0 0 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?

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?

(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 increase. 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 cm 2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m 2 . What is the linear magnification of the projector-screen arrangement?

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

= 1.55/1.75 x 10 4

= 8.857x 10 3

Linear Magnification = √Arial magnification

= √8.857x 10 3

2.10 State the number of significant figures in the following : (a) 0.007 m 2 (b) 2.64 × 10 24 kg (c) 0.2370 g cm –3 (d) 6.320 J (e) 6.032 N m –2 (f) 0.0006032 m 2

(a) 0.007 m 2

The given value is 0.007 m 2 .

Only one significant digit. It is 7

(b) 2.64 × 10 24 kg

The value is 2.64 × 10 24 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.

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.

Area of the rectangular sheet = length x breadth

= 4.234 x 1.005 = 4.255 m 2 = 4.3 m 2

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 m 3 .

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?

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 = a 3 b 2 c d 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?

( Δ P P frac{Delta P}{P} P Δ P x 100 ) % = ( 3 x Δ a a frac{Delta a}{a} a Δ a x 100 + 2 x Δ b b frac{Delta b}{b} b Δ b x 100 + 1 2 frac{1}{2} 2 1 Δ c c frac{Delta c}{c} c Δ c x 100 + Δ d d frac{Delta d}{d} d Δ d x 100 ) %

= 3 x 1 + 2 x 3 + 1 2 frac{1}{2} 2 1 x 4 + 2

= 3 + 6 + 2 + 2 = 13 %

= 13 P 100 frac{13P}{100} 1 0 0 1 3 P

= 13 × 4.235 100 frac{13 imes 4.235}{100} 1 0 0 1 3 × 4 . 2 3 5

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 ( 2 π t T frac{2pi t}{T} T 2 π t )

(b) y = a sin vt

(d) y = a 2 asqrt{2} a 2 ( sin 2 π t T frac{2pi t}{T} T 2 π t + cos 2 π t T frac{2pi t}{T} T 2 π t )

(a) y = a sin 2 π t T frac{2pi t}{T} T 2 π t

Dimension of y = M 0 L 1 T 0

The dimension of a = M 0 L 1 T 0

Dimension of sin 2 π t T frac{2pi t}{T} T 2 π t = M 0 L 0 T 0

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.

Dimension of t T frac{t}{T} T t = M 0 L 0 T 0

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 = m 0 1 – ν 2 frac{m_{0}}{sqrt{1 – u ^{2}}} 1 – ν 2 m 0

Guess where to put the missing c.

The relation given is m 0 1 – ν 2 frac{m_{0}}{sqrt{1 – u ^{2}}} 1 – ν 2 m 0

We can get, m 0 m frac{m_{0}}{m} m m 0 = 1 − ν 2 sqrt{1- u ^{2}} 1 − ν 2 m 0 m frac{m_{0}}{m} m m 0 is dimensionless. Therefore, the right hand side should also be dimensionless.

To satisfy this, 1 − ν 2 sqrt{1- u ^{2}} 1 − ν 2 should become 1 − ν 2 c 2 sqrt{1-frac{ u ^{2}}{c^{2}}} 1 − c 2 ν 2 .

Thus, m = m 0 1 − ν 2 c 2 m_{0}sqrt{1-frac{ u ^{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 m 3 of a mole of hydrogen atoms?

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

Volume = 4 3 π r 3 frac{4}{3}pi r^{3} 3 4 π r 3

= 4 3 frac{4}{3} 3 4 x 22 7 frac{22}{7} 7 2 2 x (0.5 x 10 -10 ) 3

= 0.524 x 10 -30 m 3

1 hydrogen mole contains 6.023 x 10 23 hydrogen atoms.

Volume of 1 mole of hydrogen atom = 6.023 x 10 23 x 0.524 x 10 -30

= 3.16 x 10 -7 m 3 .

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?

Radius = 0.5 A = 0.5 x 10 -10 m

= 4 3 frac{4}{3} 3 4 x 22 7 frac{22}{7} 7 2 2 x ( 0.5 x 10 -10 ) 3

= 3.16 x 10 -7 m 3

V m = 22.4 L = 22.4 x 10 -3 m 3

The molar volume is 7.1 x 10 4 times 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).

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 × 10 11 m. 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?

Diameter of Earth’s orbit = 3 × 10 11 m

Radius of Earth’s orbit r = 1.5 × 10 11 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 × 10 11 /4.847 × 10 –6 = 0.309 x 10 17

Hence 1 parsec ≈ 3.09 × 10 16 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?

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 10 15 m

Therefore, distance travelled by light in 4.29 light years = 4.29 x 9.46 x 10 15 = 4.058 x 10 16 m

Parsec is also a unit of distance 1 parsec = 3.08 x 10 16 m Therefore, the distance travelled by light in parsec is given as

4.29 light years =4.508 x 10 16 /3.80 x 10 16 = 1.318 parsec = 1.32 parsec.

Using the relation, θ = d / D here, d is the diameter of Earths orbit, d = 3 × 10 11 m D is the distance of the star from the earth, D = 405868.32 × 10 11 m ∴ θ = 3 × 10 11 / 405868.32 × 10 11 = 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 r adar 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.

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 10 7 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 × 10 30 kg, radius of the Sun = 7.0 × 10 8 m.

Mass = 2 x 10 30 kg

Radius = 7 x 10 8 m

Volume V = 4 3 π r 3 frac{4}{3}pi r^{3} 3 4 π r 3

= 4 3 frac{4}{3} 3 4 x 22 7 frac{22}{7} 7 2 2 x (7 x 10 8 ) 3

= 88 21 frac{88}{21} 2 1 8 8 x 512 x 10 24 m 3 = 2145.52 x 10 24 m 3

Density = M a s s V o l u m e frac{Mass}{Volume} V o l u m e M a s s = 3 × 1 0 30 2145.52 × 1 0 24 frac{3 imes 10^{30}}{2145.52 imes 10^{24}} 2 1 4 5 . 5 2 × 1 0 2 4 3 × 1 0 3 0 = 1.39 x 10 3 kg/m 5 .

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.

Distance of the planet Jupiter from Earth, D= 824.7 million kilometres = 824.7 x 10 6 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 -6 x 824.7 x 10 6 km =142871 = 1.43 x 10 5 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.

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 [L 1 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?

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?

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)πr 3

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

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

1 mole of sodium contains 6.023 x 10 23 atoms

Therefore, the mass of one sodium atom, M= 23 x 10 -3 /6.023 x 10 23 = 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 10 4 = 4669.2 kg m -3 The density of sodium in its solid state is 4669.2 kg m -3 but 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 = r 0 A 1/3 where r is the radius of the nucleus, A its mass number, and r 0 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 the sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

Radius of the nucleus

r = r 0 A 1/3

r o = 1.2 f = 1.2 x 10 -15 m

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 ?

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 10 8 m/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 10 8 )/2 = 3.84 x 10 8 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 ).

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 10 3 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?

Time taken by light from the quasar to reach the observer, t = 3.0 billion years = 3.0 x 10 9 years = 3.0 x 10 9 x 365 x 24 x 60 x 60 s

= 94608000 x 10 9 s

= 9.46 x 10 16 m

Speed of light = 3 x 10 8 m/s Distance of quasar from Earth = 3.0 x 10 8 x 9.46 x 10 16 m = 28.38 x 10 24 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.

From examples 2.3 and 2.4 we get the following data

Distance of the Moon from Earth = 3.84 x 10 8 m

Distance of the Sun from Earth = 1.496 x 10 11 m

Sun’s diameter = 1.39 x 10 9 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 10 8 m

Therefore, the diameter of the moon, D = θ x S = 9.31 x 10 -3 x 3.8452 x 10 8 m = 35.796 x 10 5 m

For students of Class 11 who are looking to give their best for the upcoming final exams and competitive exams, it is very important to get accustomed with the solutions to the questions given in the textbook. Thus, students are advised to have a good practice of different kinds of questions that can be framed from the chapter. Students are suggested to solve the NCERT questions. To clear all the doubts of the students, CoolGyan’S provides Solution to NCERT Class 11 Physics Chapter 2 Units and Measurement.

Topics covered in Class 11 Chapter 2 Physics Units and Measurement


2.1	Introduction
2.2	The International System Of Units
2.3	Measurement Of Length
2.4	Measurement Of Mass
2.5	Measurement Of Time
2.6	Accuracy, Precision Of Instruments And Errors In Measurement
2.7	Significant Figures
2.8	Dimensions Of Physical Quantities
2.9	Dimensional Formulae And Dimensional Equations
2.10	Dimensional Analysis And Its Applications

Scientists gather information with their senses like eyes, ears, etc. and make observations. Some observations are simple like figuring out the texture and colour while other observations may be complex for which they may need to take measurements. Measurement is one of the fundamental concepts in science. Without the ability to measure, a scientist wouldn’t be able to gather information and form a theory or conduct experiments. In this chapter, the units of physical quantities and methods of evaluating them are discussed, while the other section of the chapter deals with the errors that can occur while taking measurement and significant figures. By practising problems from NCERT Solutions Class 11 Physics Units and Measurement one gets a good understanding of measurement.

Along with Chapter 2, CoolGyan’S provides NCERT solutions for all the subjects of all the classes. CoolGyan also provides notes, study materials, numerical problems, previous year question papers, sample papers and competitive exam study materials to help you score good marks in the Class 11 examination and competitive examinations.

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

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physics class 11 chapter 2 assignment 2.3

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NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements are part of Class 11 Physics NCERT Solutions . Here we have given NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements.

Topics and Subtopics in NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements :


2	Units and Measurements
2.1	Introduction
2.2	The international system of units
2.3	Measurement of length
2.4	Measurement of mass
2.5	Measurement of time
2.6	Accuracy, precision of instruments and errors in measurement
2.7	Significant figures
2.8	Dimensions of physical quantities
2.9	Dimensional formulae and dimensional equations
2.10	Dimensional analysis and its applications

NCERT Solutions Class 11 Physics Physics Sample Papers

QUESTIONS FROM TEXTBOOK

Question 2. 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 -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 m 3 . Hence, answer is 10 -6 (b) Surface area = 2πrh + 2πr 2 = 2πr (h + r) = 2 x 22/7 x 2 x 10 (10 x 10 + 2 x 10) mm 2 = 1.5 x 10 4 mm 2 Hence, answer is 1.5 x 10 4 . (c) Speed of vehicle = 18 km/h = 18 x 1000/3600 m/s = 5 m/s ; so the vehicle covers 5 m in 1 s. = 11.3 (d) Density= 11.3 g cm -3 =11.3 x 10 3 kg m -3 [1 kg =10 3 g,1m=10 2 cm] =11.3 x 10 3 kg m -4

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Question 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. Answer: Physical quantities are called large or small depending on the unit (standard) of measurement. For example, the distance between two cities on earth is measured in kilometres but the distance between stars or inter —galactic distances are measured in parsec. The later standard parsec is equal to 3.08 x 10 16 m or 3.08 x 10 12 km is certainly larger than metre or kilometre. Therefore, the inter-stellar or intergalactic distances are certainly larger than the distances between two cities on earth. (a) The size of an atom is much smaller than even the sharp tip of a pin. (b) A Jet plane moves with a speed greater than that of a super fast train. (c) The mass of Jupiter is very large compared to that of the earth. (d) The air inside this room contains more number of molecules than in one mole of air. (e) This is a correct statement. (f) This is a correct statement.

Question 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 Sun and Earth = Speed of light in vacuum x time taken by light to travel from Sim to Earth = 3 x 10 8 m/ s x 8 min 20 s = 3 x 10 8 m/s x 500 s = 500 x 3 x 10 8 m. In the new system, the speed of light in vacuum is unity. So, the new unit of length is 3 x 10 8 m. .•. distance between Sun and Earth = 500 new units.

Question 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 of vernier callipers = 1/20 = 0.05 mm = 5 x 10 -5 m (b) Least count of screw gauge =Pitch/No. of divisions on circular scale = 1 x 10 -3 /100 = 1 x 10 -5 m (c) Least count of optical instrument = 6000 A (average wavelength of visible light as 6000 A) = 6 x 10- 7 m As the least count of optical instrument is least, it is the most precise device out of three instruments given to us.

Question 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 hair? Answer: As magnification, m =thickness of image of hair/ real thickness of hair = 100 and average width of the image of hair as seen by microscope = 3.5 mm .•. Thickness of hair =3.5 mm/100 = 0.035 mm

Question 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) Wrap the thread a number of times on a round pencil so as to form a coil having its turns touching each other closely. Measure the length of this coil, mode by the thread, with a metre scale. If n be the number of turns of the coil and l be the length of the coil, then the length occupied by each single turn i.e., the thickness of the thread = 1/n . This is equal to the diameter of the thread. (b) We know that least count = Pitch/number of divisions on circular scale When number of divisions on circular scale is increased, least count is decreased. Hence the accuracy is increased. However, this is only a theoretical idea.Practically speaking, increasing the number of ‘turns would create many difficulties. As an example, the low resolution of the human eye would make observations difficult. The nearest divisions would not clearly be distinguished as separate. Moreover, it would be technically difficult to maintain uniformity of the pitch of the screw throughout its length. (c) Due to random errors, a large number of observation will give a more reliable result than smaller number of observations. This is due to the fact that the probability (chance) of making a positive random error of a given magnitude is equal to that of making a negative random error of the same magnitude. Thus in a large number of observations, positive and negative errors are likely to cancel each other. Hence more reliable result can be obtained.

