case study questions in physics class 11

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Class 11 Physics Case Study Question

Case study question class 11 physics (cbse / ncert board).

Class 11 Physics Case Study Question and Answer: CBSE / NCERT Board Class 11 Physics Case Study Question prepared by expert Physics Teacher. Students can learn Case Based Question / Paragraph Type Question for NCERT Class 11 Physics.

There are total 15 chapter Physical World, Units and Measurement, Motion in a Straight Line, Motion in a Plane, Laws of Motion, Work, Energy and Power, Systems of Particles and Rotational Motion, Gravitation, Mechanical Properties of Solids, Mechanical Properties of Fluids, Thermal Properties of Matter, Thermodynamics, Kinetic Theory, Oscillations, Waves.

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CBSE Class 11 Physics Case Study Question

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 Systems 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

What is Case Study Question?

Ans. At case Study there will one paragraph and on the basis of that concept some question will made. Students have to solve that question.

How many marks will have at case based question?

Most of time 5 questions will made from each case. There will 1 or 2 marks for each question.

case study questions in physics class 11

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Class 11 Physics Sample Papers, Notes | NCERT Solutions

Get our premium content for CBSE class 11 Physics in the form of Sample Question Papers, Notes, Exemplar Solutions, NCERT Solutions & Question Bank as PDF tailor-made for the toppers.

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Cbse last year papers, user submitted papers, case study questions, mock tests, units and measurements, motion in a straight line, motion in a plane, laws of motion, work energy and power, system of particles and rotational motion, gravitation, mechanical properties of solids, mechanical properties of fluids, thermal properties of matter, thermodynamics, kinetic theory, oscillations.

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CBSE class 11 Physics solved sample question papers, NCERT textbook answers, revision notes, and guess papers for free download, online practice tests in MCQ format for CBSE, IIT JEE, AIPMT, NEET, and other entrance exams along with high-quality free videos.

CBSE Class 11 Physics Study Material

There is a transition from general education to a discipline-based focus on the curriculum at the senior secondary stage of school education. Hence, young learners need to pay extra attention to the preparation for their board examinations. For CBSE Class 11 Physics quality study material like NCERT exemplar solutions, the latest sample question papers, test papers, and chapter-wise quick revision notes students can check our website . Here you can also find learning videos for class 11 Physics for a better understanding of the chapter.

CBSE Class 11 Physics Syllabus

Physics as a subject in the senior secondary stage of school education aims to develop conceptual competence in the learners and make them realize and appreciate the interface of Physics with other disciplines. Therefore, it becomes very important for the students at this stage to prepare vigorously in concurrence with the updated syllabus. Those who are looking for a reliable stop to get everything for Class 11 Physics like syllabus, sample papers, notes, and NCERT Solutions may download the myCBSEguide App .

A syllabus is subject to change. This change may not be very big but there are small changes that are mostly accommodated in the syllabus. That is the reason students must know their syllabus well. Given below are the chapter names of the present CBSE Class 11 Physics syllabus.

COURSE STRUCTURE

Class XI – 2022-23 (Theory)
Unit–I Physical World and Measurement
Unit-II Kinematics
Unit–III Laws of Motion
Unit–IV Work, Energy, and Power
Unit–V Motion of System of Particles and Rigid Body
Unit-VI Gravitation
Unit–VII Properties of Bulk Matter
Unit–VIII Thermodynamics
Unit–IX Behaviour of Perfect Gases and Kinetic Theory of Gases
Unit–X Oscillations and Waves

The CBSE senior secondary Physics basically aims to expose the learners to different processes used in Physics-related industrial and technological applications. Therefore, the topics and chapters have been thoughtfully chosen. For detailed information on the syllabus, you can check Class 11 Physics Syllabus 2023 .

CBSE Class 11 Physics Revision Notes

It is impossible to get good marks without understanding the key concepts properly. Physics is a subject of theories and formulae, one has to know them well in order to answer the questions based on them. For this revision notes become important. You can download the revision notes for CBSE Class 11 Physics from myCBSEguide which has explanations of important topics in layman’s language. Students can make the best use of these notes for their preparations. The Physics revision notes are helpful in understanding the concepts easily.

