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We have provided below free printable Class 11 Physics Assignments for Download in PDF. The Assignments have been designed based on the latest NCERT Book for Class 11 Physics . These Assignments for Grade 11 Physics cover all important topics which can come in your standard 11 tests and examinations. Free printable Assignments for CBSE Class 11 Physics , school and class assignments, and practice test papers have been designed by our highly experienced class 11 faculty. You can free download CBSE NCERT printable Assignments for Physics Class 11 with solutions and answers. All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Physics Class 11. Students can click on the links below and download all Pdf Assignments for Physics class 11 for free. All latest Kendriya Vidyalaya Class 11 Physics Assignments with Answers and test papers are given below.

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We have provided below the biggest collection of free CBSE NCERT KVS Assignments for Class 11 Physics . Students and teachers can download and save all free Physics assignments in Pdf for grade 11th. Our expert faculty have covered Class 11 important questions and answers for Physics as per the latest syllabus for the current academic year. All test papers and question banks for Class 11 Physics and CBSE Assignments for Physics Class 11 will be really helpful for standard 11th students to prepare for the class tests and school examinations. Class 11th students can easily free download in Pdf all printable practice worksheets given below.

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NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements - PDF Download

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements are the best study resources. You can get to understand the main topics and to score well in your examination. This solution provides appropriate answers to the textbook questions. To get a good grip on this chapter, you can make use of the NCERT Solutions for Class 11 Physics available at our website for free in a pdf form, you can download them and use them offline.

Chapter 1 of NCERT Solutions for Class 11 Physics mainly helps understand the fundamentals of units and measurements. In our daily lives, most of the activities depend on this, and it is very important for us to learn it effectively. Everything around us depends on units and measurements, from buying milk in the morning to the pounds of bread needed for breakfast or from buying sugar for milk to the kilograms of rice needed for lunch. You can access the Physics NCERT Solutions for Class 11 to comprehend the key concepts present in this chapter.

Topics Covered in Class 11 Chapter 2 Physics Units and Measurement

Introduction.

Measurement of any kind of physical quantity involves comparison with a certain basic, arbitrarily chosen, internationally accepted reference standard we called that standard as Unit. The result of a measurement of any physical quantity is expressed by a numerical measure accompanied by a unit. However, if the number of physical quantities appears to be very large, we use only a limited number of units for expressing all the physical quantities, since all the units are interrelated with one another. The units for the fundamental or base quantities are called fundamental or base units.

The International System of Units

In past times, the scientists of different countries were using different systems of units for measurement. There are basically 3 such systems like CGS, the FPS (or British) system and the MKS system were in use extensively till recently.

 In CGS(Centimeter Gram Second system) system they were centimetre, gram and second respectively

In the FPS(foot-pound-second system of units) system they were foot, pound and second respectively.

In MKS system they were meter, kilogram and second respectively. 

Significant figure

Following are the rules for significant figure

All the non-zero digits are significant.

All the zeros between two non-zero digits are significant, no matter where the decimal point is, if at all.

If the number is less than 1, the zero(s) on the right of the decimal point but to the left of the first non-zero digit are not significant. 

The terminal or trailing zero(s) in a number without a decimal point are not significant.

The trailing zero(s) in a number with a decimal point are significant. 

Rules for Arithmetic Operations with Significant Figures

The result of a calculation involving approximate measured values of quantities (i.e. values with limited number of significant figures) must reflect the uncertainties in the original measured values. It cannot be more accurate than the original values that we measured themselves on which the result is based. In common, the final result should not have more significant figures than the original data from which it was picked up.

The given rules for arithmetic operations with significant figures ensure that the final result of a calculation is shown with the precision that is consistent with the precision of the input measured values :

 In multiplication or division, the final result should retain as many significant figures as there are in the original result with the least significant figures.

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

Rounding off the Uncertain Digits

The result of calculation with approximate numbers, which contain more than one uncertain digit, should be rounded off. The rules for rounding off numbers to the appropriate significant figures are very obvious in most cases. Let's consider a number 1.746 rounded off to three significant figures is 1.75, while the number 2.743 would be 2.74. The rule by convention is that the preceding digit is raised by 1 if the insignificant digit to be dropped is more than 5, and is left unchanged if the digit is less than 5. But what if the number is 3.745 in which the insignificant digit is 5. Here, the old method is that if the preceding digit is even, the insignificant digit is dropped and, if it is odd, the preceding digit is raised by 1.

Rules for Determining the Uncertainty in the Results of Arithmetic Calculations

The rules for determining the uncertainty or error in the number/measured quantity in arithmetic operations can be understood from the below examples. 

(1) If the length and breadth of a thin rectangular sheet are measured, using a meter scale as 17.2 cm and 10.1 cm respectively, there are three significant figures in each measurement. It means that the length l can be written as l = 17.2 ± 0.1 cm 

And breadth b can be written as b = 10.1 ± 0.1 cm

                                                       = 10.1 cm ± 1 %

(2) If a set of experimental data is specified to n significant figures, a result gained by combining the data will also be valid to n significant figures.

(3) The relative error of a value of number specified to significant figures depends not only on n but on the number itself too. 

Dimensions of Physical Quantities

The nature of all physical quantities is described by its dimensions. All the physical quantities represented by derived units can be expressed in terms of some combination of seven fundamental quantities or base quantities. We can call these base quantities as the seven dimensions of the physical world, which are represented in square brackets [ ]. So , length of dimension [L], , time [T], mass [M], electric current [A], thermodynamic temperature [K], luminous intensity [cd], and amount of substance [mol]. 

Dimensional Formulae and Dimensional Equations

The expression which shows how and which of the base quantities represent the dimensions of a physical quantity are known as the dimensional formula of the given physical quantity. For example, the dimensional formula of the volume is [M° L3 T°], and that of speed or velocity is [M° L T-1]. Similarly,  [M L–3 T°] that of mass density and  [M° L T–2] is the dimensional formula of acceleration.

 An equation which is obtained by equating a physical quantity with its dimensional formula is called the dimensional equation of the physical quantity. So, the dimensional equations are the equations, which represent the dimensions of a physical quantity in terms of the base quantities. 

Dimensional Analysis and Its Applications

The recognition of concepts of dimensions, which guide the description of physical behaviour is of basic importance as only those physical quantities can be added or subtracted which have the similar dimensions. A thorough understanding of dimensional analysis helps you in deducing certain relations among different physical quantities and checking the derivation, accuracy and dimensional consistency or homogeneity of various mathematical expressions.

Checking the Dimensional Consistency of Equations 

 if an equation fails this consistency test, it is proved wrong, but if it passes, it is not proved right. Thus, a dimensionally correct equation need not be actually an exact (correct) equation, but a dimensionally wrong (incorrect) or inconsistent equation must be wrong.

Benefits of the Class 11 Physics Chapter 1 Solution

The benefits of using the NCERT Solutions for Class 11 Physics Chapter 1 are –

1. Completely solved answers for all the questions given in the NCERT textbook are freely available in PDF format.

2. Simple and easy-to-understand language is used to make learning fun for you.

3. Our Subject teacher experts prepare the solutions after conducting vast research on each concept.

4. The solutions not only help you with your exam preparation but also for various competitive exams like JEE, NEET, etc.

5. PDF format of solutions is available in chapter-wise and exercise-wise formats to help students learn the concepts in a better way.

Frequently Asked Questions

Question 1 : What are the topics in chapter 1 in physics class 11th?

Answer : Following are the topics in chapter 1 physics class 11th

1.1 Introduction

1.2 The International System of Units

1.3 Significant figure

1.3.1 Measurement of Length

1.3.2 Rounding off the uncertain digits

1.3.3 Rules for determining the uncertainty in the result of Arithmetic calculation

1.4 Dimensions of Physical Quantities

1.5 Dimensional Formulae and Dimensional Equations

1.6 Dimensional Analysis and Its Applications

1.6.1 Checking the Dimensional Consistency of Equations

1.6.2 Deducing Relation among the physical quantities 

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NCERT Solution Class 11 Physics Chapter 1 Units and Measurement

Q4 :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 : 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.

Q5 : 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 the Sun and the Earth: = Speed of light x 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 x 500 = 500 units

Q7 :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 : 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.