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements Q9

Question 2. 10. State the number of significant figures in the following: (a) 0.007 m 2 (b) 2.64 x 10 4 kg (c) 0.2370 g cm -3 (d) 6.320 J (e) 6.032 N m -2 (f) 0.0006032 m 2 Answer: (a) 1 (b) 3 (c) 4 (d) 4 (e) 4 (f) 4.

Question 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: As Area = (4.234 x 1.005) x 2 = 8.51034 = 8.5 m 2 Volume = (4.234 x 1.005) x (2.01 x 10 -2 ) = 8.55289 x 10 -2 = 0.0855 m 3 .

Question 2. 12. The mass of a box measured by a grocer’s balance is 2.3 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: (a) Total mass of the box = (2.3 + 0.0217 + 0.0215) kg = 2.3442 kg Since the least number of decimal places is 1, therefore, the total mass of the box = 2.3 kg. (b) Difference of mass = 2.17 – 2.15 = 0.02 g Since the least number of decimal places is 2 so the difference in masses to the correct significant figures is 0.02 g.

Question 2. 16. The unit of length convenient on the atomic scale is known as an angstrom and is denoted by A: 1 A = 10 -10 m. The size of a hydrogen atom is about 0.5 A. What is the total atomic volume in m 3 of a mole of hydrogen atoms? Answer: Volume of one hydrogen atom = 4/3 πr3 (volume of sphere) = 4/3 x 3.14 x (0.5 x 10 -10 ) m 3 = 5.23 x 10 -31 m 3 According to Avagadro’s hypothesis, one mole of hydrogen contains 6.023 x 10 23 atoms. Atomic volume of 1 mole of hydrogen atoms = 6.023 x 1023 x 5.23 x 10 -31 = 3.15 x 10 -7 m 3 .

Question 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 A.) Why is this ratio so large? Answer: Volume of one mole of ideal gas, V g = 22.4 litre = 22.4 x 10 -3 m 3 Radius of hydrogen molecule = 1A/2 = 0.5 A = 0.5 x 10 -10 m Volume of hydrogen molecule = 4/3 πr 3 =4/3 x 22/7 (0.5 x 10 -10 ) 3 m 3 = 0.5238 x 10 -30 m 3 One mole contains 6.023 x 10 23 molecules. Volume of one mole of hydrogen, VH = 0.5238 x 10 -30 x 6.023 x 10 23 m 3 = 3.1548 x 10 -7 m 3 Now V g /V H =22.4 x 10 -3 /3.1548 x 10 -7 =7.1 x 10 4 The ratio is very large. This is because the interatomic separation in the gas is very large compared to the size of a hydrogen molecule.

Question 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). Answer: The line joining a given object to our eye is known as the line of sight. When a train moves rapidly, the line of sight of a passenger sitting in the train for nearby trees changes its direction rapidly. As a result, the nearby trees and other objects appear to run in a direction opposite to the train’s motion. However, the line of sight of distant and large size objects e.g., hill tops, the Moon, the stars etc., almost remains unchanged (or changes by an extremely small angle). As a result, the distant object seems to be stationary.

Question 2. 19. The principle of ‘parallax’ 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 x 10 n m. However, even the nearest stars are so distant that with such a long baseline, they show parallel 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: From parallax method we can say θ=b/D,where b=baseline ,D = distance of distant object or star Since, θ=1″ (s) and b=3 x 10 11 m D=b/20=3 x 10 11 /2 x 4.85 x 10 -6 m or D=3 x 10 11 /9.7 x 10 -6 m =30 x 10 16 /9.7 m = 3.09 x 10 16 m = 3 x 10 16 m.

Question 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: As we know, 1 light year = 9.46 x 10 15 m .•. 4.29 light years = 4.29 x 9.46 x 10 15 = 4.058 x 10 16 m Also, 1 parsec = 3.08 x 10 16 m .•. 4.29 light years =4.508 x 10 16 /3.80 x 10 16 = 1.318 parsec = 1.32 parsec. As a parsec distance subtends a parallax angle of 1″ for a basis of radius of Earth’s orbit around the Sun (r).In present problem base is the distance between two locations of the Earth six months apart in its orbit around the Sun = diameter of Earth’s orbit (b = 2r). .•. Parallax angle subtended by 1 parsec distance at this basis = 2 second (by definition of parsec). .•. Parallax angle subtended by the star Alpha Centauri at the given basis θ = 1.32 x 2 = 2.64″.

Question 2. 21. Precise measurements of physical quantities are a need of 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 modem science where precise measurements of length, time, mass etc., are needed. Also, wherever you can, give a quantitative idea of the precision needed. Answer: Extremely precise measurements are needed in modem science. As an example, while launching a satellite using a space launch rocket system we must measure time to a precision of 1 micro second. Again working with lasers we require length measurements to an angstrom unit (1 A° = 10 -10 m) or even a fraction of it. For estimating nuclear sizes we require a precision of 10 -15 m. To measure atomic masses using mass spectrograph we require a precision of 10 -30 kg and so on.

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements Q22

Question 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 x 10 30 kg, radius of the Sun = 7.0 x 10 8 m. Answer: Given M = 2 x 10 30 kg, r = 7 x 10 8 m .-. Volume of Sun = 4/3πr 3 x 3.14 x (7 x 10 8 ) 3 = 1.437 x 10 27 m 3 As p = M/V, .’. p = 2 x 10 30 /1.437 x 10 27 = 1391.8 kg m -3 = 1.4 x 10 3 kg m -3 Mass density of Sun is in the range of mass densities of solids/liquids and not gases.

Question 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: Given angular diameter θ = 35.72= 35.72 x 4.85 x 10 -6 rad = 173.242 x 10 -6 = 1.73 x 10 -4 rad Diameter of Jupiter D = θ x d = 1.73 x 10 -4 x 824.7 x 10 9 m =1426.731 x 10 3 = 1.43 x 10 8 m

Question 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: tanθ = v and checks that the relation has a correct limit: as v—>θ, 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 principle of homogenity of dimensional equations, Dimensions of L.H.S. = Dimensions of R.H.S. Here, v = tan θ i. e., [L 1 T -1 ] = dimensionless, which is incorrect. Correcting the L.H.S., we. get v/u= tan θ, where u is velocity of rain.

Question 2. 26. It is claimed that two cesium clocks, if allowed to run for 180 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 1 second=0.02/100 x 365 x 24 x 60 x 60 =6.34 x 10 -12 s .•. Accuracy of 1 part in 10 11 to 10 12 .

Question 2. 27. Estimate the average mass density of a sodium atom assuming its size to be about 2.5 A. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the 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: It is given that radius of sodium atom, R = 2.5 A = 2.5 x 10 -10 m Volume of one mole atom of sodium, V = NA .4/3 π R 3 V = 6.023 x 10 23 x –4/3 x 3.14 x (2.5 x 10 -10 ) 3 m 3 and mass of one mole atom of sodium, M = 23 g = 23 x 10 -3 kg .•. Average mass density of sodium atom, p = M/V =(23 x 10 -3 /6.023 x 10 23 x 4/3 x 3.14 x (2.5 x 10 -10 )) = 6.96 x 10 2 kg m -3 = 0.7 x 10 -3 kg m -3 The density of sodium in its crystalline phase = 970 kg m -3 = 0.97 x 10 3 kg m -3 Obviously the two densities are of the same order of magnitude (= 10 3 kg m -3 ). It is on account of the fact that in solid phase atoms are tightly packed and so the atomic mass density is close to the mass density of solid.

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements Q28

Question 2. 30. A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. 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: Here speed of sound in water v = 1450 m s -1 and time of echo t = 77.0 s. If distance of enemy submarine be d, then t = 2d/v .’. d=vt/2 =1450 x 77.0/2 =55825 m=55.8 x 10 3 m or 55.8 km.

Question 2. 31. The farthest objects in our Universe discovered by modem 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: The time taken by light from the quasar to the observer t = 3.0 billion years = 3.0 x 10 9 years As 1 ly = 9.46 x 10 15 m .•. Distance of quasar from the observer d = 3.0 x 10 9 x 9.46 x 10 15 m = 28.38 x 10 24 m = 2.8 x 10 25 m or 2.8 x 10 22 km.

Question 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 θ = 1920″ and S = 3.8452 x 10 8 m. During the 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. Angular diameter of the moon, θ= Angular diameter of the sun = 1920″ = 1920 x 4.85 x 10 -6 rad [1″ = 4.85 x 10 -6 rad] The earth-moon distance, S = 3.8452 x 10 8 m .’. The diameter of the moon, D = θ x S = 1920 x 4.85 x 10 -6 x 3.8452 x 10 8 m = 35806.5024 x 10 2 m = 3581 x 10 3 m 3581 km.

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements Q33

QUESTIONS BASED ON SUPPLEMENTARY CONTENTS

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements Numerical Questions Q1

NCERT Solutions for Class 11 Physics All Chapters

Chapter 1 Physical World
Chapter 2 Units and Measurements
Chapter 3 Motion in a Straight Line
Chapter 4 Motion in a plane
Chapter 5 Laws of motion
Chapter 6 Work Energy and power
Chapter 7 System of particles and Rotational Motion
Chapter 8 Gravitation
Chapter 9 Mechanical Properties Of Solids
Chapter 10 Mechanical Properties Of Fluids
Chapter 11 Thermal Properties of matter
Chapter 12 Thermodynamics
Chapter 13 Kinetic Theory
Chapter 14 Oscillations
Chapter 15 Waves

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Chapter 2 Units and Measurements

Answer The Following Question Answer

Question 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 –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 .

Solution : a. Length of edge = 1cm = 1/100 m Volume of the cube = side 3 Putting the value of side, we get Volume of the cube = (1/100 m) 3 =(10 -2 m) 3 10 -6 m 3 . The volume of a cube of side 1 cm is equal to 10 -6 m 3 b. Given, Radius, r = 2.0 cm = 20 mm (convert cm to mm) Height, h = 10.0 cm =100 mm The formula of total surface area of a cylinder S = 2πr (r + h) Putting the values in this formula, we get Surface area of a cylinder S = 2πr (r + h = 2 x 3.14 x 20 (20+100) = 15072 = 1.5 × 10 4 mm 2 The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to 1.5 × 10 4 mm 2

c. Using the conversion, Given, Time, t = 1 sec speed = 18 km h-1 = 18 km / hour 1 km = 1000 m and 1hour = 3600 sec Speed = 18 × 1000 /3600 sec = 5 m /sec Use formula Speed = distance / time Cross multiply it, we get Distance = Speed × Time = 5 × 1 = 5 m A vehicle moving with a speed of 18 km h –1 covers 5 m in 1 s. d. Density of lead = Relative density of lead × Density of water Density of water = 1 g/cm 3 Putting the values, we get Density of lead = 11.3 × 1 g / cm 3 = 11.3 g cm -3 1 cm = (1/100 m) =10 –2 m 3 1 g = 1/1000 kg = 10 -3 kg Density of lead = 11.3 g cm -3 = 11.3 Putting the value of 1 cm and 1 gram 11.3 g/cm 3 = 11.3 × 10 -3 kg (10-2m) -3 = 11.3 ×10 –3 × 106 kg m -3 =1.13 × 103 kg m –3 The relative density of lead is 11.3. Its density is 11.3 g cm -3 g cm –3 or 1.13 × 103 kg m –3 .

Question 2. Fill in the blanks by suitable conversion of units: a. 1 kg m 2 s –2 = ....g cm 2 s –2 b. 1 m =..... ly c. 3.0 m s –2 =.... km h –2 d. G= 6.67 × 10 –11 N m 2 (kg) –2 =.... (cm)3s –2 g –1 .

Solution : a. 1 kg = 10 3 g 1 m 2 = 104 cm 2 1 kg m 2 s –2 = 1 kg × 1 m 2 × 1 s –2 =103 g × 104 cm 2 × 1 s –2 = 10 7 g cm 2 s–2 1 kg m 2 s –2 = 10 7 g cm 2 s –2 b. Distance = Speed × Time Speed of light = 3 × 10 8 m/s Time = 1 year = 365 days = 365 × 24 hours = 365 × 24 × 60 × 60 sec Putting these values in above formula we get 1 light year distance = (3 × 10 8 m/s) × (365 × 24 × 60 × 60 s) = 9.46 × 10 15 m 9.46 × 10 15 m = 1 ly So that 1 m = 1/ 9.46 × 10 15 ly = 1.06 × 10 -16 ly c. 1 hour = 3600 sec so that 1 sec = 1/3600 hour 1 km = 1000 m so that 1 m = 1/1000 km 3.0 m s –2 = 3.0 (1/1000 km)( 1/3600 hour) -2 = 3.0 × 10 –3 km × ((1/3600) -2 h–2) = 3.0 × 10 –3 km × (3600) 2 h –2 = 3.88 × 10 4 km h –2 3.0 m s –2 = 3.88 × 10 4 km h –2

d. 1 N = 1 kg m s –2

1 kg = 10 –3 g –1

1 m 3 = 10 6 cm 3

∴ 6.67 × 10 –11 N m 2 kg –2 = 6.67 × 10 –11 × (1 kg m s –2 ) (1 m 2 ) (1 s –2 )

= 6.67 × 10 –11 × (1 kg × 1 m 3 × 1 s –2 )

= 6.67 × 10 –11 × (10 –3 g –1 ) × (10 6 cm 3 ) × (1 s –2 )

= 6.67 × 10 –8 cm 3 s –2 g –1

Question 3. A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kg m 2 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.

