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Edited By Safeer PP | Updated on Aug 08, 2022 06:13 PM IST

NCERT Exemplar Class 11 Physics solutions Chapter 2 offers reliable and comprehensive answers to all the questions and queries related to measurement and the units used to measure quantities. NCERT Exemplar Class 11 Physics chapter 2 solutions have been curated by experts and throw light on the internationally accepted measurement standards. The chapter deals with different types of units, the procedure to go about measuring an abstract or a physical quantity, and the plausible errors in a measuring instrument.

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This Story also Contains

- NCERT Exemplar Class 11 Physics Solutions Chapter 2: MCQ I
- NCERT Exemplar Class 11 Physics Solutions Chapter 2: MCQII
- NCERT Exemplar Class 11 Physics Solutions Chapter 2: Very Short Answer
- NCERT Exemplar Class 11 Physics Solutions Chapter 2: Short Answer
- NCERT Exemplar Class 11 Physics Solutions Chapter 2: Long Answer
- Class 11 Physics NCERT Exemplar Solutions Chapter 2 Topics:
- What will The Students Learn in NCERT Exemplar Class 11 Physics Solutions Chapter 2?
- NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
- Important Topics To cover For Exams in NCERT Exemplar Solutions For Class 11 Physics Chapter 2 Units and Measurement
- NCERT Exemplar Class 11 Solutions

NCERT Exemplar Class 11 Physics solutions chapter 2 PDF Download could be availed by the students for a better understanding of concepts and help revise the chapter for a flawless academic performance in final exams as well as competitive exams.

Question:1

The number of significant figures in is:

Answer:

The answer is the option (b) 4Question:2

The sun of the numbers and in appropriate significant figures is:

Answer:

The answer is the optionQuestion:3

Answer:

The answer to a multiplication or division is rounded off to the same number of significant figures as possessed by the least precise term used in the calculation. The final result should retain as many significant figures as are there in the original number with the least significant figures. In the given question, density should be reported to two significant figures .After rounding off the number, we get density =1.7

Question:3

Answer:

The answer is the optionWe know that Density = mass/ volume

So, we get the above no. after rounding up 2 significant figures.

Question:4

The numbers and on rounding off to 3 significant figures will give:

a) and

b) and

c) and

d) and

Answer:

The answer is the optionHere the fourth digit is five and is preceded by, viz., an even no. Hence, it becomes 2.74.

Here the fourth digit is 5, but the preceding digit is 3, viz., an odd no. Hence, it is increased by 1, and we get 2.74.

Question:6

Which of the following pairs of physical quantities does not have same dimensional formula?

a) work and torque

b) angular momentum and Planck’s constant

c) tension and surface tension

d) impulse and linear momentum

Answer:

The answer is the option (c) Tension & surface tensionWork = force displacement

& Torque = force distance

Hence, their dimensions are the same.

(b) Dimensions of-

Angular momentum

& Planck’s constant

Hence, their dimensions are the same.

(c) Dimensions of-

Tension

Surface tension

Their dimensions are not the same, hence opt (c).

(d) Dimensions of-

Impulse

Momentum

Thus, their dimensions are also the same.

Question:7

Measure of two quantities along with the precision of respective measuring instrument is

The value of AB will be

Answer:

The answer is the option (a)Thus,

Thus,

Question:8

You measure two quantities as . We should report correct value for √ AB as:

a)

b)

c)

d)

Answer:

The answer is the option (d)There are two significant figures in 1.0 & 2.0

After roundig up 1.0 &2.0 we het two sigificat features, i.e.

Now,

after rounding up

Thus, option (d) is the right answer.

Question:9

Which of the following measurements is most precise?

a)

b)

c)

d)

Answer:

The answer is the option (a)Question:10

The mean length of an object is . Which of the following measurements is most accurate?

a)

b)

c)

d)

Answer:

The answer is the option (a)Thus, it is the mmost acurate.

Question:11

Young’s modulus of steel is . When expressed in CGS units of dynes/cm^{2}, it will be equal to:

a)

b)

c)

d)

Answer:

The answer is the option (c)Explanation : Young's modulus (Y)

Converting into dyne/cm

Question:13

Answer:

The answer is the option (b) and (c)Explanation: (b) Here, v.t is an angle, whose dimensions are –

Thus, it is not true that sin vt is dimensionless.