Q10 : State the number of significant figures in the following: (a) 0.007 m 2 (b) 2.64 x 10 24 kg (c) 0.2370 g cm -3 (d) 6.320 J (e) 6.032 N m -2 (f) 0.0006032 m 2 Answer : (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 x 10 24 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.

Q11 : 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 : 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:

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.

Q12 : 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? Answer : 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.

Q18 : 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 : 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.

Q21 : 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. Answer : 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.

Q31 : The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us? Answer : Time taken by quasar light to reach Earth = 3 billion years = 3 x 10 9 years = 3 x 109 x 365 x 24 x 60 x 60 s Speed of light = 3 x 10 8 m/s Distance between the Earth and quasar = (3 x 10 8 ) x (3 x 10 9 x 365 x 24 x 60 x 60) = 283824 x 10 20 m = 2.8 x 10 22 km

NCERT Solutions for Class 11 Physics All Chapters

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 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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High School Physics

Friction Numericals – class 11 physics

Last updated on June 7th, 2023 at 02:46 pm

Here we will solve a set of Friction Numericals based on the concepts of frictional force or friction and friction coefficient for class 11 and class 12 students.

Here are the links to the posts on friction and friction coefficient , that you can visit and read for better understanding.

Important Concepts of Friction [that helps to solve the numericals]

• Friction is caused by bodies sliding over rough surfaces. • The degree of surface roughness is indicated by the coefficient of friction, μ. • The force of friction is calculated by multiplying the coefficient of friction by the normal force. • The frictional force always opposes motion. • Acceleration is caused by the net force which is found by subtracting the frictional force from the applied force.

Friction Numericals | Numericals based on friction class 11

Let’s solve some numerical problems using the concepts and formula of friction.

[Numerical Problem 1] A box weighing 2000 N is sliding across a cement floor. The force pushing the box is 500 N and the coefficient of sliding friction between the box and the floor is 0.20. What is the acceleration of the box?

In this case, the normal force for the box is its weight. Using the normal force and the coefficient of friction, we can find the frictional force. We can also find the mass of the box from its weight since we know the acceleration due to gravity. Then we can find the net force and the acceleration. Friction = F F = μF N = (0.20)(2000 N) = 400 N Mass of box = weight/g = (2000 / 9.8) Kg = 204 Kg F net = Pushing force – frictional force = 500 N – 400 N = 100 N

Acceleration = a = F net / mass = (100 / 204 ) m/s 2 = 0.49 m/s 2

[Numerical Problem 2] Two boxes are connected by a rope running over a pulley (see image). The coefficient of sliding friction between box A and the table is 0.20. (Ignore the masses of the rope and the pulley and any friction in the pulley.) The mass of box A is 5.0 kg and the mass of box B is 2.0 kg. The entire system (both boxes) will move together with the same acceleration and velocity. Find the acceleration of the system.

The force tending to move the system is the weight of box B and the force resisting the movement is the force of friction between the table and box A. The mass of the system would be the sum of the masses of both boxes. The acceleration of the system will be found by dividing the net force by the total mass. F N (box A) = mg = (5.0 kg)(9.8 m/s^2) = 49 N Friction = F F = μF N = (0.20)(49 N) = 9.8 N Weight of box B = mg = (2.0 kg)(9.8 m/s^2) = 19.6 N F net = 19.6 N – 9.8 N = 9.8 N a = F net / (total mass of A and B) = 9.8 N/7.0 kg = 1.4 m/s^2

Suggestion: If you need quick guidance on ‘ pulley-based numerical problems’ like the above one, then you can go through this post on pulley problems – how to address pulley numerical problems

Friction Numerical Extra Questions for class 11 – with Solution Link

[Extra Numerical Problem 1] A 52 N sled is pulled across a cement sidewalk at a constant speed. A horizontal force of 36 N is exerted. What is the coefficient of sliding friction between the sidewalk and the metal runners of the sled? Extra numerical Problem 1: Solution Link

[Extra Numerical Problem 2] If the coefficient of sliding friction between a 25 kg crate and the floor is 0.45, how much force is required to move the crate at a constant velocity across the floor? Extra numerical Problem 2: Answer & Solution Link

[ Extra Numerical Problem 3] A smooth wooden 40.0 N block is placed on a smooth wooden table. A force of 14.0 N is required to keep the block moving at constant velocity. (a) What is the coefficient of sliding friction between the block and the tabletop? (b) If a 20.0 N brick is placed on top of the wooden block, what force will be required to keep the block and brick moving at constant velocity? Friction extra numerical problem 3: Answer and Solution Link

[ Extra Numerical Problem 4] A 4000 kg truck is parked on a 15 degrees slope. How big is the friction force on the truck? Friction extra numerical problem 4: Solution

[ Extra Numerical Problem 5] A block of mass 0.500 kg slides on a flat smooth surface with a speed of 2.80 m/s. It then slides over a rough surface with μk and slows to a halt. While the block is slowing, (a) what is the frictional force on the block? (b) What is the magnitude of the block’s acceleration? (c) How far does the block slide on the rough part before it comes to a halt? Numerical problem 5 of friction: solution link

[ Extra Numerical Problem 6] A block slides down a rough incline sloped at an angle of 40.0 degrees from the horizontal. Starting from rest, it slides a distance of 0.800m down the slope in 0.600 s. What is the coefficient of kinetic friction for the block and surface? Solution : [ take this as your assignment for now. Will add the solution soon]

[ Extra Numerical Problem 7] A factory uses a motor and a cable to drag a 300 kg machine to the proper place on the factory floor. What power must the motor supply to drag the machine at a speed of 0.50 m/s? The coefficient of friction between the machine and the floor is 0.60. Solution of friction problem #7

[ Extra Numerical Problem 8] A 1500 kg car has a front profile that is 1.6 m wide by 1.4 m high and a drag coefficient of 0.50. The coefficient of rolling friction is 0.02. What power must the engine provide to drive at a steady 30 m/s if 25% of the power is “lost” before reaching the drive wheels? solution of problem #8 on friction & air drag

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Class 11 physics notes fbise & kpk board, 11th class physics chapter 1 measurement.

• Chapter 1 Theory   (complete)
• Chapter 1 Assignments   (complete)
• Chapter 1 Conceptual Questions  (complete)
• Chapter 1 Numerical Questions   (complete)

11th Class Physics Chapter 2 Vectors & Equilibrium

• Chapter 2 Theory   (complete)
• Chapter 2 Assignments   (complete)
• Chapter 2 Conceptual Questions   (complete)
• Chapter 2 Numerical Questions   (complete)

11th Class Physics Chapter 3 Forces & Motion

• Chapter 3 Theory   (complete)
• Chapter 3 Assignments   (complete)
• Chapter 3 Conceptual Questions   (complete)
• Chapter 3 Numerical Questions   (complete)

11th Class Physics Chapter 4 Work & Energy

• Chapter 4 Theory   (complete)
• Chapter 4 Conceptual Questions   (complete)

11th Class Physics Chapter 5 Rotational & Circular Motion

• Chapter 5 Theory   (complete)
• Chapter 5 Conceptual Questions   (complete)
• Chapter 5 Assignments  (selected)
• Chapter 5 Numerical Questions  (selected)

11th Class Physics Chapter 6 Fluid Dynamics

• Chapter 6 Theory   (complete)
• Chapter 6 Conceptual Questions   (complete)
• Chapter 6 Assignments & Numericals  (selected)

11th Class Physics Chapter 7 Oscillation

• Chapter 7 Theory   (complete)
• Chapter 7 Conceptual Questions   (complete)
• Chapter 7 Assignments  (selected)
• Chapter 7 Numerical Question s (selected)

11th Class Physics Chapter 8 Waves

• Chapter 8 Theory   (complete)
• Chapter 8 Conceptual Questions   (complete)
• Chapter 8 Assignments  (selected)
• Chapter 8 Numerical Questions  (selected)

11th Class Physics Chapter 9 Physical Optics Notes

• Chapter 9 Theory   (complete)
• Chapter 9 Conceptual Questions   (complete)
• Chapter 9 Assignments  (selected)
• Chapter 9 Numerical Questions  (selected)

11th Class Physics Chapter 10 Thermodynamics

• Chapter 10 Theory   (complete)
• Chapter 10 Conceptual Questions   (complete)
• Chapter 10 Assignments  (selected)
• Chapter 10 Numerical Questions  (selected)

99 comments:

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Assalamu alaikum

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Important long question for 2024

So you not solve the MCQs

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Please upload HSSC 1 physics chapters full notes from chapter 4 to 10 ... Plz

Yes upload full sallybus

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Fastly isn't a word Einstein .....