Solution : Given that, 1 Calorie=4.2 J = 4.2 Kg m 2 s -2 ...... (i) As new unit of mass = α Kg ∴ 1 Kg = 1/α new unit of mass Similarly, 1 m = β -1 new unit of length 1 s = γ -1 new unit of time Putting these values in (i), we get 1 calorie = 4.2 (α-1 new unit of mass) (β -1 new unit of length)2 (γ -1 new unit of time)-2 = 4.2 α -1 β -2 γ 2 new unit of energy (Proved)

Question 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.

Solution : The given statement is true because a dimensionless quantity may be large or small in comparision to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction. a. An atom is a very small object in comparison to a soccer ball. b. A jet plane moves with a speed greater than that of a bicycle. c. Mass of Jupiter is very large as compared to the mass of a cricket ball. d. The air inside this room contains a large number of molecules as compared to that present in a geometry box. e. A proton is more massive than an electron. f. Speed of sound is less than the speed of light.

Question 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?

Solution : Distance between the Sun and the Earth: = Speed of light × Time taken by light to cover the distance Given that in the new unit, speed of light = 1 unit Time taken, t = 8 min 20 s = 500 s ∴Distance between the Sun and the Earth = 1 × 500 = 500 units

Question 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

Solution : a. Least count of this vernier callipers = 1SD - 1 VD = 1 SD - 19/20 SD = 1/20 SD = 1.20 mm = 1/200 cm = 0.005 cm b. Least count of screw gauge = Pitch/Number of divisions = 1/1000 = 0.001 cm. c. Wavelength of light, λ ≈ 10-5 cm = 0.00001 cm Hence, it can be inferred that an optical instrument is the most suitable device to measure length.

Question 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 hair?

Solution : Magnification of the microscope = 100 Average width of the hair in the field of view of the microscope = 3.5 mm ∴Actual thickness of the hair is 3.5/100 = 0.035 mm.

8. Answer the following:

Question 8.1: You are given a thread and a metre scale. How will you estimate the diameter of the thread?

Solution : Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread using a metre scale. The diameter of the thread is given by the relation, Diameter = Length of thread /Number of turns

Question 8.2: 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?

Solution : It is not possible to increase the accuracy of a screw gauge by increasing the number of divisions of the circular scale. Increasing the number divisions of the circular scale will increase its accuracy to a certain extent only.

Question 8.3: 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?

Solution : A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved in the former are very less as compared to the latter.

Question 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 m 2 . What is the linear magnification of the projector-screen arrangement?

Solution : Here area of the house on slide = 1.75 cm 2 = 1.75 x 10 -4 m 2 and area of the house of projector-screen = 1.55 m 2 .•. Areal magnification =Area on screen/Area on slide = 1.55 m 2 / 1.75 x 10 -4 m 2 = 8.857 x 10 3 .•. Linear magnification

Question 10. State the number of significant figures in the following: a. 0.007 m 2 b. 2.64 x 1024 kg c. 0.2370 g cm -3 d. 6.320 J e. 6.032 N m -2 f. 0.0006032 m 2

Solution : a. Answer: 1 The given quantity is 0.007 m 2 . If the number is less than one, then all zeros on the right of the decimal point (but left to the first non-zero) are insignificant. This means that here, two zeros after the decimal are not significant. Hence, only 7 is a significant figure in this quantity. b. Answer: 3 The given quantity is 2.64 × 1024 kg. Here, the power of 10 is irrelevant for the determination of significant figures. Hence, all digits i.e., 2, 6 and 4 are significant figures. c. Answer: 4 The given quantity is 0.2370 g cm –3 . For a number with decimals, the trailing zeroes are significant. Hence, besides digits 2, 3 and 7, 0 that appears after the decimal point is also a significant figure. d. Answer: 4 The given quantity is 6.320 J. For a number with decimals, the trailing zeroes are significant. Hence, all four digits appearing in the given quantity are significant figures. e. Answer: 4 The given quantity is 6.032 Nm –2 . All zeroes between two non-zero digits are always significant. f. Answer: 4 The given quantity is 0.0006032 m 2 . If the number is less than one, then the zeroes on the right of the decimal point (but left to the first non-zero) are insignificant. Hence, all three zeroes appearing before 6 are not significant figures. All zeros between two non-zero digits are always significant. Hence, the remaining four digits are significant figures.

Question 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.

Solution : Length of sheet, l = 4.234 m

Breadth of sheet, b = 1.005 m

Thickness of sheet, h = 2.01 cm = 0.0201 m

The given table lists the respective significant figures:


l	4.234	4
b	1.005	4
h	0.0201	3

Hence, area and volume both must have least significant figures i.e., 3.

Surface area of the sheet = 2 ( l × b + b × h + h × l )

= 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234) = 2(4.25517 + 0.0202005 + 0.0851034) = 2 × 4.36 = 8.72 m 2

Volume of the sheet = l × b × h

= 4.234 × 1.005 × 0.0201

= 0.0855 m 3

This number has only 3 significant figures i.e., 8, 5, and 5.

Question 12. The mass of a box measured by a grocer's balance is 2.300 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?

Solution : Mass of grocer’s box = 2.300 kg

Mass of gold piece I = 20.15g = 0.02015 kg

Mass of gold piece II = 20.17 g = 0.02017 kg

(a) Total mass of the box = 2.3 + 0.02015 + 0.02017 = 2.34032 kg

In addition, the final result should retain as many decimal places as there are in the number with the least decimal places. Hence, the total mass of the box is 2.3 kg.

(b) Difference in masses = 20.17 – 20.15 = 0.02 g

In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places.

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.

The formula is dimensionally correct.

Question 15. A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m 0 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 = m 0 / (1-v 2 )1/2 Guess where to put the missing c.

Solution : Given the relation, m = m0 / (1-v2)1/2 Dimension of m = M 1 L 0 T 0 Dimension of m 0 = M 1 L 0 T 0 Dimension of v = M 0 L1 T –1 Dimension of v 2 = M 0 L 2 T –2 Dimension of c = M 0 L 1 T –1 The given formula will be dimensionally correct only when the dimension of L.H.S is the same as that of R.H.S. This is only possible when the factor, (1-v 2 )1/2 is dimensionless i.e., (1 – v 2 ) is dimensionless. This is only possible if v 2 is divided by c 2 . Hence, the correct relation is m = m 0 / (1 - v 2 /c 2 )1/2

Question 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 ?

Solution : Radius of hydrogen atom, r = 0.5 Å = 0.5 × 10 -10 m Volume of hydrogen atom = (4/3) π r 3 = (4/3) × (22/7) × (0.5 × 10-10) 3 = 0.524 × 10 -30 m 3 1 mole of hydrogen contains 6.023 × 10 23 hydrogen atoms. ∴ Volume of 1 mole of hydrogen atoms = 6.023 × 10 23 × 0.524 × 10 –30 = 3.16 × 10 –7 m 3

Question 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 ?

Question 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).

Solution : Line of sight is defined as an imaginary line joining an object and an observer's eye. When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line of sight changes very rapidly. On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance. As a result, the line of sight does not change its direction rapidly.

Question 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 × 10 11 m. 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 meters?

Solution : Diameter of Earth’s orbit = 3 × 10 11 m Radius of Earth’s orbit, r = 1.5 × 10 11 m Let the distance parallax angle be 1" = 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 × 10 11 /4.847 × 10 –6 = 0.309 x 10 17 = 0.309 × 10-6 ≈ 3.09 × 10 16 m Hence, 1 parsec ≈ 3.09 × 10 16 m

Question 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?

Solution : Distance of the star from the solar system = 4.29 ly 1 light year is the distance travelled by light in one year. 1 light year = Speed of light × 1 year = 3 × 108 × 365 × 24 × 60 × 60 = 94608 × 10 11 m ∴ 4.29 ly = 405868.32 × 10 11 m ∵ 1 parsec = 3.08 × 10 16 m ∴ 4.29 ly = 405868.32 × 1011 / 3.08 × 10 16 = 1.32 par sec Using the relation, θ = d / D where, Diameter of Earth's orbit, d = 3 × 10 11 m Distance of the star from the earth, D = 405868.32 × 10 11 m ∴ θ = 3 × 10 11 / 405868.32 × 10 11 = 7.39 × 10 -6 rad But, 1 sec = 4.85 × 10 –6 rad ∴ 7.39 × 10 -6 rad = 7.39 × 10 -6 / 4.85 × 10 -6 = 1.52"

Question 21. Precise measurements of physical quantities are a need of 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.

Solution : It is indeed very true that precise measurements of physical quantities are essential for the development of science. For example, ultra-shot laser pulses (time interval ∼ 10–15 s) are used to measure time intervals in several physical and chemical processes. X-ray spectroscopy is used to determine the inter-atomic separation or inter-planer spacing. The development of mass spectrometer makes it possible to measure the mass of atoms precisely.

Question 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.

Solution : a. During monsoons, a Metrologist records about 215 cm of rainfall in India i.e., the height of water column, h = 215 cm = 2.15 m Area of country, A = 3.3 × 10 12 m 2 Hence, volume of rain water, V = A × h = 7.09 × 10 12 m 3 Density of water, ρ = 1 × 103 kg m –3 Hence, mass of rain water = ρ × V = 7.09 × 10 15 kg Hence, the total mass of rain-bearing clouds over India is approximately 7.09 × 10 15 kg.

b. Consider a ship of known base area floating in the sea. Measure its depth in sea (say d 1 ).

Volume of water displaced by the ship, V b = A d 1

Now, move an elephant on the ship and measure the depth of the ship ( d 2 ) in this case.

Volume of water displaced by the ship with the elephant on board, V be = Ad 2

Volume of water displaced by the elephant = Ad 2 – Ad 1

Density of water = D

Mass of elephant = AD ( d 2 – d 1 )

c. Wind speed during a storm can be measured by an anemometer. As wind blows, it rotates. The rotation made by the anemometer in one second gives the value of wind speed. d. Area of the head surface carrying hair = A With the help of a screw gauge, the diameter and hence, the radius of a hair can be determined. Let it be r. ∴Area of one hair = πr 2 Number of strands of hair ≈ Total surface area / Area of one hair = A / πr 2 e. Let the volume of the room be V. One mole of air at NTP occupies 22.4 l i.e., 22.4 × 10 –3 m 3 volume. Number of molecules in one mole = 6.023 × 10 23 ∴Number of molecules in room of volume V = 6.023 × 10 23 × V / 22.4 × 10 -3 = 134.915 × 1026 V = 1.35 × 1028 V

Question 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 × 10 30 kg, radius of the Sun = 7.0 × 10 8 m.

Solution : Mass of the Sun, M = 2.0 × 10 30 kg Radius of the Sun, R = 7.0 × 10 8 m

Volume V = 4 3 π r 3 \frac{4}{3}\pi r^{3}

Question 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.

Solution : Distance of Jupiter from the Earth, D = 824.7 × 10 6 km = 824.7 × 10 9 m Angular diameter = 35.72" = 35.72 × 4.874 × 10 -6 rad Diameter of Jupiter = d Using the relation, θ = d/ D d = θ D = 824.7 × 10 9 × 35.72 × 4.872 × 10 -6 = 143520.76 × 10 3 m = 1.435 × 10 5 Km

Additional Exercises

Question 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.

Solution : Incorrect; on dimensional ground The relation is tan θ = ν Dimension of R.H.S = M 0 L 1 T –1 Dimension of L.H.S = M 0 L 0 T 0 (∵ The trigonometric function is considered to be a dimensionless quantity) Dimension of R.H.S is not equal to the dimension of L.H.S. Hence, the given relation is not correct dimensionally. To make the given relation correct, the R.H.S should also be dimensionless. One way to achieve this is by dividing the R.H.S by the speed of rainfall ν' Therefore, the relation reduces to tan θ = ν / ν' This relation is dimensionally correct.

Question 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?

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

Question 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 density of sodium in its crystalline phase : 970 kg m-3. Are the two densities of the same order of magnitude ? If so, why ?