CBSE Class11 Physics NCERT Exemplar Solutions

A complete knowledge of NCERT is a must for any CBSE subject. Students of senior secondary science need to focus more on the NCERT textbook solutions and Exemplar Solutions. The best part is our class NCERT solutions are available free of cost on our students’ dashboard with a simple registration. Get the Class 11 Physics NCERT Solutions with detailed answers. Students can also learn from the video solution for better understanding. The Physics exemplar solutions can also be accessed which are available for download.

CBSE Class 11 Physics Question Bank

There is a dynamic range of questions that the students need to be aware of. These questions can be multiple choice questions, case study questions, competency-based objective questions, and description subjective questions. The question bank of myCBSEguide is very rich in this matter and has numerous questions for every type. To get the best and latest question type for CBSE Class 11 Physics you must check our website. Our question bank is revived even and accommodates even a subtle change in the syllabus.

CBSE Class 11 Physics Sample Papers

Based on the CBSE question paper pattern, the myCBSEguide model question papers are one of our top-selling products. Students are advised to solve as many sample papers as they can because sample papers reflect the original question paper. Practicing sample papers drives away stress and boosts the confidence of the students. The class 11 Physics sample papers with the latest blueprint are now available on the myCBSEguide app. For more information check out Class 11 Physics Sample Paper 2023 .

Class 11 Physics Previous Year Questions & Practice Papers

You can find previous years’ question papers for class 11 Physics on our student dashboard absolutely free. In fact, you can check CBSE Previous Years Paper for all the subjects here. These are the most important question for senior secondary students from the board exam perspective.

The CBSE has released practice papers for the main subjects. Students who would be sitting for the upcoming boards must thoroughly revise these papers before sitting for their exams. You can find the class 11 Physics Practice papers 2023 and begin your preparation right away.

CBSE Class 11 Physics Case Study Questions

The senior secondary Physics curriculum is designed to promote problem-solving abilities and creative thinking in learners. The CBSE has incorporated the concept of competency-based education (CBE) by including a new model of questions called Case-based questions or simply case study questions. Our CBSE Class 11 Physics case study question has been prepared by senior CBSE teachers and is sure to give efficient outcomes.

Our Class 11 Physics Case Study Questions will help to develop conceptual competence in the learners and make them realize and appreciate the interface of Physics with other disciplines.

CBSE Class 11 Physics Test Papers & Online Tests

The chapter-based test can be a good way to achieve small targets. The test papers of class 11 Physics help in providing logical sequencing of units of the subject matter and proper placement of concepts with their linkage for better learning.

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Class 11 Physics Case Study Questions

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CBSE Class 11 Physics Syllabus 2022-23

CBSE Class 11 Physics Syllabus 2022-23 includes Units and Measurements, Motion in a Straight Line, Work, Energy and Power, System of Particles and Rotational Motion, Thermal Properties of Matter etc for the session 2022 – 2023. Here is the detailed syllabus. To download class 11 Physics CBSE latest sample question …

CBSE Term-1 MCQ Sample Papers 2021-22

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CBSE to Conduct Two Term Exams in 2021-22

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CBSE, New Delhi has reduced the syllabus for classes 9 to 12 by 30%. The CBSE Revised Syllabus 2020-21 is available in CBSE official website and myCBSEguide mobile app. Here is the complete circular regarding CBSE Revised Syllabus 2020-21: CBSE Reduced the Syllabus The prevailing health emergency in the country …

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CBSE Sample Papers Class 11 Physics 2024

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Important Questions for CBSE Class 11 Physics Chapter 15 - Waves

CBSE Class 11 Physics Chapter-15 Important Questions - Free PDF Download

Free PDF download of Important Questions with solutions for CBSE Class 11 Physics Chapter 15 - Waves prepared by expert Physics teachers from latest edition of CBSE(NCERT) books . Register online for Physics tuition on Vedantu.com to score more marks in your Examination.