Please upload full syllabus notes

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PLZ UPLOAD FULL NOTES PLZ PLZ PLZ PLZ PLZ

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Please upload full notes

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Complete 1st year course sir notes and lectures

Plz upload solutions of mcqs

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khud nots bana lo bachy mery pass time nahi hy

beta sab yaad karo

Please upload full notes plz

sir plz upload full syllabus notes of class 11

Comprehensive question nai ha in notes ma

Sir u r notes r outstanding in how much time more notes will be prepare

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English doesn't matters

Plz upload full syllabus plzzzz

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SIR NOTES OF COMPREHENSIVE QUESTIONS

Very good notes but not complete please complete the notes

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Lajawab ha sir

exams are coming sir , it would be alot more easy to prepare if u share the full syllabus. we all would be obliged. thank u

Kindly upload full syllabus notes from chapter 4 to 10

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AOA sir your notes are exellent and usefull but kindly upload full syllabus

Chapter 10 Conceptual Questions And Chapter 9 Conceptual Questions Are Full Syllabus Not Reduced

please post full notes as soon as possible and DONT forget to include answers of quizes and and pont to ponder boxes ,, Waiting , JazakAllah

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Pehla pta hota to parh leta ab akhri din muh marna Aya Hun🥷

where are numerical question of c.h.p 4

Notes are very good

Very Good Notes..............

Chapter 4 numerical

Chapter 4 numericals

aoa plz upload the comprehension questions answers of chapter no 1 measurement

numericals of chapter no 4 please

thanks for your notes

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How to download ?

IT WAS A GREAT HELP

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جزاك اللهُ بہت مہربانی

I write your conceptual answers in sendup exams and I got fail😢

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sir can u mark the imp questions for fbise ,

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• Chapter 15: Waves

NCERT Solutions for Class 11 Physics Chapter 15: Waves

Ncert solutions class 11 physics chapter 15 – free pdf download.

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 14.

NCERT Solutions for Class 11 Physics Chapter 15 Waves are created by the subject-matter experts at BYJU’S with the objective of helping students in their exam preparations. By practising questions from the NCERT textbook using the solutions PDF, students can gain more information about the chapter and can also have a quick review before their exam. NCERT Solutions for Class 11 Physics, created according to the latest CBSE Syllabus 2023-24, provide a strong foundation of basic concepts which will help them in their higher levels of education.

Wave is basically a quivering disruption that passes through a medium due to the repeated motion of particles. Chapter 15 of NCERT Solutions prepares students to build a good base for their higher studies in fields such as Medical Science and Engineering. Students will be introduced to new concepts like the Doppler effect, wave types and their interrelationships in this chapter. It will provide a clear insight into the important topics, which will help the students score good marks in the Class 11 exam. Get the NCERT Solutions for Class 11 Physics PDF from the links below.

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Access Answers to NCERT Solutions for Class 11 Physics Chapter 15 Waves

Q1. A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

Given, Mass of the string, M = 2.50 kg Tension in the string, T = 200 N Length of the string, l = 20.0 m

Mass per unit length, μ = M /l = 5/40 = 0.125 kg / m

Velocity  of the transverse wave , v = \(\begin{array}{l}\sqrt{\frac{T}{\mu }}\end{array} \)

= \(\begin{array}{l}\sqrt{\frac{200}{0.125 }}\end{array} \) = 40 m/s

Therefore, the time taken by the transverse wave to reach the other side is 40 m/s.

Q2. A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top, given that the speed of sound in air is 340 m s –1 ? (g = 9.8 m s –2 )

Height of the bridge, s = 300 m The initial velocity of the stone, u = 0 Acceleration, a = g = 9.8 m/s 2 Speed of sound in air = 340 m/s

Let t be the time taken by the stone to hit the water’s surface We know, s = ut + ½ gt 2 300 = 0 + ½ x 9.8 x t 2 Therefore, t = 7.82 s

The time taken by the sound to reach the bridge, t’= 300/340 = 0.88 s

Therefore, from the moment the stone is released from the bridge, the sound of it splashing the water is heard after = t +t’ = 7.82 s + 0.88 s = 8.7s.

Q3. Steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s –1 .

Given, Length of the steel wire, l = 12 m Mass of the steel wire, m = 2.0 kg Velocity of the transverse wave, v = 343 m/s

Mass per unit length, μ = M /l = 2.10/12 = 0.175 kg / m

Velocity of the transverse wave , v = \(\begin{array}{l}\sqrt{\frac{T}{\mu }}\end{array} \)

Therefore, T = v 2 μ

= 343 2 x 0.175 = 2.06 x 10 4 N

Therefore, the tension in the wire is 343 2 x 0.175 = 2.06 x 10 4 N.

Q4. Using the formula v = \(\begin{array}{l}\sqrt{\frac{\gamma P}{\rho }}\end{array} \) , explain why the speed of sound in air (a) does not depend upon pressure. (b) increases with temperature and humidity. (c) increases with humidity.

Given, v = \(\begin{array}{l}\sqrt{\frac{\gamma P}{\rho }}\end{array} \)

We know, PV = nRT ( for n moles of ideal gas ) => PV = RT (m/M ) Where, m is the total mass, and M is the molecular mass of the gas. Therefore, P  = m (RT/M) => P = \(\begin{array}{l}\frac{\rho RT}{M}\end{array} \) => \(\begin{array}{l}\frac{P}{\rho} = \frac{ RT}{M}\end{array} \)

(a) For a gas at a constant temperature, \(\begin{array}{l}\frac{P}{\rho}\end{array} \) = constant Thus, as P increases ρ and vice versa. This means that the P/ρ ratio always remains constant, meaning

v = \(\begin{array}{l}\sqrt{\frac{\gamma P}{\rho }}\end{array} \) = constant, i.e. velocity of sound does not depend upon the pressure of the gas.

(b) Since, \(\begin{array}{l}\frac{P}{\rho} = \frac{ RT}{M}\end{array} \) v = \(\begin{array}{l}\sqrt{\frac{\gamma P}{\rho }}\end{array} \)    = v = \(\begin{array}{l}\sqrt{\frac{\gamma RT}{M }}\end{array} \)

We can see that v \(\begin{array}{l}\propto \sqrt{T}\end{array} \) , i.e. speed of sound increases with temperature.

(c) When humidity increases, the effective density of the air decrease. This means \(\begin{array}{l}v \propto \frac{1}{\sqrt{\rho }}\end{array} \) , thus velocity increases.

Q5.We know that the function y = f (x, t) represents a wave travelling in one direction, where x and t must appear in the combination x + vt or x– vt or, i.e. y = f (x ± vt). Is the converse true? Can the following functions for y possibly represent a travelling wave: (i) (x – vt) 2 (ii) log [ ( x + vt)/ x 0 ] (iii) 1 / (x + vt )

No, the converse is not true because it is necessary for a wave function representing a travelling wave to have a finite value for all values of x and t. As none of the above functions satisfies the given condition, none of the options represents a travelling wave.

Q6.  A bat emits the ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound and (b) the transmitted sound? The speed of sound in air is 340 m s –1 , and in water, 1486 m s –1 .

Given, Frequency of the ultrasonic sound, ν = 1000 kHz = 10 6 Hz Speed of sound in air, v A = 340 m/s We know,

(a) The wavelength ( λ R ) of the reflected sound is: λ R = v A /v = 340/10 6 = 3.4 x 10 -4 m

(b) Speed of sound in water, v W = 1486 m/s Therefore, the wavelength (λ T )of the transmitted sound is: λ T = 1486 / 10 6 = 1.49 x 10 -3 m.

Q7. An ultrasonic scanner operating at 4.2 MHz is used to locate tumours in tissues. If the speed of sound is 2 km/s in a certain tissue, calculate the wavelength of sound in this tissue.