Solution : Diameter of sodium atom = Size of sodium atom = 2.5 Å Radius of sodium atom, r = (1/2) × 2.5 Å = 1.25 Å = 1.25 × 10 -10 m Volume of sodium atom, V = (4/3) π r 3 = (4/3) × 3.14 × (1.25 × 10 -10 ) 3 = V Sodium According to the Avogadro hypothesis, one mole of sodium contains 6.023 × 10 23 atoms and has a mass of 23 g or 23 × 10 –3 kg. ∴ Mass of one atom = 23 × 10 -3 / 6.023 × 10 23 Kg = m 1 Density of sodium atom, ρ = m 1 / V Sodium Substituting the value from above, we get Density of sodium atom, ρ =4.67 × 10 -3 Kg m -3 It is given that the density of sodium in crystalline phase is 970 kg m –3 . Hence, the density of sodium atom and the density of sodium in its crystalline phase are not in the same order. This is because in solid phase, atoms are closely packed. Thus, the inter-atomic separation is very small in the crystalline phase.

Question 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 = r 0 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.

Solution : Radius of nucleus r is given by the relation, r = r 0 A 1/3 r 0 = 1.2 f = 1.2 × 10 -15 m Volume of nucleus, V = (4 / 3) π r 3 = (4 / 3) π (r 0 A1/3) 3 = (4 / 3) π r 0 A ..... (i) Now, the mass of a nuclei M is equal to its mass number i.e., M = A amu = A × 1.66 × 10 –27 kg Density of nucleus, ρ = Mass of nucleus / Volume of nucleus = A X 1.66 × 10 -27 / (4/3) π r 0 3 A = 3 X 1.66 × 10-27 / 4 π r0 3 Kg m -3

his relation shows that nuclear mass depends only on constant r0. Hence, the nuclear mass densities of all nuclei are nearly the same. Density of sodium nucleus is given by, ρ Sodium = 3 × 1.66 × 10 -27 / 4 × 3.14 × (1.2 × 10 -15 ) 3 = 4.98 × 10 18 / 21.71 = 2.29 × 10 17 Kg m -3

Question 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?

Solution : Time taken by the laser beam to return to Earth after reflection from the Moon = 2.56 s Speed of light = 3 × 10 8 m/s Time taken by the laser beam to reach Moon = 1 / 2 × 2.56 = 1.28 s Radius of the lunar orbit = Distance between the Earth and the Moon = 1.28 × 3 × 10 8 = 3.84 × 10 8 m = 3.84 × 10 5 km

Question 30. A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. 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 ).

Solution : Let the distance between the ship and the enemy submarine be ‘S’. Speed of sound in water = 1450 m/s Time lag between transmission and reception of Sonar waves = 77 s In this time lag, sound waves travel a distance which is twice the distance between the ship and the submarine (2S). Time taken for the sound to reach the submarine = 1/2 × 77 = 38.5 s ∴ Distance between the ship and the submarine (S) = 1450 × 38.5 = 55825 m = 55.8 km

Question 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 ?

Solution : Time taken by quasar light to reach Earth = 3 billion years

= 3 × 10 9 years

= 3 × 10 9 × 365 × 24 × 60 × 60 s

Speed of light = 3 × 10 8 m/s

Distance between the Earth and quasar

= (3 × 10 8 ) × (3 × 10 9 × 365 × 24 × 60 × 60)

= 283824 × 10 20 m

= 2.8 × 10 22 km

Question 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.

From examples 2.3 and 2.4 we get the following data

Distance of the Moon from Earth = 3.84 x 10 8 m

Distance of the Sun from Earth = 1.496 x 10 11 m

Sun’s diameter = 1.39 x 10 9 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 10 8 m

Therefore, the diameter of the moon, D = θ x S = 9.31 x 10 -3 x 3.8452 x 10 8 m = 35.796 x 10 5 m

Question 33. A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (~15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?

The values of different fundamental constants are given below:

We have to try to make permutations and combinations of the universal constants and see if there can be any such combination whose dimensions come out to be the dimensions of time. One such combination is:

NCERT Solutions For Class-11 Physics Chapter Wise

Chapter 1 Physical World

Chapter 3 Motion In A Straight Line

Chapter 4 Motion In A Plane

Chapter 5 Laws of Motion

Chapter 6 Work, Energy and Power

Chapter 7 System of Particles and Rotational Motion

Chapter 8 Gravitation

Chapter 9 Mechanical Properties of Solid

Chapter 10 Mechanical Properties of Fluids

Chapter 11 Thermal Properties of Matter

Chapter 12 Thermodynamics

Chapter 13 Kinetic Theory

Chapter 14 Oscillations

Chapter 15 Waves

Related Chapters

chapter-1 Physical World
chapter-2 Units And Measurements
chapter-3 Motion In A Straight Line
chapter-4 Motion In A Plane
chapter-5 Laws of Motion
chapter-6 Work, Energy And Power
chapter-7 System of Particles and Rotational Motion
chapter-8 Gravitation
chapter-9 Mechanical Properties Of Solids
chapter-10 Mechanical Properties of Fluids
chapter-11 Thermal Properties Of Matter
chapter-12 Thermodynamics
chapter-13 Kinetic Theory
chapter-14 Oscillations
chapter-15 Waves

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NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements

Subject specialists have created NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements, which includes thorough solutions for reference. All of the questions from the textbook’s exercises are answered here. Students can use these answers to help them prepare for their exams. The NCERT Solutions for Class 11 provide useful solutions for improving conceptual knowledge.

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It is critical for students in Class 11 who want to do their best in the next term – I exams and competitive exams to become familiar with the textbook solutions to the questions. As a result, students should practice numerous types of questions that can be framed from the chapter. It is recommended that students answer the NCERT questions. Our specialists provide NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements to clear all of the students’ worries.

The topics and subtopics covered in Class 11 Chapter 2 Physics Units and Measurement are:


	Introduction
	The International System Of Units
	Measurement Of Length
	Measurement Of Mass
	Measurement Of Time
	Accuracy, Precision Of Instruments And Errors In Measurement
	Significant Figures
	Dimensions Of Physical Quantities
	Dimensional Formulae And Dimensional Equations
	Dimensional Analysis And Its Applications

Scientists use their senses, such as their eyes and ears, to gather data and make observations. Some observations are straightforward, such as determining the texture and colour, while others are more complex, necessitating the use of measurements. One of the most fundamental notions in science is measurement. A scientist would be unable to gather data, construct a theory, or conduct experiments without the ability to measure. The units of physical quantities and methods for evaluating them are described in this chapter, whereas the other portion of the chapter deals with measurement mistakes and significant figures. One can gain a strong grasp of measurement by studying problems from Physics NCERT Solutions for Class 11.

Our specialists also provide NCERT Solutions for all subjects in all classes, in addition to Chapter 2. It also includes notes, study materials, numerical problems, previous year question papers, sample papers, and competitive exam preparation tools to help you do well in the first and second terms of Class 11.

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Chapter 2 Units and Measurements NCERT Solutions for Class 11 Physics All of the questions in the textbook have diagrams and answers in PDF format. Every question is answered with the understanding capacities of the pupils in mind. To help students face the first term exams without worry, the solutions created closely comply with the most recent CBSE Syllabus 2021-22 and exam format. It also enhances their time management abilities, which are critical for exam success.

Following are the benefits of using the NCERT Solutions for Class 11 Physics Chapter 2:

All of the problems in the NCERT textbook have fully answered answers available in PDF format.
To make learning enjoyable for students, simple and easy-to-understand language is employed.
After completing extensive research on each notion, subject matter experts prepare solutions.
The solutions assist students in preparing for not only their term – I exams, but also for a variety of competitive exams such as JEE, NEET, and others.
Solutions are supplied in PDF format in chapter and practice style to assist students to grasp the ideas quickly.

The International System Of Units (SI) is a metric system that is widely used as a measurement standard. In scientific and technological research and development, SI units are extremely important.

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NCERT solutions for Physics Class 11 chapter 2 - Units and Measurements [Latest edition]

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NCERT solutions for Physics Class 11 chapter 2 - Units and Measurements - Shaalaa.com

Solutions for chapter 2: units and measurements.

Below listed, you can find solutions for Chapter 2 of CBSE NCERT for Physics Class 11.

NCERT solutions for Physics Class 11 Chapter 2 Units and Measurements Exercises [Pages 35 - 38]

The volume of a cube of side 1 cm is equal to ______ m 3

The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ______ (mm) 2 .

A vehicle moving with a speed of 18 km h –1 covers ______ m in 1 s.

The relative density of lead is 11.3. Its density is ______ g cm –3 or ______ kg m –3 .

Fill in the blank by suitable conversion of unit:

1 kg m 2 s –2 = ______ g cm 2 s –2

Fill in the blank by suitable conversion of unit:

3.0 m s –2 = ______ km h –2

Fill in the blank by suitable conversion of unit:

G= 6.67 × 10 –11 N m 2 (kg) –2 = ______ (cm) 3 s –2 g –1

A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kg m 2 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.

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:

Atoms are very small objects
A jet plane moves with great speed
The mass of Jupiter is very large
The air inside this room contains a large number of molecules
A proton is much more massive than an electron
The speed of sound is much smaller than the speed of light.

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?

Which of the following is the most precise device for measuring length:

a vernier callipers with 20 divisions on the sliding scale
a screw gauge of pitch 1 mm and 100 divisions on the circular scale
an optical instrument that can measure length to within a wavelength of light?

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 hair?

You are given a thread and a metre scale. How will you estimate the diameter of the thread?

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?

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?

The photograph of a house occupies an area of 1.75 cm 2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m 2 . What is the linear magnification of the projector-screen arrangement?

State the number of significant figures in the following:

0.2370 g cm –3

State the number of significant figures in the following:

6.032 N m –2

0.0006032 m 2

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.

The mass of a box measured by a grocer’s balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is

What is the total mass of the box?
What is the difference in the masses of the pieces to correct significant figures?

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

`P=(a^3b^2)/((sqrtcd))`

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?

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 `(2pit)/T`

(b) y = a sin vt

d) y = `(a/sqrt2) (sin 2πt / T + cos 2πt / T )`

(a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds.

A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m 0 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 = m_0/(1-v^2)^(1/2)`

Guess where to put the missing c.

The unit of length convenient on the atomic scale is known as an angstrom and is denoted by `Å: 1Å = 10^(-10)m`. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m 3 of a mole of hydrogen atoms?

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 1Å). Why is this ratio so large?

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).

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 × 10 11 m. 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 meters?

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?

Precise measurements of physical quantities are a need of 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.

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):

the total mass of rain-bearing clouds over India during the Monsoon

the wind speed during a storm

the number of strands of hair on your head

the number of air molecules in your classroom.

The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 10 7 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 × 10 30 kg, radius of the Sun = 7.0 × 10 8 m.

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

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.

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?

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 density of sodium in its crystalline phase: 970 kg m –3 . Are the two densities of the same order of magnitude? If so, why?

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 = r_0A^(1/3)` where r is the radius of the nucleus, A its mass number, and r 0 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

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?

A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. 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 ).

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?

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.

A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics ( c , e , mass of electron, mass of proton) and the gravitational constant G , he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (~15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?

NCERT solutions for Physics Class 11 chapter 2 - Units and Measurements

Shaalaa.com has the CBSE Mathematics Physics Class 11 CBSE solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. NCERT solutions for Mathematics Physics Class 11 CBSE 2 (Units and Measurements) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.

Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.

Concepts covered in Physics Class 11 chapter 2 Units and Measurements are International System of Units, Measurement of Length, Measurement of Mass, Measurement of Time, Accuracy, Precision and Least Count of Measuring Instruments, Significant Figures, Dimensions of Physical Quantities, Dimensional Formulae and Dimensional Equations, Dimensional Analysis and Its Applications, Introduction of Units and Measurements, Errors in Measurements, Need for Measurement, Units of Measurement, Fundamental and Derived Units, Length, Mass and Time Measurements, International System of Units, Measurement of Length, Measurement of Mass, Measurement of Time, Accuracy, Precision and Least Count of Measuring Instruments, Significant Figures, Dimensions of Physical Quantities, Dimensional Formulae and Dimensional Equations, Dimensional Analysis and Its Applications, Introduction of Units and Measurements, Errors in Measurements, Need for Measurement, Units of Measurement, Fundamental and Derived Units, Length, Mass and Time Measurements.

Using NCERT Physics Class 11 solutions Units and Measurements exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in NCERT Solutions are essential questions that can be asked in the final exam. Maximum CBSE Physics Class 11 students prefer NCERT Textbook Solutions to score more in exams.

Get the free view of Chapter 2, Units and Measurements Physics Class 11 additional questions for Mathematics Physics Class 11 CBSE, and you can use Shaalaa.com to keep it handy for your exam preparation.

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Important Questions for CBSE Class 11 Physics Chapter 2 - Units and Measurement

Class 11 Important Question
Chapter 2: Units And Measurement

Crucial Practice Problems for CBSE Class 11 Physics Chapter 2: Units and Measurement

The Vedantu team of science experts precisely design the Class 11 Physics Chapter 2 extra questions set to enhance the students' practice and preparations. Class 11th is the foundation for the class 12th syllabus and curriculum and is also essential for the competitive level examinations. For the students of Maths or biology streaming, physics carries a higher weightage in all types of examinations. Thus the students must refer to extra resources like Units and Measurements Class 11 important questions for better preparations.