Download CBSE Class 11 Physics Important Questions 2024-25 PDF

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

Study Important Questions for Class 11 Physics Chapter 15 – Waves

Very Short Answer Type Questions (1 Mark)

1. Explosions on other planets are not heard on earth. Why?

Ans: Explosions on other planets are not heard on earth because there is no material medium between the earth and the planets over a long distance, and without a material medium for propagation, sound waves cannot travel.

2. Why longitudinal waves are called pressure waves?

Ans: Longitudinal waves are called pressure waves because the propagation of longitudinal waves through a medium consists of the variations in the volume and the pressure of the air, these variations in volume and air pressure result in the formation of compressions and rarefactions.

3. Why do tuning forks have two prongs?

Ans: The tuning fork has two prongs because the two prongs of a tuning fork produce resonant vibrations that help to keep the vibrations going for longer.

4. Velocity of sound increases on a cloudy day. Why?

Ans: Velocity of sound increases on a cloudy day, because the air is wet on a cloudy day, it contains a lot of moisture, the density of air is lower, and because velocity is inversely proportional to density, velocity increases.

5. Sound of maximum intensity is heard successively at an interval of 0.2 second on sounding two tuning fork to gather. What is the difference of frequencies of two tuning forks?

Ans: The beat period is 0.2 second so that the beat frequency is ${\text{fb}} = \dfrac{1}{{0.2}} = 5\;{\text{Hz}}$ Therefore, the difference of frequencies of the two tuning forks is $5{\text{HZ}}$ .

6. If two sound waves have a phase difference of ${60^0}$, then find out the path difference between the two waves?

Ans, Phase difference, $\phi = {{60}^{0}} = \dfrac{\pi }{3}{\text{rad}}$

Now, in general for any phase difference, $(\Phi )$ , the path difference (x) :-

$\phi = \dfrac{{2\pi }}{\pi }x$

Given $\phi = \dfrac{\pi }{3},x = ?$

$\dfrac{\pi }{3} = \dfrac{{2\pi }}{\pi } \times x$

$x = \dfrac{\pi }{3} + \dfrac{{2\pi }}{\pi }$

$x = \dfrac{\pi }{3} \times \dfrac{\pi }{{2\pi }}$

$x = \dfrac{{\pi m}}{6}$

7. If the displacement of two waves at a point is given by: -

${{\text{Y}}_1} = asin{\text{wt}}$

${{\text{Y}}_2} = {\text{an}}\sin \left( {wt + \dfrac{\pi }{2}} \right)$

Calculate the resultant amplitude?

Ans: Given data :

If ${{\text{a}}_1} = $ amplitude of first wave

${a_2} = $ amplitude of second wave

${a_r} = $ resultant amplitude

$\phi = $ phase difference between 2 waves

${{\mathbf{a}}_t} = \sqrt {{a_1}^2 + {a_1}^2 + 2{a_1}{a_2}\cos \phi } $

In our case,

${{\text{a}}_1} = a;{{\text{a}}_2} = a;\phi = \dfrac{\pi }{2}50$

${{\text{a}}_t} = \sqrt {{a^2} + {a^2} + 2a \times a\operatorname{Cos} \left( {\dfrac{\pi }{2}} \right)} $

$ = \sqrt {2{a^2}} \quad \left( {\because \operatorname{Cos} \left( {\dfrac{\pi }{2}} \right) = 0} \right)$

${a_t} = \sqrt {2a} $

8.A hospital uses an ultrasonic scanner to locate tumors in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is $1.7\;{\text{km}}_5^{ - 1}$ ? The operating frequency of the scanner is $4.2{\text{MHz}}$.

Speed of sound in the tissue, $v = 1.7\;{\text{km}}/{\text{s}} = 1.7 \times {10^3}\;{\text{m}}/{\text{s}}$

Operating frequency of the scanner, ${\text{v}} = 4.2{\text{MHz}} = 4.2 \times {10^5}\;{\text{Hz}}$

The wavelength of sound in the tissue is given as:

$\lambda = \dfrac{v}{V}$

$ = \dfrac{{1.7 \times {{10}^2}}}{{4.2 \times {{10}^8}}} = 4.1 \times {10^{ - 4}}\;{\text{m}}$

9. Given below are some functions of $x$ and $t$ to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a traveling wave, (ii) a stationary wave or (iii) none at all:

(a) $y = 2\cos (3x)\sin (10t)$

(b) $y = 2\sqrt {x - 13} $

(d) $y = \cos x\sin t + \cos 2x\sin 2t$

Ans: (a) It is a stationary wave because the harmonic terms are present individually in the equation, the preceding equation indicates a stationary wave $kx$ and $\omega t$ appear separately in the equation.