Given, Speed of sound in the tissue, v T = 2 km/s = 2 × 10 3 m/s Operating frequency of the scanner, ν = 4.2 MHz = 4.2 × 10 6 Hz

Therefore, the wavelength of sound: λ = v T / v = (2 × 10 3 )/ (4.2 × 10 6 ) = 4.76 x 10 -4 m.

Q8. A transverse harmonic wave on a wire is expressed as: y( x, t ) = 3 sin ( 36t +0.018x + π / 4 ). (i) Is it a stationary wave or a travelling one? (ii) If it is a travelling wave, give the speed and direction of its propagation. (iii) Find its frequency and amplitude. (iv) Give the initial phase at the origin. (v) Calculate the smallest distance between two adjacent crests in the wave. [X and y are in cm and t in seconds. Assume the left to the right direction as the positive direction of x]

Given, y(x, t) =3 sin (36t +0.018x + π/4)                 . . . . . . . . .  . . ( 1 )

(i) We know the equation of a progressive wave travelling from right to left is: y (x, t) = a sin (ωt + kx + Φ)         . . . . . . . . . . . . . . . . . . .  ( 2 )

Comparing equation (1) to equation (2), we see that it represents a wave travelling from right to left and also we get: a = 3 cm,  ω = 36 rad/s , k = 0.018 cm and ϕ = π/4.

(ii) Therefore, the speed of propagation, v = ω/k = 36/ 0.018 = 20 m/s.

(iii) Amplitude of the wave, a = 3 cm Frequency of the wave v = ω / 2π = 36 /2π = 5.7 hz. (iv) Initial phase at the origin = π/4 (v) The smallest  distance between two adjacent crests in the wave, λ = 2π/ k = 2π / 0.018 = 349 cm.

Q9. For the wave in the above question (Q8), plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm.

(i) Give the shapes of these plots. (ii) With respect to which aspects (amplitude, frequency or phase) does the oscillatory motion in a travelling wave differ from one point to another?

(i) Given, y(x, t) =3 sin (36t +0.018x + π/4)                . . . . . . . . .  . . ( 1 )

For x= 0, the equation becomes: y( 0, t ) =3 sin ( 36t +0 + π/4 )       . . . . . . . . .  . . ( 2 ) Also, ω = 2 π/t = 36 rad/s => t = π/18 secs.

Plotting the displacement (y) vs (t) graphs using different values of t listed below:

(ii) Similarly, graphs are obtained for x = 0, x = 2 cm, and x = 4 cm. The oscillatory motion in the travelling wave is different from each other only in terms of phase. Amplitude and frequency are invariant for any change in x. The y-t plots of the three waves are shown in the given figure:

Q10. A travelling harmonic wave is given as: y (x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35). What is the  phase difference between the oscillatory motion of two points separated by a distance of: (i) 8 m (ii) 1 m (iii) λ /2 (iv) 6λ/4 [ X and y are in cm, and t is in secs ].

Given, The equation for a travelling harmonic wave: y (x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35) = 2.0 cos (20πt – 0.016πx + 0.70 π) Where, Propagation constant, k = 0.0160 π Amplitude, a = 2 cm Angular frequency, ω= 20 π rad/s

We know, Phase difference Φ = kx = 2π/ λ

(i) For x = 8 m = 800 cm Φ = 0.016 π × 800 = 12.8 π rad.

(ii) For x =  1 m = 100 cm Φ = 0.016 π × 100 = 1.6 π rad.

(iii) For x = λ /2 Φ = (2π/ λ)  x  ( λ/2) =  π rad.

(iv) For x = 6λ /4 Φ = (2π/ λ)  x  ( 6λ/4) = 3 π rad.

Q11. The transverse displacement of a wire (clamped at both its ends) is described as: y (x, t) = \(\begin{array}{l}0.06sin(\frac{2\pi}{3}x)cos(120\pi t)\end{array} \) The mass of the wire is 6 x 10 -2 kg, and its length is 3m. Provide answers to the following questions: (i) Is the function describing a stationary wave or a travelling wave? (ii) Interpret the wave as a superposition of two waves travelling in opposite directions. Find the speed, wavelength and frequency of each wave. (iii) Calculate the wire’s tension. [X and y are in meters and t in secs]

We know, The standard equation of a stationary wave is described as: y (x, t) = 2a sin kx cos ωt

The given equation y (x, t) = \(\begin{array}{l}0.06sin(\frac{2\pi}{3}x)cos(120\pi t)\end{array} \) is similar to the general equation. (i) Thus, the given function describes a stationary wave.

(ii) We know a wave travelling in the positive x-direction can be represented as: y 1 = a sin( ωt – kx ) Also, A wave travelling in the negative x-direction is represented as: y 2 = a sin( ωt + kx )

Super-positioning these two waves gives: y = y 1 + y 2 = a sin( ωt – kx ) – a sin( ωt + kx ) = asin(ωt)cos(kx) – asin(kx)cos(ωt) – asin(ωt)cos(kx) – asin(kx)cos(ωt) = – 2asin(kx)cos(ωt) = \(\begin{array}{l}-2asin(\frac{2\pi }{\lambda }x)cos(2\pi vt)\end{array} \) . . . . . . . . . . . . . . ( 1 )

The transverse displacement of the wires is described as: \(\begin{array}{l}0.06sin(\frac{2\pi}{3}x)cos(120\pi t)\end{array} \)             . . . . . . . . . . . . . . . . .( 2 )

Comparing equations ( 1 ) and ( 2 ), we get 2π/ λ = 2π/ 3 Therefore, wavelength  λ = 3m

Also, 2πv/λ = 120π Therefore, speed v = 180 m/s

And, Frequency = v/λ = 180/3 = 60 Hz

(iii)Given, The velocity of the transverse wave, v = 180 m / s The string’s mass, m = 6 × 10 -2 kg String length, l = 3 m Mass per unit length of the string, μ = m/l = (6 x 10 -2 )/3 = 2 x 10 -2 kg/m Let the tension in the wire be T Therefore, T = v 2 μ = 180 2 x 2 x 10 -2 = 648 N.

Q12. Considering the wave described in the above question (Q11), answer the following questions: (a) Are all the points in the wire oscillating at the same values of (i) frequency, (ii) phase, and (iii) amplitude? Justify your answers. (b) Calculate the amplitude of a point 0.4 m away from one end.

(a) As the wire is clamped at both ends, the ends behave as nodes, and the whole wire vibrates in one segment. Thus, (i) Except at the ends, which have zero frequency, all the particles in the wire oscillate with the same frequency. (ii) All the particles in the wire lie in one segment; thus, they all have the same phase. Except for the nodes. (iii) Amplitude, however, is different for different points.

(b) Given the equation, y (x, t) = \(\begin{array}{l}0.06sin(\frac{2\pi}{3}x)cos(120\pi t)\end{array} \)

For x = 0.4m and t =0

Amplitude = displacement = \(\begin{array}{l}0.06sin(\frac{2\pi}{3}x)cos0 \end{array} \) = \(\begin{array}{l}0.06sin(\frac{2\pi}{3}\times 0.4)1\end{array} \) = 0.044 m.

Q13. Given below are functions of x and t to describe the displacement (longitudinal or transverse) of an elastic wave. Identify the ones describing  (a) a stationary wave, (b) a travelling wave and (c) neither of the two: (i) y = 3 sin( 5x – 0.5t ) + 4cos( 5x – 0.5t ) (ii) y = cosxsint + cos2xsin2t. (iii) y = 2 cos (3x) sin (10t) (iv) y = \(\begin{array}{l}2\sqrt{x – vt}\end{array} \)

(i) This equation describes a travelling wave, as the harmonic terms ωt and kx, are in the combination of kx – ωt. (ii) This equation describes a stationary wave because the harmonic terms ωt and kx appear separately in the equation. In fact, this equation describes the superposition of two stationary waves. (iii) This equation describes a stationary wave because the harmonic terms ωt and kx appear separately. (iv) This equation does not contain any harmonic term. Thus, it is neither a travelling wave nor a stationary wave.