Units and measurements chapter also carries a weightage in the higher education of science stream students. It carries some essential concepts that are very helpful in the lives of the students. Referring to the class 11 Physics Chapter 2 important questions ease up with the help of Vedantu, and assures an in-depth clarity of all the concepts for the students.

Download CBSE Class 11 Physics Important Questions 2024-25 PDF

Also, check CBSE Class 11 Physics Important Questions for other chapters:

CBSE Class 11 Physics Important Questions
Sl.No	Chapter No	Chapter Name
1	Chapter 1
2	Chapter 2	Units and Measurement
3	Chapter 3
4	Chapter 4
5	Chapter 5
6	Chapter 6
7	Chapter 7
8	Chapter 8
9	Chapter 9
10	Chapter 10
11	Chapter 11
12	Chapter 12
13	Chapter 13
14	Chapter 14
15	Chapter 15

Boost Your Performance in CBSE Class 11 Physics Exam Chapter 2 with Important Questions

Very short questions and answers (1 marks questions).

1. What is the difference between $\overset{{}^\circ }{\mathop{A}}\,$ and A.U.?

Ans: Angstrom (\[\overset{{}^\circ }{\mathop{A}}\,\]) and astronomical unit (A.U.) are both the units of distance.

However, their values are very different. Their values in SI unit of distance are: $1{{A}^{0}}={{10}^{-10}}m$ and $1A.U.=1.496\times {{10}^{11}}m$.

2. Define S.I. Unit of Solid Angle.

Ans: The SI unit of solid angle is steradian. One steradian is defined as the angle made by a spherical plane of unit square meter area at the centre of a sphere with radius of unit length.

3. Name Physical Quantities Whose Units are Electron Volt and Pascal.

Ans: The physical quantities whose units are electron volt and pascal are energy and pressure respectively.

4. Fill ups.

$3.0m/{{s}^{2}}=$ ………….$km/h{{r}^{2}}$

We have, $1m={{10}^{-3}}km$

$1hr=3600s$

$\Rightarrow 1{{s}^{2}}={{\left( \frac{1}{3600} \right)}^{2}}h{{r}^{2}}$

Then,

$3.0m/{{s}^{2}}=\frac{3\times {{10}^{-3}}}{{{\left( \frac{1}{3600}h \right)}^{2}}}km/h{{r}^{2}}$

$\therefore 3.0m/{{s}^{2}}=3.9\times {{10}^{4}}km/h{{r}^{2}}$

\[6.67\times {{10}^{-11}}N{{m}^{2}}/k{{g}^{2}}=\] ……….. \[{{g}^{-1}}c{{m}^{3}}{{s}^{-2}}\]

$1N=1kgm{{s}^{-2}}$

$1kg={{10}^{-3}}g$

$1{{m}^{3}}={{10}^{6}}c{{m}^{3}}$

\[\Rightarrow 6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}=6.67\times {{10}^{-11}}\times \left( 1kgm{{s}^{-2}} \right)\left( 1{{m}^{2}} \right)\left( 1{{s}^{-2}} \right)\]

\[=6.67\times {{10}^{-11}}\times \left( 1kg\times 1{{m}^{3}}\times 1{{s}^{-2}} \right)\]

\[=6.67\times {{10}^{-11}}\times \left( {{10}^{-3}}{{g}^{-1}} \right)\left( {{10}^{6}}c{{m}^{3}} \right)\left( 1{{s}^{-2}} \right)\]

$\therefore 6.67\times {{10}^{-11}}N{{m}^{2}}/k{{g}^{2}}=6.67\times {{10}^{-8}}c{{m}^{3}}{{s}^{-2}}{{g}^{-1}}$

Very Short Questions and Answers (2 Marks Questions)

1. When a planet X is at a distance of 824.7 million kilometres from earth its angular diameter is measured to be $35.72''$ of arc. Calculate the diameter of planet X.

Distance between planet X and earth, $r=824.7\times {{10}^{6}}km$.

The angular diameter $\theta $ is given to be,

$\theta =35.72''$

$\theta =\frac{35.72}{60\times 60}\times \frac{\pi }{180}radian$

Diameter $l=?$

We have the relation,

$l=r\theta $

$\Rightarrow l=824.7\times {{10}^{6}}\left( \frac{35.72}{60\times 60}\times \frac{\pi }{180} \right)$

$\therefore l=1.429\times {{10}^{5}}km$

2. A Radar Signal Is Beamed Towards a Planet from the Earth and Its Echo is Received Seven Minutes Later. Calculate the Velocity of the Signal, If the Distance Between the Planet and the Earth is \[6.3\times {{10}^{10}}m\].

Time after which the echo is received, $t=7\min =7\times 60s$.

Distance between the planet and earth, $x=6.3\times {{10}^{10}}m$.

The net distance covered while the radar signal reaches the planet and echo to reach back to earth $2x$.

We know that the velocity is defined as the net distance covered per total time taken. So,

$c=\frac{2x}{t}$

$\Rightarrow c=\frac{2\times 6.3\times {{10}^{10}}}{7\times 60}$

$\therefore c=3\times {{10}^{8}}m/s$

3. Find the Dimensions of Latent Heat and Specific Heat.

It is known that:

Latent Heat $=\frac{Q\left( \text{Heat energy} \right)}{m\left( \text{mass} \right)}$

Dimension of Latent Heat $=\frac{M{{L}^{2}}{{T}^{-2}}}{M}=[{{M}^{0}}{{L}^{2}}{{T}^{-2}}]$

Specific Heat:

$S=\frac{Q}{m\Delta T}$

$\Rightarrow \left[ S \right]=\frac{\left[ Q \right]}{\left[ m\Delta T \right]}=\frac{M{{L}^{2}}{{T}^{-2}}}{M\times K}$

$\therefore \left[ S \right]=[{{M}^{0}}{{L}^{2}}{{T}^{-2}}{{K}^{-1}}]$

4. What are the dimensions of ‘a’ and ‘b’ in Vander Waals equation $\left( P+\frac{a}{{{V}^{2}}} \right)\left( V-b \right)=RT$ ?

Ans: We know that physical quantities undergoing addition or subtraction should be of the same dimension.

So, $\frac{a}{{{V}^{2}}}$will have the same dimensions as $P$ and $b$ will have the same dimensions as $V$.

$\left[ P \right]=\left[ \frac{a}{{{V}^{2}}} \right]\Rightarrow \left[ a \right]=\left[ P{{V}^{2}} \right]$

$\Rightarrow \left[ a \right]=\left[ \frac{F}{A}\times {{V}^{2}} \right]$

$\Rightarrow \left[ a \right]=\frac{\left[ ML{{T}^{-2}} \right]}{\left[ {{L}^{2}} \right]}\times {{\left[ {{L}^{3}} \right]}^{2}}$

$\Rightarrow \left[ a \right]=\frac{ML{{T}^{-2}}{{L}^{6}}}{{{L}^{2}}}$

Therefore, the dimension of ‘a’ would be,

$\therefore \left[ a \right]=\left[ M{{L}^{5}}{{T}^{-2}} \right]$

Also, $\left[ b \right]=\left[ V \right]$

Therefore, the dimension ‘b’ would be,

$\therefore \left[ b \right]=\left[ {{M}^{0}}{{L}^{3}}{{T}^{0}} \right]$

5. If $E,m,l$ and $G$ denote energy, mass, angular momentum and gravitational constant respectively, determine the dimensions of $\frac{E{{L}^{2}}}{{{m}^{5}}{{G}^{2}}}$ .

We have, dimensions of:

$E=\left[ M{{L}^{2}}{{T}^{-2}} \right]$

$L=\left[ M{{L}^{2}}{{T}^{-1}} \right]$

$m=\left[ M \right]$

$G=\left[ {{M}^{-1}}{{L}^{3}}{{T}^{-2}} \right]$

Now, the dimensions of $\frac{E{{L}^{2}}}{{{m}^{5}}{{G}^{2}}}$could be written as,

$=\frac{\left[ M{{L}^{2}}{{T}^{-2}} \right]{{\left[ M{{L}^{2}}{{T}^{-1}} \right]}^{2}}}{{{\left[ M \right]}^{5}}{{\left[ {{M}^{-1}}{{L}^{3}}{{T}^{-2}} \right]}^{2}}}$

$=\frac{{{M}^{3}}{{L}^{6}}{{T}^{-4}}}{{{M}^{3}}{{L}^{6}}{{T}^{-4}}}=1$

Therefore, the given term is dimensionless.

6. Calculate the time taken by light to pass through a nucleus of diameter $1.56\times {{10}^{-16}}m$. (Take the speed of light to be $c=3\times {{10}^{8}}m/s$).

We know that speed could be defined as the total distance covered per unit time. Mathematically,

$c=\frac{x}{t}$

$\Rightarrow t=\frac{x}{c}$

Here, the net distance covered is the diameter of the nucleus. Now, on substituting the given values, we get,

$t=\frac{1.56\times {{10}^{-16}}}{3\times {{10}^{8}}}$

$\therefore t=5.2\times {{10}^{-25}}s$

Therefore, we found that light takes $t=5.2\times {{10}^{-25}}s$ to cross the given nucleus.

7. Express the dimension of energy if F, A and T are considered as the base quantities. (Where, F stands for force, A for acceleration and T for time).

We know that, dimension of acceleration, $\left[ A \right]=\left[ L{{T}^{-2}} \right]$…….. (1)

By Newton's second law of motion,

$\Rightarrow M=F{{A}^{-1}}$……. (2)

Now, we know that the dimension of energy is generally given by, $\left[ E \right]=\left[ M{{L}^{2}}{{T}^{-2}} \right]$.

Substituting (1) and (2) in the above expression,

$\left[ E \right]=\left[ F{{A}^{-1}}{{A}^{2}}{{T}^{4}}{{T}^{-2}} \right]$

$\therefore \left[ E \right]=\left[ FA{{T}^{2}} \right]$

Therefore, we found the dimension of energy in powers of F, A and T to be, $\left[ FA{{T}^{2}} \right]$.

8. If Velocity, Time and Force Were Chosen as the Base Quantities, Find the Dimensions of Mass.

From Newton’s second law of motion,

$\text{force}=\text{mass}\times \text{acceleration}$

\[\Rightarrow \text{force}=\text{mass}\times \frac{\text{Velocity}}{\text{Time}}\]

\[\Rightarrow \frac{\text{Time}\times \text{force}}{\text{Velocity}}=\text{mass}\]

\[\Rightarrow \left[ \text{mass} \right]=\left[ \frac{FT}{V} \right]\]

\[\therefore \left[ M \right]=\left[ FT{{V}^{-1}} \right]\]

Therefore, we found the dimension of mass in terms of velocity, time and force to be \[\left[ M \right]=\left[ FT{{V}^{-1}} \right]\].

9. A calorie is a unit of heat or energy and is equivalent to 4.2 J where $1J=1kg{{m}^{2}}{{s}^{-2}}$. Suppose we employ a system of units in which the unit of mass equals $\alpha \text{ kg}$, the unit of length equals $\beta $ 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.

Given that,

\[1\text{ }Calorie=4.2\left( 1kg \right)\left( 1{{m}^{2}} \right)\left( 1{{s}^{-2}} \right)\]

New unit of mass $=\alpha \text{ kg}$

Hence, one kilogram in terms of the new unit, $1\text{ kg}=\frac{1}{\alpha }={{\alpha }^{-1}}$

One meter in terms of the new unit of length, $\text{1m}=\frac{1}{\beta }={{\beta }^{-1}}$ or $\text{1}{{\text{m}}^{2}}={{\beta }^{-2}}$

And, one second in terms of the new unit of time,

$1\text{ s}=\frac{1}{\gamma }={{\gamma }^{-1}}$

$1\text{ }{{\text{s}}^{2}}={{\gamma }^{-2}}$

$1\text{ }{{\text{s}}^{-2}}={{\gamma }^{2}}$

\[\therefore 1\text{ }Calorie=4.2\left( 1{{\alpha }^{-1}} \right)\left( 1{{\beta }^{-2}} \right)\left( 1{{\gamma }^{2}} \right)=4.2{{\alpha }^{-1}}{{\beta }^{-2}}{{\gamma }^{2}}\]

10. 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:

Ans: The given statement is true because a dimensionless quantity may be large or small, but there should be some standard reference to compare that. For example, the coefficient of friction is dimensionless but we could say that the coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction.

A. Atoms are Very Small Objects.

Ans: An atom is very small compared to a soccer ball.

B. A Jet Plane Moves With Great Speed.

Ans: A jet plane moves with a speed greater than that of a bicycle.

C. The Mass of Jupiter Is Very Large.

Ans: Mass of Jupiter is very large compared to the mass of a cricket ball.

D. The Air Inside This Room Contains a Large Number of Molecules.

Ans: The air inside this room contains a large number of molecules as compared to that contained by a geometry box.

E. A Proton Is Much More Massive Than an Electron.

Ans: A proton is more massive than an electron.