(b) There is no harmonic term in the provided equation. As a result, it doesn't represent either a moving or stationary wave.

(c) The harmonic terms in the preceding equation describe a moving wave and are in the combination of $kx$ and $\omega t$ are in the combination of $kx - \omega t$ .

(d) The given equation represents a stationary wave because the harmonic terms $kx$ and $\omega t$ appear separately in the equation. This equation actually represents the superposition of two stationary waves.

10. A narrow sound pulse (for example, a short pap by a whistle) is sent across a medium. (a) Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation? (b) if the pulse rate is 1 after every $20\;{\text{s}}$, (that is the whistle is blown for a split of second after every $20\;{\text{s}}$ ), is the frequency of the note produced by the whistle equal to $\dfrac{1}{{20}}$ or $0.05{\text{Hx}}$ ?

Ans: (a) No, the pulse doesn’t have a definite .

(ii) No, It doesn’t have frequency.

(iii) Yes, It have a wavelength.

(b) No, the frequency of the note produced by the whistle is not equal .

Explanation:

(a) There is no defined wavelength or frequency for the narrow sound pulse. However, the sound pulse's speed remains constant, i.e., it is equal to the speed of sound in that medium.

(b) The short pip produced after every 20 s does not mean that the frequency of the whistle is $\dfrac{1}{{20}}$ or $0.05{\text{H}}$ . It means that $0.06{\text{H}}$ s is the frequency of the repetition of the pip of the whistle.

2 Marks Questions

1. A pipe $20\;{\text{cm}}$ long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a $430\;{\text{Hz}}$ source? will this source be in resonance with the pipe if both the ends are open?

Length of pipe $ = {\text{L}} = 20\;{\text{cm}} = 0.2\;{\text{m}}$

Frequency of ${{\text{n}}^{{\text{th }}}}$ node $ = {v_{\text{n}}} = 430\;{\text{Hz}}$

Velocity of sound $ = {\text{v}} = 340\;{\text{m}}\mid {\text{s}}$

Now, ${v_n}$ of closed pipe is :-

${v_n} = \dfrac{{(2n - 1)v}}{{4L}}$

$430 = \dfrac{{(2n - 1) \times 340}}{{4 \times 0.2}}$

$2n - 1 = \dfrac{{430 \times 4 \times 0.2}}{{340}}$

$2n - 1 = 1.02$

$2n = 1.02 + 1$

$2n = 2.02$

Hence, it will be the first normal mode of vibration, ln a pipe, open at both ends We have ,

${v_n} = \dfrac{{H \times {\text{V}}}}{{2L}} = \dfrac{{{\text{A}} \times 340}}{{2 \times 0.2}} = 430\quad {\text{So}},430 = \dfrac{{{\text{A}} \times 340}}{{2 \times 0.2}}$

$n = \dfrac{{430 \times 2 \times 0.2}}{{340}}$

As ${\text{n}}$ has to be an integer, open organ pipe cannot be in resonance with the source.

2. Can beats be produced in two light sources of nearly equal frequencies?

Ans: No, because light emission is a random and fast occurrence, and we obtain uniform intensity instead of beats.

3 . A person deep inside water cannot hear sound waves produces in air. Why?

Ans: Person deep inside water cannot hear sound produces in air because the speed of sound in water is roughly four times that of sound in air, refractive index is four times that of sound in air.

index \[u = \dfrac{{{\text{ Sin }}i}}{{\sin r}} = \dfrac{{Va}}{{Vw}} = \dfrac{1}{4} = 0.25\]

For, refraction ${r_{\max }} = {90^0}$ , ${i_{\max }} = {90^0}$ , ${i_{\max }} = {14^0}$ .