Q14. A string clamped at both ends is stretched out, it is then made to vibrate in its fundamental mode at a frequency of 45 Hz. The linear mass density of the string is 4.0 × 10 -2 kg/ m, and its mass is 2 × 10 -2 kg. Calculate: (i) The velocity of a transverse wave on the string (ii) The tension in the string

Given, Mass of the string, m = 2 x 10 -2 kg Linear density of the string = 4 x 10 -2 kg Frequency, v F = 45 Hz

We know the length of the wire = m/µ = (2 x 10 -2 )/ (4 x10 -2 ) = 0.5 m

We know, λ = 2l/n Where, n = number of nodes in the wire.

For the fundamental node, n =1 => λ = 2l = 2 x 0.5 = 1m

(i) Therefore, the speed of the transverse wave, v = λ v F = 1 x 45 = 45 m/s. (ii) Tension in the string = µ v 2 = 4 x10 -2 x 45 = 81

Q15. A 1 m long pipe with a movable piston at one end and an opening at the other will be in resonance with a tuning fork vibrating at 340 Hz, if the length of the pipe is 79.3 cm or 25.5 cm. Calculate the speed of sound in the air. Neglect the edge effects.

Given, Frequency of the turning fork, ν F = 340 Hz Length of the pipe, l 1 = 0.255 m

As the given pipe has a piston at one end, it will behave as a pipe with one end closed and the other end open, as depicted in the figure below:

This kind of system creates odd harmonics. We know that the fundamental note in a closed pipe is written as: l 1 = λ / 4 0.255 x 4 = λ = 1.02m Therefore, the speed of sound, v = λ v F = 340 × 1.02 = 346.8 m/s.

Q16. A steel bar of length 200 cm is nailed at its midpoint. The fundamental frequency of the longitudinal vibrations of the rod is 2.53 kHz. At what speed will the sound be able to travel through steel?

Given, Length , l = 200 cm = 2 m Fundamental frequency of vibration, ν F = 2.53 kHz = 2.53 × 10 3 Hz

The bar is then plucked at its midpoint, forming an antinode (A) at its centre and nodes (N) at its two edges, as depicted in the figure below:

The distance between two successive nodes is   λ / 2 => l =  λ / 2 Or,   λ = 2 x 2 = 4m

Thus, sound travels through steel at a speed of v = νλ v = 4 x 2.53 x 10 3 =10.12 km/s. Q17. One end of A 20 cm long tube is closed. Find the harmonic mode of the tube that will be resonantly excited by a source of frequency 430 Hz. If both ends are open, can the same source still produce resonance in the tube? (Sound travels in air at 340 m /s).

Given, Length of the pipe, l = 20 cm = 0.2 m Frequency of the source = n th the normal mode of frequency, ν N = 430 Hz Speed of sound, v = 340 ms -1

We know that in a closed pipe, the n th normal mode of frequency v N   = ( 2n -1 )v/4l where n is an integer = 0, 1, 2, 3, 4, . . . . .

430 = ( 2n – 1 )( 340/4 x0.2) 2n = 2.01 n ≈ 1 Thus, the given source resonantly excites the first mode of vibration frequency

Now, for a pipe open at both ends, the n th mode of vibration frequency: V R = nv/2l n = V R 2l/v n = (2 x 0.2 x 430)/340 = 0.5 As the mode of vibration (n) has to be an integer, this source is not in resonance with the tube.

Q18. Guitar strings X and Y striking the note ‘Ga’ are a little out of tune and give beats at 6 Hz. When the string X is slightly loosened, and the beat frequency becomes 3 Hz. Given that the original frequency of X is 324 Hz, find the frequency of Y.

Given, Frequency of  X, f X = 324 Hz Frequency of  Y = f Y Beat’s frequency, n = 6 Hz Also, n = \(\begin{array}{l}\left | f_{X}\pm f_{Y} \right |\end{array} \) 6 = \(\begin{array}{l}324 \pm f_{Y}\end{array} \) => f Y = 330 Hz or 318 Hz As frequency drops with a decrease in tension in the string, thus f Y cannot be 330 Hz => f Y = 318 Hz.

Q19. Explain how: (i) A sound wave’s pressure antinode is a displacement node and vice versa. (ii) The Ganges river dolphin, despite being blind, can manoeuvre and swim around obstacles and hunt down prey. (iii) A guitar note and violin note are being played at the same frequency, however, we can still make out which instrument is producing which note. (iv) Both transverse and longitudinal waves can propagate through solids, but only longitudinal waves can move through gases. (v) In a dispersive medium, the shape of a pulse propagating through it gets distorted.

(i) An antinode is a point where pressure is the minimum, and the amplitude of vibration is the maximum. On the other hand, a node is a point where pressure is the maximum, and the amplitude of vibration is the minimum. (ii) The Ganges river dolphin sends out click noises which return back as a vibration, informing the dolphin about the location and distances of objects in front of it. Thus, allowing it to manoeuvre and hunt down prey with minimum vision. (iii) The guitar and the violin produce overtones of different strengths. Thus, one can differentiate between the notes coming from a guitar and a violin, even if they are vibrating at the same frequencies. (iv) Both solids and fluids have a bulk modulus of elasticity. Thus, they both allow longitudinal waves to propagate through them. However, unlike solids, gases do not have a shear modulus. Thus, transverse waves cannot pass through gases. (v) A pulse is a combination of waves of un-similar wavelengths. These waves move at different velocities in a dispersive medium. This causes distortion in its shape.

Q20.A train, standing at the outer signal of a railway station, blows a whistle of frequency 400 Hz in still air. (i) What is the frequency of the whistle for a platform observer  when the train (a) approaches the platform with a speed of 10 m s –1 , (b) recedes  from the platform with a speed of 10 m s –1 ? (ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 m s –1 .

Frequency of the whistle = 400 Hz

Speed of sound in still air = 340 m/s

(a) Train approaches the platform at speed, v s = 10 m/s

Frequency of the whistle for a platform observer

(b) Train recedes from the platform at speed, v s = 10 m/s

(ii) The speed of the sound will not change. It will remain at 340 m/s.

Q21. A train, standing in a station yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a  speed of 10 m s –1 . What are the frequency, wavelength, and speed of sound for an  observer standing on the station’s platform? Is the situation exactly identical to  the case when the air is still, and the observer runs towards the yard at a speed of 10 m s –1 ? The speed of sound in still air can be taken as 340 m s –1 .

Speed of wind, v w = 10 m/s

Speed of sound in still air,v= 340 m/s

The effective speed of the sound for an observer standing on the platform

v’ = v + v w  = 340 + 10 = 350 m/s

There is no relative motion between the source and the observer; therefore, the frequency of the sound heard by the observer will be the same.

Therefore, f = 400 Hz

Wavelength of the sound heard by the observer = speed of wave/frequency = 350/400 = 0.875 m

When the air is still, and the observer runs towards the yard at a speed of 10 m s –1, then there is a relative motion between the observer and the source with respect to the medium.

The medium is at rest. Therefore,

v’ = v = 340 m/s

The change in frequency, f ‘ =[ (v + v 0 )/ v ] x f

= [(340 + 10)/340] x 400 = 411.76 Hz

Wavelength = 340/411.76 =  0.826  m

Obviously, the situations in the two cases are entirely different.

Q22. A travelling harmonic wave on a string is described by  y(x, t) = 7.5 sin (0.0050x +12t + π/4). (a) What are the displacement and velocity of oscillation of a point at  x = 1 cm and t = 1 s? Is this velocity equal to the velocity of wave propagation? (b) Locate the points of the string which have the same transverse displacements and velocity as the x = 1 cm point at t = 2 s, 5 s and 11 s.