F. The Speed of Sound Is Much Smaller Than the Speed of Light.

Ans: Speed of sound is less than the speed of light.

11. Which of the Following Is the Most Precise Device for Measuring Length:

Ans: A device which has the minimum least count is considered to be the most precise device to measure length.

A vernier caliper with 20 divisions on the sliding scale.

Least count of a vernier caliper is given by,

\[LC=1\text{ }standard\text{ }division\left( SD \right)-1\text{ }vernier\text{ }division\left( VD \right)\]

$\Rightarrow L.C=1-\frac{9}{10}=\frac{1}{10}=0.01cm$

A Screw Gauge of Pitch 1 Mm and 100 Divisions on the Circular Scale

Least count of screw gauge $=\frac{\text{Pitch}}{\text{No of divisions}}$

$\Rightarrow L.C=\frac{1}{1000}=0.001cm$

An Optical Instrument That Can Measure Length to Within a Wavelength of Light?

Ans: Least count of an optical device $=\text{Wavelength of light}\sim \text{1}{{\text{0}}^{-5}}cm$

$\Rightarrow L.C=0.00001cm$

Hence, among the given three options, it can be inferred that the optical instrument with the minimum least count that can measure length to within a wavelength of light is the most suitable device to measure length.

12. Answer the following:

You are given a thread and a meter scale. How will you estimate the diameter of the thread?

Ans: Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length that is wound by the thread using a metre scale. The diameter of the thread is given by the relation,

Diameter $=\frac{\text{Length of thread}}{\text{Number of turns}}$

Ans: Increasing the number of divisions of the circular scale will increase its accuracy to a negligible extent only.

The mean diameter of a thin brass rod is to be measured by Vernier calipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only

Ans: A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved will be reduced on increasing the number of measurements.

13. The mass of a box measured by a grocer's balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is:

The Total Mass of the Box?

We are given:

Mass of grocer’s box $=2.300kg$

Mass of gold piece $I=20.15g=0.02015kg$

Mass of gold piece $II=20.17g=0.02017kg$

Total mass of the box $=2.3+0.02015+0.02017=2.34032kg$

In addition, the final result should retain as many decimal places as there are in the number with the least decimal places. Hence, the total mass of the box is $2.3kg$.

The Difference in the Masses of the Pieces to Correct Significant Figures?

Difference in masses $=20.17-20.15=0.02g$

During subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places.

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

$P=\frac{{{a}^{3}}{{b}^{2}}}{\left( \sqrt{cd} \right)}$

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?

We are given the relation,

$P=\frac{{{a}^{3}}{{b}^{2}}}{\left( \sqrt{cd} \right)}$

The error could be calculated using the following expression,

$\frac{\Delta P}{P}=\frac{3\Delta a}{a}+\frac{2\Delta b}{b}+\frac{1}{2}\frac{\Delta c}{c}+\frac{\Delta d}{d}$

$\left( \frac{\Delta P}{P}\times 100 \right)%=\left( 3\times \frac{\Delta a}{a}\times 100+2\times \frac{\Delta b}{b}\times 100+\frac{1}{2}\times \frac{\Delta c}{c}\times 100+\frac{\Delta d}{d}\times 100 \right)%$

$ =3\times 1+2\times 3+\frac{1}{2}\times 4+2$

$=3+6+2+2=13%$

Percentage error in $P=13%$

Value of P is given as $3.763$.

By rounding off the given value to the first decimal place, we get $P=3.8$.

15. The unit of length convenient on the atomic scale is known as an angstrom and is denoted by \[\overset{{}^\circ }{\mathop{A}}\,\] and we know that \[\left( 1\overset{{}^\circ }{\mathop{\text{A}}}\,={{10}^{-10}}m \right)\]. The size of a hydrogen atom is about \[0.5\overset{{}^\circ }{\mathop{\text{A}}}\,\]. What is the total atomic volume in ${{m}^{3}}$ of a mole of hydrogen atoms?

Radius of hydrogen atom, \[r=0.5\overset{{}^\circ }{\mathop{\text{A}}}\,=0.5\times {{10}^{-10}}m\]

Volume of hydrogen atom $V=\frac{4}{3}\pi {{r}^{3}}$

$\begin{align}

& \Rightarrow V=\frac{4}{3}\times \frac{22}{7}\times {{\left( 0.5\times {{10}^{-10}} \right)}^{3}} \\

& \Rightarrow V=0.524\times {{10}^{-30}}{{m}^{3}} \\

\end{align}$

1 mole of hydrogen contains $6.023\times {{10}^{23}}$ hydrogen atoms.

$\therefore $ Volume of 1 mole of hydrogen atoms $V'=6.023\times {{10}^{23}}\times 0.524\times {{10}^{-30}}$

$\Rightarrow V'=3.16\times {{10}^{-7}}{{m}^{3}}$

16. 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).

Ans: Line-of-sight is defined as an imaginary line joining an object and an observer's eye. When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line-of-sight changes very rapidly.

On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance. As a result, the line-of-sight does not change its direction rapidly.

17. Precise measurements of physical quantities are a need of 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 \[\mathbf{II}\]. 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.

Ans: It is indeed very true that precise measurements of physical quantities are essential for the development of science. Some examples are:

Ultrashort laser pulses (time interval $\sim {{10}^{-15}}s$) are used to measure time intervals in several physical and chemical processes.

X-ray spectroscopy is used to determine the interatomic separation or inter-planar spacing.

The development of the mass spectrometer makes it possible to measure the mass of atoms precisely.

18. 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.

Distance of Jupiter from the earth, $D=824.7\times {{10}^{6}}km=824.7\times {{10}^{9}}m$

Angular diameter $=35.72''=35.72\times 4.874\times {{10}^{-6}}rad$

Diameter of Jupiter $=d$

Using the relation,

\[\begin{align}

& \theta =\frac{d}{D} \\

& \Rightarrow d=\theta D=824.7\times {{10}^{9}}\times 35.72\times 4.874\times {{10}^{-6}}=143520.76\times {{10}^{3}} \\

& \therefore d=1.435\times {{10}^{5}}km \\

\end{align}\]

19. A LASER is a source of a 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 \[\mathbf{2}.\mathbf{56}\text{ }\mathbf{s}\] to return after reflection at the Moon's surface. How much is the radius of the lunar orbit around the Earth?

We are given the time taken by the laser beam to return to Earth after reflection from the moon $=2.56s$

We know that speed of light $=3\times {{10}^{8}}m/s$

Time taken by the laser beam to reach moon $=\frac{1}{2}\times 2.56=1.28s$

Radius of the lunar orbit = Distance between the Earth and the Moon $=1.28\times 3\times {{10}^{8}}=3.84\times {{10}^{8}}m=3.84\times {{10}^{5}}km$

20. 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 \[\mathbf{77}.\mathbf{0}\text{ }\mathbf{s}\]. What is the distance of the enemy submarine? (Speed of sound in water \[=\mathbf{1450}m{{s}^{-1}}\]).

Let the distance between the ship and the enemy submarine be $'S'$.

We are given,

Speed of sound in water \[=1450\text{ }m/s\]

Time lag between transmission and reception of Sonar waves \[=77\text{ }s\]

In this time lag, sound waves travel a distance which is twice the distance between the ship and the submarine \[\left( 2S \right)\] .

So, the time taken for the sound to reach the submarine $=\frac{1}{2}\times 77=38.5s$

Therefore, the distance between the ship and the submarine is given by \[S=1450\times 38.5=55825\text{ }m=55.8\text{ }km\]

Short Questions and Answers (3 Marks Questions)

1. 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):

The Total Mass of Rain-Bearing Clouds Over India During the Monsoon.

For estimating the total mass of rain-bearing clouds over India during the Monsoon:

During monsoons, a meteorologist records about 215 cm of rainfall in India i.e., the height of the water column, \[h=215\text{ }cm=2.15\text{ }m\]

We have the following information,

Area of country, \[A=3.3\times {{10}^{12}}{{m}^{2}}\]

Hence, the volume of rainwater, \[V=A\times h=7.09\times {{10}^{12}}{{m}^{3}}\]

Density of water, \[\rho =1\times {{10}^{3}}kg\text{ }{{m}^{-3}}\]

We can find the mass from the given value of density and volume as,

\[M=\rho \times V=7.09\times {{10}^{15}}kg\]

Hence, the total mass of rain-bearing clouds over India is approximately found to be \[7.09\times {{10}^{15}}kg\].

The Mass of an Elephant.

For estimating the mass of an elephant:

Consider a ship floating in the sea whose base area is known. Measure its depth at sea (say ${{d}_{1}}$).

Volume of water displaced by the ship would be, ${{V}_{b}}=A{{d}_{1}}$

Now one could move an elephant on the ship and then measure the depth of the ship $\left( {{d}_{2}} \right)$.

Let the volume of water displaced by the ship with the elephant on board be given as ${{V}_{be}}=A{{d}_{2}}$.

Then the volume of water displaced by the elephant $=A{{d}_{2}}-A{{d}_{1}}$ .

If the density of water $=D$

Mass of an elephant would be $M=AD\left( {{d}_{2}}-{{d}_{1}} \right)$.

The Wind Speed During a Storm.

Ans: Estimation of wind speed during a storm:

Wind speed during a storm can be measured by using an anemometer. As wind blows, it rotates and the number of rotations in one second as recorded by the anemometer gives the value of wind speed.

The Number of Strands of Hair on Your Head.

Ans: Estimation of the number of strands of hair on your head:

Let the area of the head surface carrying hair be \[A\].

The radius of a hair can be determined with the help of a screw gauge and let it be \[r\] .

$\therefore $ Area of one hair strand $=\pi {{r}^{2}}$

Number of strands of hair $\approx \frac{\text{Total surface area}}{\text{Area of one hair}}=\frac{A}{\pi {{r}^{2}}}$

The Number of Air Molecules in Your Classroom.

Estimation of the number of air molecules in your classroom:

Let the volume of the room be \[V\].

We know that:

One mole of air at NTP occupies \[22.4\text{ }l\] i.e., $22.4\times {{10}^{-3}}{{m}^{3}}$ volume.

Number of molecules in one mole ${{N}_{A}}=6.023\times {{10}^{23}}$ (Avogadro number)

$\therefore $ Number of molecules in room of volume(V) could be found as,

$n=\frac{6.023\times {{10}^{23}}}{22.4\times {{10}^{-3}}}\times V$

$\Rightarrow n=134.915\times {{10}^{26}}V$

$\therefore n=1.35\times {{10}^{28}}V$

2. The unit of length convenient on the nuclear scale is a fermi: \[\mathbf{1f}={{10}^{-15}}\mathbf{m}\]. Nuclear sizes obey roughly the following empirical relation: $r={{r}_{0}}{{A}^{\frac{1}{3}}}$ , where \[\mathbf{r}\] is the radius of the nucleus, \[\mathbf{A}\] its mass number, and ${{r}_{0}}$ is a constant equal to about, \[\mathbf{1}.\mathbf{2}\text{ }\mathbf{f}\]. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus and compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

Let r be the radius of the nucleus given by the relation,

$r={{r}_{0}}{{A}^{\frac{1}{3}}}$

${{r}_{0}}=1.2f=1.2\times {{10}^{-15}}m$

Then the volume of nucleus would be, $V=\frac{4}{3}\pi {{r}^{3}}$

\[V=\frac{4}{3}\pi {{\left( {{r}_{0}}{{A}^{\frac{1}{3}}} \right)}^{3}}=\frac{4}{3}\pi {{r}_{0}}^{3}A\] …….. (1)

Now, the mass of the nuclei M is equal to its mass number that is,

$M=A\text{ amu}=A\times 1.66\times {{10}^{-27}}kg$

Density of nucleus could be given by,

$\rho =\frac{\text{Mass of nucleus}}{\text{Volume of nucleus}}$

\[\Rightarrow \rho =\frac{A\times 1.66\times {{10}^{-27}}}{\frac{4}{3}\pi {{r}_{0}}^{3}A}=\frac{3\times 1.66\times {{10}^{-27}}}{4\pi {{r}_{0}}^{3}}kg/{{\text{m}}^{3}}\]

This relation shows that nuclear mass depends only on constant ${{r}_{0}}$. Hence, we could conclude that the nuclear mass densities of all nuclei are nearly the same.

Density of sodium nucleus could now be given by,

${{\rho }_{sodium}}=\frac{3\times 1.66\times {{10}^{-27}}}{4\times 3.14\times {{\left( 1.2\times {{10}^{-15}} \right)}^{3}}}$

$\Rightarrow \rho =\frac{4.98}{21.71}\times {{10}^{18}}$

$\therefore \rho =2.29\times {{10}^{17}}kg{{\text{m}}^{-3}}$

3. P.A.M. Dirac, a great physicist of this century loved playing with numerical values of fundamental constants of nature. This led him to an interesting observation that from the basic constants of atomic physics (\[c,e\], mass of electron, mass of proton) and the gravitational constant $G$ , one could arrive at a number with the dimension of time. Further, it was a very large number whose magnitude was close to the present estimate on the age of the universe (\[\sim\mathbf{15}\] billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?