Since ${i_{\max }} \ne {r_{\max }}$

Thus, Sounds are only reflected in the air, and people deep in the water are unable to hear them.

4. If the splash is heard 4.23 seconds after a stone is dropped into a well. 78.4 meters deep, find the velocity of sound in air?

Ans: Given data,

depth of $well = s = 78.4m$

Total time after which splash is heard $ = 4.23\;{\text{s}}$

Assume that,

If ${t_1}$ = time taken by stare to hit the water surface is the well

${t_2}$ = time taken be splash of sound to reach the top of well.

then ${t_1} + {t_2} = 4.23{\text{sec}}$ .

Now, for downward journey of stone;

$u = 0,a = 9.8m\mid {s^2},s = 78.4m$

${\text{t}} = {\text{t}}1 = 2$

$As,s = ut + \dfrac{1}{2}a{t^2}$

$\therefore 78.4 = 0 + \dfrac{1}{2} \times 9.8 \times t_1^2$

$78.4 = 4.9t_1^2$

$t_1^2 = \dfrac{{78.4}}{{4.9}}$

$t_1^2 = 16$

${t_1} = \sqrt {16} = 4{\text{sec}}$

Now, $t1 + t2 = 4.23$

$4 + {t_2} = 4.23$

${t_2} = 4.23 - 4.00$

${t_2} = 0.23\;{\text{s}}$

If ${\text{V}} = $ velocity of sound in ${\text{air}}$ .

${\text{v}} = \dfrac{{{\text{ dis }}\tan ce(t)}}{{\operatorname{tim} (t)}} = \dfrac{{78.4}}{{0.23}} = 340.87\;{\text{m}}\mid {\text{s}}$

5. How roar of a lion can be differentiated from bucking of a mosquito?

Ans: The roaring of a lion produces a low-pitched, high-intensity sound, but the buzzing of mosquitoes produces a high-pitched, low-intensity sound, and therefore the two noises may be distinguished.

6. The length of a sonometer wire between two fixed ends is $110\;{\text{cm}}$. Where the two bridges should be placed so as to divide the wire into three segments whose fundamental frequencies are in the ratio of 1: 2: 3?

Ans: Assume that,

$Let\,\,{{\text{l}}_1},{{\text{l}}_2}$ and ${{\text{L}}_{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{3} }}$ be the length of the three parts of the wire and

${{\text{f}}_1},{{\text{f}}_2}$ and ${t_3}$ be their respective frequencies.

T and m are fixed quantities, and 2 are constant

$f = \dfrac{1}{{2t}}\sqrt {\dfrac{T}{m}} $

$f = \alpha \dfrac{1}{l}$

$or\,{{\text{f}}_1} = $ constant

$50,{{\text{f}}_1}{{\text{l}}_1} = $ Constant $ \to (1)$

${{\text{f}}_2}{{\text{l}}_2} = $ Constant $ \to (2)$

${{\text{f}}_3}{{\text{l}}_3} = $ Constant $ \to (3)$

Equating equation 1), 2) & 3)

${{\text{f}}_1}{{\text{l}}_1} = {{\text{f}}_2}{{\text{l}}_2} = {{\text{f}}_3}{{\text{l}}_3}$

${\text{No}}{{\text{w}}_n}{{\text{l}}_2} = \dfrac{{{f_1}}}{{{f_2}}}{l_{\text{I}}}$

${l_2} = \dfrac{1}{2}{l_1} \Rightarrow (4)\left( {\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{1}{2}} \right)G$

Also, ${l_3} = \dfrac{{{f_1}}}{{{f_i}}}{l_1}$

${l_1} = \dfrac{1}{3}{l_1}\left( {\dfrac{{{f_1}}}{{{f_1}}} = \dfrac{1}{3}(} \right.$ given $\left. ) \right)$