(a) The travelling harmonic wave is y(x, t) = 7.5 sin (0.0050x +12t + π/4)

At x = 1 cm and t = 1 s

y (1,1) = 7.5 sin (0.0050 (1)  +12 (1) + π/4)

= 7.5 sin (12.0050 + π/4)  ——-(1)

= 7.5 sin θ

θ = (12.0050 + π/4) x (180/3.14)

= 12.0050 + (3.14/4) = 12.79 x (180/3.14)

θ  = 733.18 0

y (1,1) = 7.5 sin (733.18 0 )

= 7.5 sin (90 x 8 + 13.18 0 )

= 7.5 sin (13.18 0 )

= 7.5 x 0.228

The velocity of oscillation,

θ = (12.005 + π/4)

= (12.005 + π/4) x 180/π

= 12.79 x 180/3.14

θ = 12.79 x 57.32 = 733.18

v = 90 cos (733.18)

= 90 cos (720 + 13.18)

= 90 cos 13.18

= 90 x 0.97 cm/s

= 87.63 cm/s

The standard equation is given as

We get  \(\begin{array}{l}y(x,t)= asin(kx +\omega t+\phi )\end{array} \)

here, k = 2π/λ

and ω = 2πv

Comparing the given equation with the standard equation,

ω = 12 rad/s

⇒ 2πf = 12 rad/s

Therefore, f = 12/2π = 6/π

k = 0.0050 cm -1

2π/λ = 0.0050 cm -1

⇒λ = 2π/0.0050 = 400 π

The velocity of the wave propagation, v = fλ

= (6/π) x 400 π = 2400 cm/s

Velocity at x = 1 cm and t = 1s is not equal to the velocity of wave propagation

(b) Propagation constant , k=2π/λ

here λ is the wavelength

⇒ λ=2π/k=(2×3.14)/0.0050 =1256cm=12.56 m

All the points at a distance ±λ,±2λ,… from x = 1 cm will have the same transverse displacement and velocity.  As λ = 12.56 m, the points ±12.56m,±25.12m,… and so on for x = 1 cm, will have the same displacement as the x= 1 cm points at t = 2s, 5s and 11s.

Q23. A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) Does the pulse have a definite (i) frequency, (ii) wavelength, and (iii) speed of propagation? (b) If the pulse rate is 1 after every 20 s (that is, the whistle is blown for a split of seconds after every 20 s), is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz?

(a) The speed of propagation is definite, it is equal to the speed of the sound in air. The wavelength and frequency will not be definite.

(b) The frequency of the note produced by a whistle is not 1/20 = 0.05 Hz. However, 0.05 Hz is the frequency of repetition of the short pip of the whistle.

Q24. One end of a long string of linear mass density 8.0 × 10 –3 kg m –1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacements (y = 0) and is moving along the positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as a function of x and t that describes the wave on the string.

Linear mass density of the string, μ= 8.0 × 10 –3 kg m –1

Frequency of the tuning fork = 256 Hz

Mass on the pan = 90 kg

Tension on the string, T = 90 x 9.8 = 882 N

Amplitude, A = 0.05 m

For a transverse wave, the velocity is given by the relation.

Angular frequency, ω = 2πf

= 2 x 3.14 x 256 = 1608.5 rad/sec

Wavelength, λ = v/f = 332/256 = 1.296 m

Propagation constant, k = 2π/λ  = (2 x 3.14)/ 1.296

= 4.845 m -1 ​The general equation of the wave is

y (x,t) = A sin (ωt – kx)

Substituting all the values, we get

y(x,t) = A sin (1608.5t -4.845x)

x and y are in metres, and t is in seconds.

Q25. A SONAR system fixed in a submarine operates at a frequency of 40.0 kHz. An enemy  submarine moves towards the SONAR with a speed of 360 km h –1 . What is the  frequency of sound reflected by the submarine? Take the speed of sound in water  to be 1450 m s –1 .

Frequency of the SONAR system, f=40 kHz = 40 x 10 3 Hz

Speed of sound in water, v =1450m/s Speed of the enemy submarine, v 0 =360km/h= 360 x (5/18) = 100m/s

The SONAR is at rest, and the energy submarine moves towards it. Therefore, the apparent frequency is given by the relation f’ = [(v+v 0 )/v]f

=  [(1450 + 100)/1450] x 40 x 10 3

= 42.75 x 10 3 This frequency (f’) is reflected by the enemy submarine, and it is observed by the SONAR. Therefore, v s = 360 km/s  = 100m/s

f” = (v/(v – v s )) x f’

= (1450/(1450 – 100)) x 42.75 x 10 3

= (1450/1350) x 42.75 x 10 3

= 45.91 x 10 3 .

Q26.​ Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of the S wave is about 4.0 km s –1 , and that of the P wave is 8.0 km s –1 . A seismograph  records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in a straight line, at what distance does the  earthquake occur?

Let the speeds of S and P be v 1 and v 2, respectively. The time taken by the S and P waves to reach the position of the seismograph is t 1 and t 2, respectively

l = v 1 t 1 = v 2 t 2

The speed of S wave, v 1 =4.0 km s –1

The speed of P wave, v 2 = 8.0 km s –1

4t 1 = 8t 2

The first P wave arrives 4 min before the S wave.

t 1 – t 2 = 4 min = 4 x 60 s = 240 s

2t 2  – t 2 = 240 s

t 2 = 240 s

t 1 = 2t 2 = 2 x 240 = 480 s

Therefore, the distance at which the earthquake occurs, l = v 1 t 1 = 4 x 480 = 1920 km.

Q27. A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the  sound emission frequency of the bat is 40 kHz. During one fast swoop directly  toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in the air.  What frequency does the bat hear reflected off the wall?

The sound emission frequency of the bat = 40 kHz

The velocity of the bat, v b = 0.03 v

Where v is the velocity of the sound in the air

The apparent frequency of the sound hitting the wall

= 40/0.97 kHz

The frequency gets reflected by the wall and is received by the bat moving towards the wall

= (1.03 x 40 /0.97) = 42.47 kHz.

Q28. A man standing at a certain distance from an observer blows a horn of frequency 200 Hz in still air. (a) Find the horn’s frequency for the observer when the man (i) runs towards him at 20 m/s  (ii) runs away from him at 20 m /s. (b) Find the speed of sound in both cases. [Speed of sound in still air is 340 m/ s]

Given, Frequency of the horn, ν H = 200 Hz Velocity of the man, v T = 20 m/ s Velocity of sound, v = 340 m/ s

(a) We know, (i) The apparent frequency of the horn as the man approaches the observer is: v’ = v H [ v/(v – v T ) ] = 200 [ 340 /(340 – 20) ] = 212.5 Hz. (ii) The apparent frequency of the horn as the man runs away from  the observer is: v’’ = v H [ v/(v + v T ) ] = 200 [ 340 /(340 + 20) ] = 188.88 Hz. (b) The speed of sound is 340 m/s in both cases. The apparent change in frequency is a result of the relative motions of the observer and the source.

Q29. A truck parked outside a petrol pump blows a horn of frequency 200 Hz in still air.  The wind then starts blowing towards the petrol pump at 20 m/s. Calculate the wavelength, speed, and frequency of the horn’s sound for a man standing at the petrol pump. Is this situation completely identical to a situation when the observer moves towards the truck at 20 m/s and the air is still?

For the standing observer: Frequency, ν H = 200 Hz Velocity of sound, v = 340 m/s Speed of the wind, v W = 20 m/s

The observer will hear the horn at 200 Hz itself because there is no relative motion between the observer and the truck.

Given that the wind blows in the observer’s direction at 20 m/s. Effective velocity of the sound, v E = 340 + 20 = 360 m/s

The wavelength ( λ ) of the sound : λ = v E /v H  = 360/200 λ = 1.8 m

For the observer running towards the train: Speed of the observer, v o = 20 m/s

We know, The apparent  frequency of the sound as the observers move towards the truck is: v’ =   v H [(v + v o )/v ] =  200[ (20 + 340 )/ 340 ] =  211.764 Hz

As the air is still, the effective velocity of sound is still 340 m/s. As the truck is stationary, the wavelength remains 1.8 m.

Thus, the two cases are not completely identical.

These solutions consist of answers to extra questions prepared by subject experts at BYJU’S along with exemplary problems, worksheets, short and long answer questions, MCQs, Tips and tricks to prepare for CBSE exams with NCERT Solutions .

This chapter comprises comprehensive questions and solutions on a very important topic of Physics, such as questions on wave dynamics. Questions from this chapter repeatedly appear in exams and will guide you through every topic and type of wave, such as tension on strings, the speed of sound in air, transverse waves and dependence of the speed of sound in the air on factors like pressure, humidity and temperature. This chapter also has questions on concepts like the wavelength of ultrasonic sound, transverse harmonic wave and their frequency, etc.