One relation that consists of some fundamental constants to give the age of the Universe could be given by:

$t=\left( \frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}} \right)\times \frac{1}{{{m}_{p}}{{m}_{e}}^{2}{{c}^{3}}G}$

$t=$ Age of universe

\[e=\] Charge of electrons \[=1.6\times {{10}^{-19}}C\]

${{\varepsilon }_{0}}=$ Absolute permittivity

${{m}_{p}}=$ Mass of protons $=1.67\times {{10}^{-27}}kg$

${{m}_{e}}=$ Mass of electrons $=9.1\times {{10}^{-31}}kg$

$c=$ Speed of light $=3\times {{10}^{8}}m/s$

$G=$ Universal gravitational constant $=6.67\times {{10}^{11}}N{{m}^{2}}k{{g}^{-2}}$

We also have,

$\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}$

Substituting all these values into the above equation, we would get,

$t=\frac{{{\left( 1.6\times {{10}^{-19}} \right)}^{4}}\times {{\left( 9\times {{10}^{9}} \right)}^{2}}}{{{\left( 9.1\times {{10}^{-31}} \right)}^{2}}\times 1.67\times {{10}^{-27}}\times {{\left( 3\times {{10}^{8}} \right)}^{3}}\times 6.67\times {{10}^{11}}}$

$\Rightarrow t=\frac{{{\left( 1.6 \right)}^{4}}\times 81}{9.1\times 1.67\times 27\times 6.67\times 365\times 24\times 3600}\times {{10}^{-76+18+62+27-24+11}}years$

$\Rightarrow t\approx 6\times {{10}^{-9}}\times {{10}^{18}}years$

$\therefore t=6\text{ billion years}$

Long Questions and Answers (4 Marks Questions)

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

(\[a=\]maximum displacement of the particle, $v=$ speed of the particle. $T=$ time period of motion). Rule out the wrong formulas on dimensional grounds.

$y=a\sin \left( \frac{2\pi t}{T} \right)$

Ans: Correct.

$y=a\sin \left( \frac{2\pi t}{T} \right)$

Dimensions of $y={{M}^{0}}{{L}^{1}}{{T}^{0}}$

Dimensions of $a={{M}^{0}}{{L}^{1}}{{T}^{0}}$

Dimensions of $\sin \left( \frac{2\pi t}{T} \right)={{M}^{0}}{{L}^{0}}{{T}^{0}}$

Since the dimensions on the RHS are equal to LHS, the given formula is dimensionally correct.

$y=a\sin vt$

Ans: I ncorrect.

$y=a\sin vt$

Dimensions of $vt={{M}^{0}}{{L}^{1}}{{T}^{-1}}\times {{M}^{0}}{{L}^{0}}{{T}^{1}}={{M}^{0}}{{L}^{1}}{{T}^{0}}$

Since the dimensions on the RHS are not equal to that of LHS, the given formula is dimensionally incorrect.

$y=\left( \frac{a}{T} \right)\sin \frac{t}{a}$

Ans: Incorrect.

Dimensions of $\frac{a}{T}={{M}^{0}}{{L}^{1}}{{T}^{-1}}$

Dimensions of $\frac{t}{a}={{M}^{0}}{{L}^{1}}{{T}^{1}}$

Since the dimensions on the RHS are not equal to LHS, the given formula is dimensionally incorrect.

$y=\left( a\sqrt{2} \right)\left( \sin \frac{2\pi t}{T}+\cos \frac{2\pi t}{T} \right)$

Ans: Correct.

Dimensions of \[a={{M}^{0}}{{L}^{1}}{{T}^{0}}\]

Dimensions of $\frac{t}{T}={{M}^{0}}{{L}^{0}}{{T}^{0}}$

2. One mole of an ideal gas at standard temperature and pressure occupies $22.4L$ (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of the hydrogen molecule to be about $1\overset{{}^\circ }{\mathop{\text{A}}}\,$). Why is this ratio so large?

Volume of hydrogen atom, $V=\frac{4}{3}\pi {{r}^{3}}$

$\Rightarrow V=\frac{4}{3}\times \frac{22}{7}\times {{\left( 0.5\times {{10}^{-10}} \right)}^{3}}=0.524\times {{10}^{-30}}{{m}^{3}}$

Now, 1 mole of hydrogen contains $6.023\times {{10}^{23}}$ hydrogen atoms.

$\therefore $ Volume of 1 mole of hydrogen atoms, ${{V}_{a}}=6.023\times {{10}^{23}}\times 0.524\times {{10}^{-30}}=3.16\times {{10}^{-7}}{{m}^{3}}$

Molar volume of 1 mole of hydrogen atoms at STP,

${{V}_{m}}=22.4L=22.4\times {{10}^{-3}}{{m}^{3}}$

So, the required ratio would be,

$\frac{{{V}_{m}}}{{{V}_{a}}}=\frac{22.4\times {{10}^{-3}}}{3.16\times {{10}^{-7}}}=7.08\times {{10}^{4}}$

Hence, we found that the molar volume is $7.08\times {{10}^{4}}$ times higher than the atomic volume. For this reason, the interatomic separation in hydrogen gas is much larger than the size of a hydrogen atom.

3. 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?

Distance of the star from the solar system $=4.29ly$

1 light year is the distance travelled by light in one year.

1 light year $=\text{speed of light }\times \text{ 1 year}$

$1ly=3\times {{10}^{8}}\times 365\times 24\times 60\times 60=94608\times {{10}^{11}}m$

$\Rightarrow 4.29ly=405868.32\times {{10}^{11}}m$

But we have,

$1\text{parsec}=3.08\times {{10}^{16}}m$

$\therefore 4.29ly=\frac{405868.32\times {{10}^{11}}}{3.08\times {{10}^{6}}}=1.32\text{parsec}$

We have another relation,

$\theta =\frac{d}{D}$

Diameter of Earth’s orbit, $d=3\times {{10}^{11}}m$

Distance of star from the Earth, \[D=405868\times {{10}^{11}}m\]

Substituting these values,

$\theta =\frac{3\times {{10}^{11}}}{405868.32\times {{10}^{11}}}=7.39\times {{10}^{-6}}rad$

But $1\sec =4.85\times {{10}^{-6}}rad$

$\therefore 7.39\times {{10}^{-6}}rad=\frac{7.39\times {{10}^{-6}}}{4.85\times {{10}^{-6}}}=1.52''$

4. Estimate the average mass density of a sodium atom assuming its size to be about $2.5\overset{{}^\circ }{\mathop{\text{A}}}\,$. (Use the known values of Avogadro's number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase: $970kg\text{ }{{m}^{-3}}$. Are the two densities of the same order of magnitude? If so, why?

Diameter of sodium atom = Size of sodium atom $=2.5\overset{{}^\circ }{\mathop{\text{A}}}\,$

Radius of sodium atom, \[r=\frac{1}{2}\times 2.5\overset{{}^\circ }{\mathop{\text{A}}}\,=1.25\overset{{}^\circ }{\mathop{\text{A}}}\,=1.25\times {{10}^{-10}}m\]

Volume of sodium atom, $V=\frac{4}{3}\pi {{r}^{3}}$

$\Rightarrow V=\frac{4}{3}\times 3.14\times {{\left( 1.25\times {{10}^{-10}} \right)}^{3}}$

According to the Avogadro hypothesis, one mole of sodium contains $6.023\times {{10}^{23}}$ atoms and has a mass of \[23\text{ }g\] or $23\times {{10}^{-3}}kg$.

$\therefore $ Mass of one atom $=\frac{23\times {{10}^{-3}}}{6.023\times {{10}^{23}}}kg$

Density of sodium atom,

$\rho =\frac{\frac{23\times {{10}^{-3}}}{6.023\times {{10}^{23}}}}{\frac{4}{3}\times 3.14\times {{\left( 1.25\times {{10}^{-10}} \right)}^{3}}}$

$\Rightarrow \rho =4.67\times {{10}^{-5}}kg\text{ }{{\text{m}}^{-3}}$

It is given that the density of sodium in the crystalline phase is $970kg\text{ }{{m}^{-3}}$.

Hence, the density of sodium atoms and the density of sodium in its crystalline phase are not in the same order. This is because in the solid phase, atoms are closely packed and hence the interatomic separation is very small in the crystalline phase.

5. it Is a Well-Known Fact That During a Total Solar Eclipse the Disc of the Moon Almost Completely Covers the Disc 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.

The position of the Sun, Moon, and Earth during a lunar eclipse would be as shown in the given figure.

We know that,

Distance of the Moon from the Earth $=3.84\times {{10}^{8}}m$

Distance of the sun from the Earth $=1.496\times {{10}^{11}}m$

Diameter of the sun $=1.39\times {{10}^{9}}m$

You could see that $\Delta TRS$ and $\Delta TPQ$ are similar. So,

$\frac{1.39\times {{10}^{9}}}{RS}=\frac{1.496\times {{10}^{11}}}{3.84\times {{10}^{8}}}$

$\Rightarrow RS=\frac{1.39\times 3.84}{1.496}\times {{10}^{6}}$

$\therefore RS=3.57\times {{10}^{6}}m$

Hence, the diameter of the Moon is found to be $3.57\times {{10}^{6}}m$.

Very Long Questions and Answers (5 Marks Questions)

1. 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):

Hence, mass of rainwater \[M=\rho \times V=7.09\times {{10}^{15}}kg\]

Hence, the total mass of rain-bearing clouds over India is approximately found to be\[7.09\times {{10}^{15}}kg\].

Consider a ship of known base area floating in the sea. Measure its depth at sea (say ${{d}_{1}}$).

Volume of water displaced by the ship, ${{V}_{b}}=A{{d}_{1}}$.

Now, move an elephant on the ship and measure the depth of the ship $\left( {{d}_{2}} \right)$ in this case.

Volume of water displaced by the ship with the elephant on board, ${{V}_{be}}=A{{d}_{2}}$.

Volume of water displaced by the elephant $=A{{d}_{2}}-A{{d}_{1}}$.

Density of water $=D$

Mass of elephant $=AD\left( {{d}_{2}}-{{d}_{1}} \right)$

Wind speed during a storm can be measured by an anemometer. As the wind blows, it rotates. The rotation made by the anemometer in one second gives the value of wind speed.

Area of the head surface carrying hair \[=A\]

With the help of a screw gauge, the diameter and hence, the radius of a hair can be determined. Let it be \[r\] .

$\therefore $ Area of one hair strand$=\pi {{r}^{2}}$

$\therefore $ Number of molecules in room of volume(V),

$n=\frac{6.023\times {{10}^{23}}}{22.4\times {{10}^{-3}}}\times V$

We know that the radius of the nucleus $r$ is given by the relation,

Volume of nucleus, $V=\frac{4}{3}\pi {{r}^{3}}$

Now, the mass of the nuclei $M$ is equal to its mass number i.e.,

Density of nucleus,

This relation shows that nuclear mass depends only on constant ${{r}_{0}}$. Hence, the nuclear mass densities of all nuclei are nearly the same.

Density of sodium nucleus could be given by,

One relation that is consisting of some fundamental constants to give the age of the Universe could be given by:

Substituting all these values in the above equation, we get,

4. Write S.I. Units of Luminous Intensity and Temperature.

Ans: S.I unit of luminous intensity is candela and is represented by \[cd\]. The SI unit of temperature is kelvin and is represented by $K$.

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?

Distance between the Sun and the Earth is given by:

\[\text{x= Speed of light}\times \text{Time taken by light to cover the distance}\]

We are given, in the new unit, the speed of light, \[c=1\text{ }unit\].

Time taken, \[t=8\text{ }min\text{ }20\text{ }s=500\text{ }s\]

$\therefore $ Distance between the Sun and the Earth \[x'=c\times t'=1\times 500=500\text{ }units\].

6. A student measures the thickness of a human hair using 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. Estimate the thickness of hair.

We are given:

Magnification of the microscope \[=100\]

Average width of the hair in the field of view of the microscope \[=3.5\text{ }mm\]

$\therefore $ Actual thickness of the hair would be, $\frac{3.5}{100}=0.035\text{ }mm.$

7. The photograph of a house occupies an area of $1.75c{{m}^{2}}$ on a 35 mm slide. The slide is projected onto a screen, and the area of the house on the screen is $1.55{{m}^{2}}$. What is the linear magnification of the projector-screen arrangement?

We are given,

The area of the house on the 35mm slide is found to be, ${{A}_{O}}=1.75c{{m}^{2}}$.

The area of the image of the house that is formed on the screen, ${{A}_{I}}=1.55{{m}^{2}}=1.55\times {{10}^{4}}c{{m}^{2}}$.

We know that areal magnification is given by,

${{m}_{a}}=\frac{{{A}_{I}}}{{{A}_{O}}}$

Substituting the given values,

${{m}_{a}}=\frac{1.55\times {{10}^{4}}}{1.75}$

Now, we have the expression for linear magnification as, ${{m}_{l}}=\sqrt{{{m}_{a}}}$

$\Rightarrow {{m}_{l}}=\sqrt{\frac{1.55}{1.75}\times {{10}^{4}}}$

$\therefore {{m}_{l}}=94.11$

Thus, we found the linear magnification in the given case to be, ${{m}_{l}}=94.11$.