(c) Here the dimension of amplitude a/T on the R.H.S. is equal to , viz., not equal to the dimensions of y and the angle , which means that they are not dimensionless.

Question:14

Answer:

The answer is the option and(i) The different physical quantities (P-Q) & (R+Q) in option a & e and never be added or subtracted. Thus, they will be meaningless.

(ii) In opt (d), dimensions of PR & may be equal.

(iii) in opt (b), the dimension of PQ may be equal to the dimension of R.

(iv) There is no addition or subtraction, which gives the possibility of opt (c).

Hence, opt (a) & (e).

Question:15

Photon is quantum of radiation with energy E = hv where v is frequency and h is Planck’s constant. The dimensions of h are the same as that of:

a) linear impulse

b) angular impulse

c) linear momentum

d) angular momentum

Answer:

The answer is the option (b) Angular Impulse and (d) Angular momentumHence, h (Planck’s constant)

Now, Linear Impulse = F.t

& Angular Impulse =

Now, Linear momentum = mv

& Angular momentum (L) = mvr

Thus, the dimensional formulae of Planck’s constant, Angular Impulse and Angular Momentum are the same.

Question:16

If Planck’s constant (h) and speed of light in vacuum (c) are taken as two fundamental quantities, which of the following can in addition be taken to express length, mass, and time in terms of the three chosen fundamental quantities?

a) mass of electron (m_{e})

b) universal gravitational constant (G)

c) charge of electron (e)

d) mass of proton (m_{p})

Answer:

The answer is the option (a) Mass of electron (mExplanation: Dimensions of –

h (Planck’s constant)

c (Speed of light in vacuum)

Thus, dimension of hc

Now,

Thus,

Now,

Now,

Also,

&

Therefore, in terms of chosen fundamental quantities, the physical quantities a, b & d can be used to represent L, M & T.

Question:17

Which of the following ratios express pressure?

a) Force/area

b) Energy/volume

c) Energy/area

d) Force/volume

Answer:

The answer is the option (a) Force/Area and (d) Force/VolumeWe know that Pressure(P)

..... this also verifies the opt (a).

Now let us compare it with other options-

= Dimesions of P

Thus, opt (b) is verified as well.

...............viz., not equal to Dimensions of P

.................viz., again not equal to dimensions of P

Thus, opt (c) & (d) are discarded.

Question:18

Which of the following are not a unit of time?

a) second

b) parsec

c) year

d) light year

Answer:

The answer is the option (b) Parsec and (d) Light yearQuestion:19

Why do we have different units for the same physical quantity?

Answer:

(i) Different orders of quantity are measured by the same physical quantities.(ii)The radius of a nucleus is measured in fermi. The distance between stars is measured in light-years. The length of cloth is measured in meter while the distance between two cities is measured in Kilometer or miles. Depending upon the physical quantity to be measured the units also vary.

Question:21

Name the device used for measuring the mass of atoms and molecules.

Answer:

The mass spectrograph is the device used for measuring the mass of atoms and molecules.Name the device used for measuring the mass of atoms and molecules.

Question:22

Express unified atomic mass unit in kg.

Answer:

We know that the unified atomic mass unit or amu or the mass of one atomMass of carbon atoms

Thus, the mass of one atom

Now , 1 amu

Thus, 1 amu

Question:23

A function is defined as

Why is it necessary for to be a dimensionless quantity ?

Answer:

(i) is a dimensionless physical quantity since it is represented by an angle viz. equal to arc/radius.(ii) As is dimensionless, all its powers will also be dimensionless. Here the first term is 1 viz. also dimensionless.

Hence, the L.H.S. must be dimensionless so that all the terms in R.H.S. are also dimensionless.

Question:24

Why length, mass and time are chosen as base quantities in mechanics?

Answer:

Length, mass & time is chosen as base quantities in mechanics because unlike other physical quantities, they cannot be derived from any other physical quantities.Question:25

(a) The earth-moon distance is about 60 earth radius. What will be the diameter of the earth (approximately in degrees) as seen from the moon?

(b) Moon is seen to be of diameter from the earth. What must be the relative size compared to the earth?