Class 11 Physics NCERT Solutions for  Chapter 15 Waves

In the NCERT Solutions for Class 11 Physics, we have taken a wire as a reference for signifying a wave motion, and we have talked about the transverse displacement of a wire which is clamped on both sides and in which, we will be talking about finding the amplitude of a point at a certain distance in the wire. We also have questions regarding calculating the speed of sound in a wire with pistons at the end and the other end in resonance with a tuning fork. These topics are very important for competitive exams. Also, waves are one of the important topics in Physics and are repeatedly asked in every sort of exam. You can check out NCERT Solutions for Class 11 Physics for chapter-wise solutions here.

We can gain a lot of knowledge about waves by solving questions on the speed of sound in a steel medium. There are questions on how your guitar strings produce sounds and what kind of frequency each string of guitar has. We get the explanation on how a sound wave’s pressure antinode is a displacement node and vice versa. We will be seeing how a dolphin, being blind, can manoeuvre through obstacles in a river and hunt prey. If a man is standing at a certain distance from an observer and he blows a horn, we can get to know the horn’s frequency depending on whether the man is running towards or away from the observer. A similar example can be found in a truck blowing the horn at a man in a petrol pump, finding its frequency, speed and wavelength.

Waves characteristics comprise the frequency, amplitude, wavelength, phase, resonance, and displacement of waves, such as longitudinal and transverse motion in various mediums like materials with high and low density, air, steel wires in guitars, sea waves, etc.

Subtopics of Class 11 Physics Chapter 15 Waves

• Introduction
• Transverse and longitudinal waves
• Displacement relation in a progressive wave
• The speed of a travelling wave
• The principle of superposition of waves
• Reflection of waves
• Doppler effect

BYJU’S provides chapter-wise NCERT Solutions for all the classes to help the students prepare well for the annual examination. All the solutions in this study material are explained in simple language and in an interactive manner for students to learn better and effectively.

These NCERT Solutions for Class 11 Physics, Chapter 15, Waves, provide complete information regarding the topic along with the definitions and examples. These NCERT Solutions are available in downloadable PDF for free. Students can download the BYJU’S – The Learning App to avail effective and interactive lessons of CBSE Class 11 subjects.

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Question and Answer forum for K12 Students

MCQ Questions for Class 11 Physics Chapter 6 Work, Energy and Power with Answers

We have compiled the NCERT MCQ Questions for Class 11 Physics Chapter 6 Work, Energy and Power with Answers Pdf free download covering the entire syllabus. Practice MCQ Questions for Class 11 Physics with Answers on a daily basis and score well in exams. Refer to the Work, Energy and Power Class 11 MCQs Questions with Answers here along with a detailed explanation.

Work, Energy and Power Class 11 MCQs Questions with Answers

Multiple Choice Type Questions

Question 1. A bullet is fired horizontally and gets embedded in a block kept on a table. If the table is frictionless, then (a) only momentum is conserved. (b) only potential energy is conserved. (c) only K.E. is conserved. (d) both (a) and (b).

Answer: (d) both (a) and (b).

Question 2. A shell is fired from a canon with a velocity v ms -1 at an angle 6 with the horizontal direction. At the highest point in its path, it explodes into two pieces of equal mass. One of the pieces retraces its path to the canon and the speed of the other piece (in ms -1 ) just after the explosion is (a) v cos θ (b) 3v cos θ (c) 2v cos θ (d) \(\frac {v}{2}\) cos θ

Answer: (b) 3v cos θ

Question 3. The work performed on an object does not depend upon (a) the displacement. (b) the force applied. (c) the angle at which the force is applied to the displacement (d) initial velocity of the object.

Answer: (d) initial velocity of the object.

Question 4. A ball is dropped from a height of 15 m. It gets embedded in sand by 10 mm and then stops. Which of the following is conserved? (a) Temperature (b) Momentum (c) Kinetic energy (d) Both (a) and (c)

Answer: (b) Momentum

Question 5. A chain of length L and mass M is held on a smooth table with its \(\frac {1}{n}\) th part hanging over the edge. The work done in pulling the chain is directly proportional to (a) n -3 (b) √n (c) n (d) n -2

Answer: (d) n -2

Question 6. A pump delivers water at the rate of V cubic meter per second. By what factor its power should be raised so that it delivers water at the rate of nV cubic metre per second. (a) n³ (b) √n (c) n (d) n²

Answer: (a) n³

Question 7. Liquid of density ρ flows along a horizontal pipe of uniform area of cross-section ‘a’ with a velocity v through a right angled bend. What force should be applied to the bend to hold it in equilibrium? (a) \(\frac {av^2ρ}{2}\) (b) \(\frac {av^2ρ}{√2}\) (c) 2av²ρ (d) √2 av²ρ

Answer: (c) 2av²ρ

Question 8. A body of mass M accelerates uniformly from rest to a velocity v in time t. What is the instantaneous power delivered to the body at time T. (a) \(\frac {mv}{t}\)T (b) \(\frac {mv^2}{t}\)T (c) \(\frac {m^2v^2}{t^2}\)T (d) \(\frac {mv^2}{t^2}\)T

Answer: (d) \(\frac {mv^2}{t^2}\)T

Question 9. A man weighing 50 kgf carries a load of 10 kgf to the top of the building in 5 minutes. The work done by him is 10 5 J. If he carries the same load in 10 minutes, the work done by him will be: (a) 10 5 J (b) 5 × 10 5 J (c) 12 × 10 5 J (d) 2.5 × 10 5 J

Answer: (a) 10 5 J

Question 10. If v be the instantaneous velocity of a body dropped from the top of a tower, when it is located at a height h, then which of the following remains constant? (a) gh + \(\frac {1}{2}\) v² (b) gh – \(\frac {1}{2}\) v² (c) gh + v² (d) gh – v²

Answer: (a) gh + \(\frac {1}{2}\) v²

Question 11. A ball B 1 of mass m is moving with a velocity v along north. It collides with another ball B 2 of same mass moving with a velocity v along east. After the collision, both the balls stick together and move along north east. The velocity of the combination is (a) \(\frac {v}{√2}\) (b) √2v (c) 2v (d) v

Answer: (a) \(\frac {v}{√2}\)

Question 12. A ball B 1 of mass m moving with velocity u, collides head on with another ball B 2 of the same mass at rest. Given that the coefficient of restitution is e. Then the ratio of velocities of two balls after collision will be (a) \(\frac {1+e}{1-e}\) (b) e (c) \(\frac {1+e}{2}\) (d) \(\frac {1-e}{2}\)

Answer: (a) \(\frac {1+e}{1-e}\)

Question 13. A vehicle of mass m is moving on a rough horizontal road with a momentum p. If the coefficient of friction between the tyres and the road be µ, then the stopping distance is given by (a) \(\frac {p}{2µm^2g}\) (b) \(\frac {p}{2µmg}\) (c) \(\frac {p^2}{2µmg}\) (d) \(\frac {p^2}{2µm^2g}\)

Answer: (d) \(\frac {p^2}{2µm^2g}\)

Question 14. A uniform chain of length l and mass m is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g be the acceleration due to gravity, then the work required to pull the hanging part on the table is : (a) \(\frac {mg l}{3}\) (b) \(\frac {mgl}{9}\) (c) \(\frac {mgl}{18}\) (d) mg l

Answer: (c) \(\frac {mgl}{18}\)

Answer: (d) \(\frac {2mg(h+x)}{x^2}\)

Question 16. The slope of the potential energy versus position vector gives (a) momentum (b) force (c) work done (d) power

Answer: (b) force

Question 17. The slope of the kinetic energy versus position vector gives the time rate of change of: (a) momentum (b) force (c) work done (d) power

Answer: (a) momentum

Question 18. A bus weighing 100 quintals moves on a rough road with a constant speed of 72 kmh -1 . The friction of the road is 9% of its weight and that of air is 1 % of its weight. What is the power of the engine? Take g = 10 ms -2 . (a) 100 kW (b) 150 kW (c) 200 kW (d) 50 kW

Answer: (c) 200 kW

Question 19. A metre rod is pivoted at its end and stands vertically. If it is displaced through 60° with the vertical, what will be the ratio of its potential energy in this position to its maximum potential energy? (a) 0.25 (b) 0.75 (c) 0.80 (d) 0.50

Answer: (d) 0.50

Question 20. A bullet is fired into a block of sand and its velocity decreases by 50% when it penetrates through 9 cm. What will be the total distance penetrated by the bullet? (a) 9 cm (b) 10 cm (c) 12 cm (d) 18 cm

Answer: (c) 12 cm

Fill in the blanks

Question 1. When a body is dropped from a certain height on the ground, the work is done on the body ……………. and is ……………….