8. 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?

We are given, time taken by quasar light to reach Earth, \[t=3\text{ }billion\text{ }years\].

$\Rightarrow t=3\times {{10}^{9}}years$

$\Rightarrow t=3\times {{10}^{9}}\times 365\times 24\times 60\times 60s$

Speed of light, $c=3\times {{10}^{8}}m/s$

Distance between the Earth and quasar,

$x=c\times t$

$\Rightarrow x=\left( 3\times {{10}^{8}} \right)\times \left( 3\times {{10}^{9}}\times 365\times 24\times 60\times 60 \right)$

$\Rightarrow x=283824\times {{10}^{20}}m$

$\therefore x=2.8\times {{10}^{22}}km$

Therefore, we found the distance (in km) of a quasar from which light takes 3.0 billion years to reach us to be $x=2.8\times {{10}^{22}}km$.

Chapter 2 Physics Class 11 Important Questions – PDF Download

Class 11 Physics Ch 2 important questions are easily available on Vedantu website in an easy and free download PDF version. The set of questions are useful for the revision and practice of the students. They brief the students with the type of questions asked in the examinations, how to handle all the questions, and get an enhanced practice for all the topics and sub-topics involved in the curriculum.

CBSE class 11th is a comparatively more challenging section, and the entire syllabus of the class is more challenging. The students must get an extra load of a practice resource to get a crystal clearance of all the concepts and know different ways of tackling and solving various problems. Using the PDF version of the CBSE class 11 physics chapter 2 important questions serves the purpose best and assists all the students during their preparations and whenever they require more study and revision resources.

Physics Class 11 Ch 2 Important questions are available in free download PDF version on Vedantu, making sure that every student can access it quickly and whenever required. Furthermore, there is no device limitation for accessing the questions, and the students can practice for their examinations whenever they wish to do so. Class 11 Physics Chapter 2 extra questions contain a list of problems and the solutions for them, making sure that the students are not required to have such multiple resources whenever they are stuck.

Important Questions for Class 11 Physics Chapter 2 – Related Essential Concepts

There are several topics associated with the syllabus from which the questions are asked in the examinations. Here are some essential concepts related to the Units and Measurement Class 11 important concepts.

Unit and S.I. Unit

Unit is the measuring term for the different types of quantities. The most common systems of the unit for measuring any quantity are as follows:

S.I. System

S.I. unit is the international measuring system, used universally for the scientific and technical research to avoid confusion. This is the standard measuring system and is useful because it helps the entire World understand the measurements in a single unit system. Physics has different quantity measurements, and thus other S.I. units.

The Quantities Mainly Group As

S.I. Base Unit – Base unit is the fundamental unit and the building block of any system. They are used for deriving the other units—for example, Kilogram.

S.I. Derived Unit – The units derived from the base units are known as S.I. derived units.

Distance and Its Units

Distance is the measurement of the path that a body covers between the initial and the final positions. It is the result of the product of speed and time, and is represented as:

D = S * T. Here,

D is the distance that the body travelled; its unit is metre.

T is the time that the body took for covering the said distance; its unit is second.

S is the speed with which the body covered the said distance; its unit is metre /second.

The most commonly used units are Kilometre, Metre, Centimetre, and millimetre for measuring the distance.

Mass and Weight

Mass means the amount of matter present in an object. It is the intrinsic quantity of anybody and helps find out the parameters dependent on the mass. It helps determine the strength of the object's gravitational attraction concerning the other bodies, the item's resistance to acceleration due to any force, inertia, and many more. It also helps in deriving the Theory of Relativity, using E = mc 2 .

Weight of the body is different from its mass. It refers to the force exerted over the mass of anybody due to the gravitational pull. Mass is the universal measurement, but the body's weight is varying, as it depends on the gravity of the planet. It is described as W = mg.

Atomic Mass Unit

The Atomic Mass unit is helpful while dealing with the smaller atoms and molecules with tiny mass such that measuring them in Kilogram gets a lot inconvenient. One atomic mass unit refers to the 1/ 12 th of the mass of the C-12 Carbon atom. It is equal to 1.22 x 10 -27 .

Other Important Concepts Related to CBSE Class 11 Physics Chapter 2 Important Questions

Apart from these concepts, there are many other important topics involved in Chapter Units and Measurements. Time measurement is also included in the chapter. The unit also contains the study of various measurement errors, like absolute error, relative error, random error, systematic error, and gross error. All the concepts carry some questions of different marks in the examinations and are essential for both school curriculum and practice for the competitive level exams.

Important Questions for Class 11 Physics Chapter 2

Practising the questions enhances the practice for the concepts and makes the students confident about them. Here is the list of top ten problems from the set of Units and Measurements Class 11 important questions:

Differentiate between A0 and A.U.

What is the S.I. unit of a solid angle?

Mention the physical quantities having the S.I. unit as Pascal and electron volt.

Suppose there is a planet X which is at a distance of 824.7 kilometres from Earth. The angular diameter of planet X is 35.7211 arcs. Calculate the diameter of the planet X.

There is a radar signal that beams towards the planet X from the Earth, and then the echo for the same is received after seven minutes. Calculate the signal's velocity, if we have the distance between the Earth and the planet X to be 6.3 x 1010 m.

Calculate the dimensions of the specific heat and the latent heat.

G, E, m, and l denote Gravitational constant, Energy, mass, and angular momentum, respectively. Determine the dimension for the given term: El 2 / m 3 G 2 .

Calculate the maximum time that a light beam will take to pass through a nucleus of diameter 1.56 x 10 -16 m. (The speed of light is 3 x 10 8 m/s).

If we chose the time, force, and velocity as the basic quantities, calculate the dimension of mass accordingly.

A calorie refers to the standard unit of energy or heat and is equal to 4.2 Joules. Here, 1J = 1 Kg m 2 / s 2 . Suppose we get a unit system where the mass unit is equal to a kg, and length equals to bm, the time is ys. Now show that the magnitude of a calorie is 4.2 a -1 b -2 y 2 , according to the new units taken.

Benefits of Important Questions for Class 11 Physics Chapter 2

The set of Class 11 Physics Chapter 2 important questions are very important for the students' preparation and practice and help them pass the examinations with flying colours. Here are some key benefits of having the set of important questions handy:

Important questions for Class 11 Physics Chapter 2 are essential as they provide the practice material for all the types of questions to the students, thus helping them be confident while attempting the examinations' questions.

Class 11 Physics Chapter 2 extra questions provide the students with a brief idea about the various problem sets asked in the questions and multiple approaches to solving them.

The experts of Vedantu explicitly design the question set, and thus the solutions are available in an easily understandable language and with sufficient sets of explanations.

Important Related Links for CBSE Class 11

CBSE Class 11 Study Materials

The Class 11 Physics Chapter 2 Units and Measurements is an essential chapter that lays the foundation for the study of Physics. It introduces students to the fundamental concepts of measurements, units, and dimensions, which are crucial for understanding the laws of Physics. The chapter covers topics such as the International System of Units (SI), dimensional analysis, significant figures, and errors in measurements. Students should pay close attention to this chapter and practice solving numerical problems to gain a thorough understanding of the concepts. Vedantu provides comprehensive study materials, including notes, sample papers, and solutions to previous year question papers, to help students prepare for their exams. By mastering the concepts covered in this chapter, students can build a strong foundation for their future studies in Physics and related fields.

FAQs on Important Questions for CBSE Class 11 Physics Chapter 2 - Units and Measurement

Q1. Where can I get the important questions of Chapter 2 of Class 11 Physics?

Ans: S tudents can get the important questions of Chapter 2 of Class 11 Physics on Vedantu app and website. By following the given steps they can download them:

Visit the page-Important questions of Chapter 2 of class 11 Physics.

The official website of Vedantu (vedantu.com) will appear on the screen of your device after clicking the link.

Students will discover the PDF file of important questions of Chapter 2 of Class 11 Physics.

Above this PDF file, there is the option of "Download PDF".

Click this option and the file will get downloaded free of cost.

Students can also visit the Vedantu app to download the PDF of important questions.

Q2. Which concepts are discussed in Chapter 2 of Class 11 Physics?

Ans: Chapter 2 "Units And Measurement" of Class 11 Physics contains the following topics:

Introduction

The International System Of Units

Measurement Of Length

Measurement of Large Distances

Estimation of Very Small Distances: Size of a Molecule

Range of Lengths

Measurement of Mass

Range of Masses

Measurement of Time

Accuracy, Precision of Instruments and Errors in Measurement

Absolute Error, Relative Error and Percentage Error

Combination of Errors

Significant Figures

Rules For Arithmetic Operations With Significant Figures

Rounding Off the Uncertain Digits

Rules For Determining the Uncertainty in the Results of Arithmetic Calculations

Dimensions of Physical Quantities

Dimensional Formulae and Dimensional Equations

Dimensional Analysis and its Applications

Checking the Dimensional Consistency of Equations

Deducing Relation among the Physical Quantities

Q3. What is the system of units mentioned in Chapter 2 of Class 11 Physics?

Ans: Following is the system of units mentioned in Chapter 2 of Class 11 Physics :

CGS System - The CGS stands for Centimetre, Gram and Second. It is used to express length, mass and time respectively.

FPS System - Length, mass and time are expressed by Foot, Pound and Second respectively.

MKS System - Metre is used to express length, mass is expressed by the kilogram and time is expressed in seconds.

SI Units - Units like metre, second, ampere, etc. are included in this system.

Q4. Name the two supplementary base units.

Ans: There are seven fundamental units in the system of units but besides these, there are two supplementary units. These are radian and steradian.

Radian (rad) - The unit defined for the plane angle is known as radian. The unit radian is described as the angle subtended at the centre of the circle by the arc. The length of the arc is equal to the radius of the circle.

Steradian (sr) - This unit is for solid angles. The angle subtended at the centre of the sphere by its surface whose area is equal to the square of the radius of the sphere.

Q5. Write some applications of Dimensions.

Ans: The uses of dimensions are listed below:

The dimensions are used in checking the obtained results.

One system of units can be converted into another by using dimensions.

These dimensions can be used to derive relationships between physical quantities.

Studying and scaling models can be done using dimensions.

The principle of homogeneity of dimensions is used in these applications. This principle says that the net dimensions of the different physical quantities on both sides of a permissible physical relation must be the same.

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Assignment 2.3 Class 11 Physics Chapter 2
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NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements
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Also working on Class 11 Physics Chapter 2 Units and Measurements NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time. ... the area of the country, which is about A = 3.3 x 10 6 km 2 = 3.3 x 10 6 x 10 6 = 3.3 x 10 12 m 2 Therefore, volume of the rain water, V = A h = 3.3 x 10 12 x 1 = 3.3 x 10 ...
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NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements
Chapter 2 Units and Measurement Class 11 Physics NCERT Solutions. 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.
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NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements are part of NCERT Solutions for Class 11 Physics. ... then A = 3.3 million sq. km = 3.3 x 10 6 (km)2 = 3.3 x 10 6 x 10 6 m 2 = 3.3 x 10 12 m 2 Mass of rain-bearing clouds = area x height x density = 3.3 x 10 12 x 1 x 1000 kg = 3.3 x 10 15 kg. (b) Measure the depth of an empty ...
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Below listed, you can find solutions for Chapter 2 of CBSE NCERT for Physics Class 11. Exercises. Exercises [Pages 35 - 38] NCERT solutions for Physics Class 11 Chapter 2 Units and Measurements Exercises [Pages 35 - 38] Exercises | Q 1.1 | Page 35. The volume of a cube of side 1 cm is equal to _____ m 3.
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NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurements. 2.1 Fill in the blanks. The volume of a cube of side 1 cm is equal to .....m3. The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to...(mm)2. A vehicle moving with a speed of 18 km h-1 covers....m in 1 s. The relative density of lead is 11.3.
NCERT Solutions for Class 11 Physics (Updated for 2023
NCERT Solutions for Class 11 Physics (Chapter-wise) Chapter 1: Units and Measurements. Chapter 2: Motion in a Straight Line. Chapter 3: Motion in a Plane. Chapter 4: Laws of Motion. Chapter 5: Work, Energy, and Power. Chapter 6: System of Particles and Rotational Motion. Chapter 7: Gravitation.

NCERT Solutions for Class 11 Physics Chapter 2 Free PDF Download

CBSE Class 11 Physics Chapter 15 Units and Measurement NCERT Solutions

Subtopics covered under NCERT Solutions for Class 11 Physics Chapter 15

2.3 Measurement of Length

2.4 Measurement of Mass

2.5 Measurement of Time

2.6 Accuracy, Precision of Instruments and Errors in Measurement

2.7 Significant Figures

2.8 Dimensions of Physical Quantities

2.9 Dimensional Formulae and Dimensional Equations

2.10 Dimensional Analysis and its Applications

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Important Questions for Class 11 Physics Chapter 2 – Related Essential Concepts

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