(c) From parallax measurement, the sun is found to be at a distance of about 400 times the earth-moon distance. Estimate the ratio of sun-earth diameters.

Answer:

(a) Here we know that, arc/radiusNow, the angle from the moon to the diameter of the earth will be-

(b) If the moon is seen from the earth diametrically, the angle will be

& if the earth is seen from the moon, the angle will be

Thus, the size of the moon/ size of the earth =

Therefore, as compared to the earth, the size of the moon is the size(diameter) of the earth.

(c)Let r

Now, the distance between the sun and the earth (r

On a solar eclipse, the sun is completely covered by the moon; also the angle formed by the diameter of the sun, moon and the earth are equal.

Thus,

......... ( the angle from the earth to the moon is & angle from sun to earth is

Now since

Thus,

Thus,

Therefore,

Question:26

Which of the following time measuring devices is most precise?

(a) A wall clock.

(b) A stopwatch.

(c) A digital watch.

(d) An atomic clock.

Give reason for your answer.

Answer:

The least counts of the devices are as follows-Wall clock = 1 sec

Stopwatch = sec

Digital watch = sec &

Atomic clock = sec.

Hence, the atomic clock is the most precise.

Question:31

Give an example of

(a) a physical quantity which has a unit but no dimensions.

(b) a physical quantity which has neither unit nor dimensions.

(c) a constant which has a unit.

(d) a constant which has no unit.

Answer:

(a) A plane angle is an example of a physical quantity which has unit but no dimension since, plane angle = arc/radius in radian & solid angle.(b) Relative density is a physical quantity which neither has neither units nor dimension since µr (relative density) is a ratio of same physical quantities.

(c) Gravitational constant (G) and Dielectric constant (k) are an example of constant that have a unit.

(d) Examples of the constant that has no units are Reynolds no. & Avogadro’s no.

Question:38

Answer:

As per the information given in the question, percentage error inAs per the information given in the question, Percentage error in

As per the information given in the question, Percentage error in

As per the information given in the question, Percentage error in

Percentage error in

Now, let us calculate the mean absolute error

(after rounding up)

Therefore, we get ‘2.8’ after rounding X i.e.

Question:40

Answer:

Let us first write down the dimensions of-(i) Let us consider that Mass m is directly proportional to

Comparing the powers on R.H.S. & L.H.S.

Thus,

Substituting the value of a in eq (i) & (ii)

By adding (iii) & (iv), we get,

Thus,

Substituting these values in eq. (iii)

Thus,

Now, let

Thus,

Equating the powers on both sides

Thus,

Now, since

becomes

& eq. (ii)

becomes

Subtracting eq (iv) from eq (iii) we get,

Thus,

And eq (iv) will be

Therefore,

Let us consider

Thus,

On adding eq (ii) & (iii) we get,

From eq (i) &(iv)

Thus,

Thus,

Question:41

Answer:

According to Kepler’s 3rd law of planetary motion,Now, R & g also affects T

Thus,

Now, comparing M, L & T according to the principle of homogeneity

Thus,

i.e.,

An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis, that where k is a dimensionless constant and g is acceleration due to gravity.

Question:42

In an experiment to estimate the size of a molecule of oleic acid, 1mL of oleic acid is dissolved in 19mL of alcohol. Then 1mL of this solution is diluted to 20mL by adding alcohol. Now, 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film we can calculate the thickness of the film which will give us the size of oleic acid molecule.

Read the passage carefully and answer the following questions:

a) Why do we dissolve oleic acid in alcohol?

b) What is the role of lycopodium powder?

c) What would be the volume of oleic acid in each mL of solution prepared?

d) How will you calculate the volume of n drops of this solution of oleic.

e) What will be the volume of oleic acid in one drop of this solution?

Answer:

(a) We can reduce the concentration of oleic acid by dissolving it in a proper solvent to get a molecular level. It can be dissolved in organic solvent alcohol and not ionic solvent water since it is an organic compound.(b) Mixing of oleic acid in water is prevented by Lycopodium when a drop of oleic acid is poured on water. Thus, if we spread lycopodium powder on the water surface, the layer of oleic acid will be formed on the surface of the powder,

(c) ml is the concentration of oleic acid in an alcohol solution. Its required conc. in 1 ml solution is 1/400 ml.