Answer: by the force of gravity, positive

Question 2. When it is lifted from the ground to a certain height, the work is done on the body ……………. and is ……………

Answer: against the force of gravity, negative

Question 3. The statement that inertial mass of a body is equal to the gravitational mass is called ……………

Answer: principle of equivalence

Question 4. When a vertical force equal to the weight of a body is applied to it, then its acceleration is ……………..

Answer: zero.

Question 5. When work done is zero, then the speed of a body is ……………..

Answer: uniform

Question 6. Coefficient of restitution is one for ……………… collision.

Answer: perfectly elastic collision

Question 7. The coefficient of restitution is zero (e = 0) for ……………… collision which means that ……………….. of the colliding body is lost and is changed to other forms like ………………. or …………………..

Answer: perfectly inelastic, kinetic energy, sound energy, or heat energy

Question 8. The mass of a body increases with the increase in the velocity of the particle according to the relation ………………..

Answer: m = \(\frac {m_0}{\sqrt{1-\frac{v^2}{c^2}}}\)

Question 9. Mass of the particle becomes infinite if it moves with the ……………….. and the acceleration produced in the body by a given force F becomes ………………

Answer: velocity of light c = 3 × 10 8 ms -1 , zero (∵ a = \(\frac {F}{m}\) = \(\frac {F}{∞}\) = 0)

Question 10. E = mc² shows that mass and energy are …………….

Answer: inter convertible

Energy is transferred from one form to another. Name the transformations of energy in the following in the form ……………….. to ……………….. Question 11. lifting of a book by a boy ………………..

Answer: Lifting of a book by a boy transforms energy from chemical to gravitational.

Question 12. burning of candle ………………..

Answer: Burning of candle transforms energy from chemical to light and heat.

Question 13. a moving truck on a road ………………..

Answer: A moving truck on a road transforms energy from chemical to kinetic energy of motion.

Question 14. emission of light by Sun ………………..

Answer: Emission of light by Sun transforms energy from nuclear to electromagnetic energy.

Question 15. The work done against friction is ……………….. proportional to the force with which the surfaces are pressed together.

Answer: directly

Question 16. If the momentum of a body is doubled then its K.E. becomes ………………..

Answer: four times

Question 17. The sum of K.E. and P.E. for an isolated system is always ………………..

Answer: constant

Question 18. If the range of a projectile be R, then its K.E. is minimum when horizontal distance covered by it is ………………..

Answer: \(\frac {R}{2}\)

Question 19. ………………. is not conserved during an inelastic collision.

Answer: K.E.

Question 20. When a bullet striking a block gets embedded in it, then the nature of collision is ……………….

Answer: perfectly inelastic

Question 21. ………………. is a non-conservative force.

Answer: friction

Question 22. For a perfectly inelastic collision, the coefficient of restitution is ……………….

Answer: zero

Question 23. The work done …………….. updn the time taken in doing it.

Answer: does not depend

Question 24. The work done by the conservative force 1s always ……………..

Answer: positive

Question 25. The P.E. gained by the body raised from the surface of Earth by a height equal to radius of Earth is ……………..

Answer: mg \(\frac {R}{2}\)

Question 26. K.E. of a projectile is …………….. at the highest point of its trajectory.

Answer: minimum

Question 27. The work done on a body by a resultant external force is always ………………. to the change in its K.E.

Answer: equal

Question 28. A body is suspended by vertical string. The work done on the body by the tension in the string is …………….

Question 29. Power is ………………. of force and velocity.

Answer: scalar product

Question 30. Work is ………………. of force and displacement.

Question 31. 1 kwh is the energy consumed by an appliance of power I kw in ………….

Answer: 1 hour

Question 32. The work done by a conservative force moving an object on a closed path is …………….

Question 33. In a perfectly inelastic collision, the two bodies ………………… after collision.

Answer: stick together

Question 34. For an inelastic collision, the coefficient of restitution lies between ………………

Question 35. Kinetic energy increases during a ……………….

Answer: super elastic collision

Question 36. ………………. energy results from the chemical bounding between the atoms.

Answer: Chemical

Question 37. ………………. energy results due to the separation between two objects in the gravitational field.

Answer: Gravitational

Question 38. ………………… energy between two nucleons is due to nuclear force.

Answer: Nuclear

Question 39. Chemical, gravitational and nuclear energies are the ……………….. for different types of forces in nature.

Answer: Potential energies

True/False Type Questions

1. Which of the following statement is True/False? (a) Work done can be positive, negative and zero. (b) A teacher sitting in a chair is dong work. (c) Work is the cross product of force and displacement. (d) Work done by a coolie in carrying a bag on his head is zero.

Answer: (a) True (b) False (c) False (d) True

2. Which of the following statement is True/False? (a) The principle of conservation of linear momentum can be strictly applied during a collision between two particles provided the time of impact is extremely small. (b) In an inelastic collision, the linear momentum is conserved but not the kinetic energy. (c) The collision is said to be elastic if two bodies stick together after collision. (d) The unit of energy is increased by 16 times if the unit of force and length be each increased by 4 times.

Answer: (a) True (b) True (c) False (d) True

3. Tell which of the following statement is True/False? (a) Momentum is conserved in elastic collision but not in inelastic collision. (b) Total kinetic energy is conserved in elastic collision but momentum is not conserved in elastic collision. (c) Both K.E. and momentum are conserved in all types of collision. (d) Total energy, K.E. and the momentum is conserved in elastic collision.

Answer: (a) False (b) False (c) False (d) True

4. Which of the following statement is True/False? (a) Both momentum and energy are conserved in an inelastic collision. (b) Neither momentum nor energy is conserved in an inelastic collision. (c) Momentum is conserved but not the K.E. in an inelastic collision. (d) Momentum is not conserved but the K.E. is conserved in an inelastic collision.

Answer: (a) False (b) False (c) True (d) False

5. Which of the following statement is True/False? There will be an increase in potential energy of the system if work is done upon the system by (a) a conservative force. (b) a non-conservative force. (c) any conservative or non-conservative force.

Answer: (a) True (b) False (c) False

6. Which of the following statement is True/False? (a) The gain in the P.E. of an object of mass m raised from the surface of earth to a height equal to the radius R is \(\frac {mgR}{2}\). (b) 1 eV = 1.6 × 10 -19 J (c) Power is the cross-product of force and velocity. (d) Work is the dot product of force and displacement.

7. Which of the following statement is True/False? (a) A teacher rubbing the blockboard by a duster is doing no work. (b) A light and a heavy body have equal K.E. of translation. The heavier body has larger momentum. (c) A bullet is fired from a rifle. If the rifle recoils freely, then the K.E. of the rifle is greater than that of the bullet. (d) The energy produced when 10 gram of coal is burnt is 9 × 10 4 J.

8. Tell which one of the following statement is True/False? (a) A truck and car having same K.E. cover equal distances before stopping after applying equal retarding force through brakes. (b) 1 kg m = 9.8 J (c) 1 J = 10 7 erg (d) For an elastic collision, the velocity of separation after collision is always equal to the velocity of approach before collision.

Answer: (a) True (b) True (c) True (d) True

9. Tell which of the following statement is True/False? (a) Friction is non-conservative force. (b) Gravity is a conservative force. (c) No work is done against gravity while moving an object along horizontal. (d) Kilowatt-hour is the unit of energy.

10. Tell which of the following statements are True/False? (a) The forces involved during the elastic collision are conservative in nature. (b) K.E. lost in an inelastic collision appears in some other form of energy such as heat, sound etc. (c) Mechanical energy is not converted into any other form of energy in an elastic collision. (d) Coefficient of restitution is zero (e = 0) for a perfectly inelastic collosion.

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