Now, Total length $ = 110\;{\text{cm}}$

1.e ${{\text{l}}_1} + {{\text{l}}_2} + {{\text{l}}_3} = 110\;{\text{cm}}$

${l_1} + \dfrac{1}{2}{l_1} + \dfrac{1}{3}{l_1} = 110$

$\dfrac{{11\,{l_1}}}{6} = 110$

i. e. ${{\text{l}}_1} = 60\;{\text{cm}}$

${l_2} = \dfrac{{{l_1}}}{2}$

${l_2} = 30\;{\text{cm}}$

${l_2} = \dfrac{{60}}{2}$

${l_1} = \dfrac{{{l_1}}}{3}\quad {l_3} = \dfrac{{60}}{3}$

${l_3} = 20\;{\text{cm}}$

7. If string wires of same material of length $l$ and $2i$ vibrate with frequencies $100{\text{H}}$ and$150\;{\text{Hz}}$. Find the ratio of their tensions.

Ans: Given data ,

Since frequency $ = {\text{f}}$ at a vibrating string of mass and

Tension $ = {\text{T}}$ is given by:

$f = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $

Let for first case,

${{\text{f}}_1} = 100\;{\text{Hz}};{{\text{l}}_1} = 1;{{\text{T}}_1} = $ Initial Tension.

For second case, ${t_2} = 150Hz;{1_2} = 21;{{\text{T}}_2} = $ Final Tension

${f_1} = \dfrac{1}{{2l}}\sqrt {\dfrac{{{T_1}}}{m}} $

$100 = \dfrac{1}{{2l}}\sqrt {\dfrac{{{T_1}}}{m}} \to ({\text{l}})$

and ${f_2} = \dfrac{1}{{2{l_2}}}\sqrt {\dfrac{{{T_1}}}{m}} $

$150 = \dfrac{1}{{{2_1}l}}\sqrt {\dfrac{{{T_2}}}{m}} \to (2)$

Divide equation 1) by equation 2)

$\dfrac{{100}}{{150}} = \dfrac{{\dfrac{1}{{2l}}\sqrt {\dfrac{{{T_1}}}{m}} }}{{\dfrac{1}{{4l}}}}\sqrt {\dfrac{{T2}}{m}} $

$\dfrac{{1\phi \times 2}}{{153}} = 2\sqrt {\dfrac{{{T_1}}}{{{T_2}}}} $

$\dfrac{2}{3} = 2\sqrt {\dfrac{{{T_1}}}{{{T_2}}}} $

$\dfrac{{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{Z} }}{{3 \times {2^\prime }}} = \sqrt {\dfrac{{{{\bar T}_1}}}{{{T_2}}}} $

$\dfrac{1}{3} = \sqrt {\dfrac{{{T_1}}}{{{T_2}}}} \to $ Squaring both sides

or $\dfrac{1}{9} = \dfrac{{{T_1}}}{{{T_2}}}$

Hence, the ratio of tensions is 1: 9 .

8. Two similar sonometer wires of the same material produces 2 beats per second. The length of one is $50\;{\text{cm}}$ and that of the other is $50.1{\text{em}}$. Calculate the frequencies of two wives?

Ans: Given data,

The frequency (f) of a Sonometer wire of length (L), mass (m) and

Tension (T) is given by

$k = \dfrac{1}{2}\sqrt {\dfrac{T}{m}} $

$f = \dfrac{k}{l}$

In first case;

$\left. {{{\text{f}}_1} = \dfrac{k}{{{l_1}}} \to 1} \right$ and

$f2 = \dfrac{k}{{{l_2}}} \to 2$

Subtract equation $1$ & $2$

${{\text{f}}_1} - {{\text{f}}_2} = \dfrac{k}{{{l_1}}} = \dfrac{k}{{{l_3}}} = k\left( {\dfrac{1}{{{l_1}}} - \dfrac{1}{{{l_1}}}} \right)$

Now, given ${l_1} = 50\;{\text{cm}};{{\text{l}}_2} = 50.1\;{\text{cm}}$

${{\text{f}}_1} \cdot {f_2} = 2$

$50,2 = k\left[ {\dfrac{1}{{50}} = \dfrac{1}{{50.1}}} \right]$

\[2=k\dfrac{50.1{}^\circ -50{}^\circ }{50\times 50.1}\]