(d) By dropping one ml of solution drop by drop in a beaker through burette and counting its no., can be used to calculate the volume of n drop solution. If there is 1ml of n drop, then its volume will be 1/n ml.

(e) If there is 1ml of n drop, then its volume will be 1/n ml. The concentration of oleic acid in one drop of the solution will be

of oleic acid.

Question:43

a) How many astronomical units (AU) make 1 parsec?

b) Consider the sun like a star at a distance of 2 parsec. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears to be (1/2) degree from the earth. Due to atmospheric fluctuations, eye cannot resolve objects smaller than 1 arc minute.

c) Mars has approximately half of the earth’s diameter. When it is closer to the earth it is at about ½ AU from the earth. Calculate at what size it will disappear when seen through the same telescope.

Answer:

(a)1 A.U. long arc subtends the angle of 1s or 1 arc sec at distance of 1 parsec.Thus, angle 1 sec = parsec

Thus, 1 parsec = arc sec

Thus, 1 parsec=

(b)Angle of the sun’s diameter is subtended by 1 A.U. since the distance from the sun increases angle subtended in the same ratio.

Now,

will form an angle of , since the diameter is the same angle subtended on earth by 1 parsec will be same.

If the sunlike star is at 2 parsec the angle becomes half

Thus, angle min

When it is seen with a telescope that has a magnification of 100, the angle formed will be min, viz., less than a minute.

Hence, it can’t be observed by a telescope.

Question:44

Einstein’s mass-energy relation emerging out of his famous theory of relativity relates mass (m) to energy (E) as E = mc^{2}, where c is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in MeV where , the masses are measured in unified equivalent of 1u is 931.5 MeV.

a) Show that the energy equivalent of 1 u is 931.5 MeV.

b) A student writes the relation as . The teacher points out that the relation is dimensionally incorrect. Write the correct relation.

Answer:

(a) Given :By the formula

(b) 1 u mass converted into total energy will be released by 931.5 MeV.

However, 1 amu = 931.5 MeV is dimensionally incorrect.

, will be dimensionally correct.

Main Subtopics in NCERT Exemplar Class 11 Physics Solutions Chapter 2 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

**Also, Read NCERT Solution subject wise -**

- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Physics
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Biology

**Check NCERT Notes subject wise -**

- NCERT Notes for Class 11 Maths
- NCERT Notes for Class 11 Physics
- NCERT Notes for Class 11 Chemistry
- NCERT Notes for Class 11 Biology

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Download EBookStudents would be able to understand the dynamics behind the change in measuring trends, learn about the basic units, and try to bring all that knowledge to life by using these concepts in actual measuring instruments. NCERT Exemplar Class 11 Physics solutions chapter 2 would also introduce the students to utilise the system of precise measurement to answer real-life questions related to the mass of rain-bearing clouds and ascertain the speed of an aircraft through dimensional analysis and determine the distance between two stars.

· NCERT Exemplar Class 11 Physics solutions chapter 2 introduces the student to various measurement units across the world and the standard unit that is accepted internationally. It talks about measuring different physical quantities like length and mass and abstract quantities like time.

· The dimensional formulae and equations are also introduced in this chapter, which helps to utilise dimensional analysis in real-life applications. Students are also made aware of the importance of significant figures of measurement.

· NCERT Exemplar Class 11 Physics solutions chapter 2 also helps students understand the accuracy and precision in making measurements and the errors that one stumbles on along the way. It also highlights the different types of errors found in instruments and the measuring procedure to quickly identify and rectify them.

Chapter 1 | Physical world |

Chapter 2 | Units and Measurement |

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 |

1. How can these NCERT Exemplar Class 11 Physics Solutions Chapter 2 help?

These solutions from Chapter 2 can help better understand the basic concepts and prepare for your academic exams. Questions of multiple choice type, short answer and long answer types are solved.

2. How to download these Solutions?

These Class 11 physics NCERT exemplar solutions chapter 1 can be downloaded from the website and the solution page directly in PDF format.

3. What are the essential topics of this chapter?

This chapter's essential topics are the International System of Units, Measurement of quantities, accuracy and precision, dimensional formulae, and dimensional analysis.

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