$2 = K\left[ {\dfrac{{0.1}}{{2505.0}}} \right]$

$\dfrac{{2 \times 2505}}{{0.1}} = k$

$\dfrac{{5010 \times 10}}{{01.1}} = k$

$50100 = k$

so, ${f_1} = \dfrac{k}{{{l_1}}} = \dfrac{{501009}}{{5\phi }} = 1002HZ$

9. Why are all stringed instruments provided with hollow boxes?

Ans: The sound box is a hollow box that comes with stringed instruments. Forced vibrations are produced in the sound box when the strings are set into vibration. The enormous surface area of the sound box causes a big amount of air to vibrate. The result is a loud sound with the same frequency as the string.

10. Two waves have equations:

${X_1} = a\operatorname{Sin} (wt + \phi )$ ${X_2} = a\operatorname{Sin} \left( {wt + {\phi _2}} \right)$

If in the resultant wave, the amplitude remains equal to the amplitude of the super posing waves. Calculate the phase difference between ${{\text{x}}_1}$ and ${{\text{x}}_2}$ ?

The first wave $ \to {{\text{x}}_1} = a\sin (wt + \phi )$

The second wave $ \to {X_2} = a\operatorname{Sin} \left( {wt + {\phi _2}} \right)$

${{\text{x}}_1}$ = wave function first wave

${{\text{X}}_2} = $ wave function in second wave

a = amplitude

$w = $ Angular frequency

${\text{t}} = $ time

${\phi _1} = $ phase difference of first wave

${\phi _2} = $ phase difference of second wave.

The resultant amplitude $={{\text{a}}^{\prime }}$ and phase difference$= \phi s0$,

$a = \sqrt {a_1^2 + a_2^2 + 2a,a2\operatorname{Cos} \phi } $

${a_1} = $ amplitude of first wave

$\phi = $ phase difference between two waves.

Now, in oar case,

so, $a = \sqrt {{a^2} + {a^2} + 2a \times a\operatorname{Cos} \phi } $

$a = \sqrt {2{a^2} + 2{a^2}\operatorname{Cos} \phi } $

$a = \sqrt {2{a^2}(1 + \operatorname{Cos} \phi )} $

$a = \sqrt {2{a^2} \times 2\operatorname{Cos} \dfrac{\phi }{2}} \quad \left( {\because 1 + \operatorname{Cos} \theta = 2\operatorname{Cos} \dfrac{\theta }{2}} \right)$

$a = \sqrt {4{a^2}\operatorname{Cos} \dfrac{\phi }{2}} $

$a = 2a\operatorname{Cos} \dfrac{\theta }{2}$

$\not x = 2,a\operatorname{Cos} \dfrac{\phi }{2}$

$\dfrac{1}{2} = \operatorname{Cos} \dfrac{\phi }{2}$

$\frac{\theta }{2}= \cos ^{-1}(\frac{1}{2})\\z = 6 \theta ^4 \frac{\theta}{2} \\ =60^{\circ}$

$\phi \text{ }\!\!~\!\!\text{ }=2\times 60{}^\circ $

$\Phi \text{ }\!\!~\!\!\text{ }=120{}^\circ $

So, the phase difference between ${{\text{X}}_1}$ and ${{\text{X}}_2}$ is $120{}^\circ $.

11 A Tuning fork of frequency $300\;{\text{Hz}}$ resonates with an air column closed at one end at ${27^9}{\text{C}}$. How many beats will be heard in the vibration of the fork and the air column at ${0^ \circ }{\text{c}}$ ?

Ans: According to the question,

Frequency of air column at ${27^ \circ }{\text{C}}$ is $300\;{\text{Hz}}$

Let 1 = length of air column and speed of sound $ = {{\text{V}}_{27}}$

For a pipe, closed of one end, the frequency of ${{\text{n}}^{{\text{th }}}}$ harmonic is:

${f_n} = n\left( {\dfrac{v}{{4l}}} \right)$

${\text{n}} = 1,3,5,7$

1 = length of air column

${\text{v}} = $ velocity

So, Let at ${0^9}{\text{C}}$ , the speed of sound $ = {{\text{v}}_{\text{a}}}$ then,

${f_1} = {r_1}\left( {\dfrac{{{v_1}}}{{4t}}} \right)$ and ${f_1} = {n_2}\left( {\dfrac{{{v_2}}}{{4i}}} \right)$

${f_1} = $ frequency at ${{27}^{0} }{\text{C}} = 300{\text{HZ}}$

${{\text{f}}_2} = $ frequency at ${{0}^{0} }{\text{C}} = ?$

${\text{So}},{{\text{f}}_1} = \dfrac{{{V_2}}}{4}$

$\left. {300 = \dfrac{{{V_{27}}}}{{4!}} \Rightarrow 1} \right)$

$\left. {{f_2} = \dfrac{{Vo}}{{4!}} \Rightarrow 2} \right)$

Now, $\dfrac{{{V_1}}}{{{V_2}}} = \sqrt {\dfrac{{{T_1}}}{{{T_2}}}} $

$\dfrac{{{V_{27}}}}{{{V_0}}}\sqrt {\dfrac{{(27 + 273)k}}{{(273)k}}} $

$\dfrac{{{V_{27}}}}{{{V_0}}} = \sqrt {\dfrac{{300}}{{273}}} $

$\dfrac{{{V_{21}}}}{{{V_0}}} = \not \rho \cdot g$

or $\dfrac{{{V_0}}}{{{V_{27}}}} = \sqrt {\dfrac{{273}}{{300}}} $

$\dfrac{{{V_O}}}{{{V_{27}}}} = 0.954$

${{\text{v}}_0} = (0.954){{\text{V}}_{27}}$

$\therefore $ frequency of air column at ${0^ \circ }{\text{C is}}: - $

${f_2} = \dfrac{{{V_0}}}{{4l}}($ equation 2)

Using ${\text{Vo}} = (0.954){{\text{v}}_2}$ , We get

${{\text{f}}_2} = \dfrac{{(0.954){V_{27}}}}{{4t}}$ Now $\dfrac{{{V_{11}}}}{{4l}} = 330($ equation $A)$

${f_2} = 0.954 \times 330 = 286{\text{HZ}}$

The frequency of tuning fork remains $300 + {\text{H}}$ . Number of beats $ = {{\text{f}}_1} - {t_2}$

$ = 300 - 286 = 14$ Per second

12. A vehicle with horn of frequency ' $^\prime {{\mathbf{n}}^\prime }$ is moving with a velocity of $30\;{\text{m}}\mid {\text{s}}$ in a direction perpendicular to the straight Line joining the observer and the vehicle. If the observer perceives the sound to have a frequency of ${\text{n}} + {{\text{n}}_1}$. Calculate ${{\text{n}}_1}$ ?

Ans: The apparent change in wave frequency caused by relative motion between the source of waves and the observer is known as the Doppler effect.

${v_2} = $ velocity of the listener

${{\text{v}}_{\text{x}}} = $ velocity of the source

$v = $ velocity of sound

$\gamma $ = frequency of sound reaching from the source to the listener.

${\gamma ^1} = $ Apparent frequency (Le. Changed frequency due to movement of source and listener)

${\gamma ^1} = \left( {\dfrac{{V - {V_2}}}{{V - {V_s}}}} \right)x\gamma $

But in our case, the source and observer more at right angles to each other. The Doppler Effect is not observed when the source of the sound and the observer are moving at right argyles to each other.

S0, if ${\text{n}} = $ original frequency of sound the observer will perceive the sound with a frequency of ${\text{n}}$ because of no Doppler effect) Hence the ${{\text{n}}_1} = $ charge frequency $ = 0$ .

13. We cannot hear echo in a room. Explain?

Ans: We all know that in order for an echo to be heard, the obstruction must be hard and vast in size. In addition, the obstruction must be at least a distance from the source. The parameters for the generation of Echo are not met since the length of the room is usually less than. As a result, there is no echo in the room.

14. Why do the stages of large auditoriums give curved backs?

Ans: The backs of big auditorium stages are curved because a speaker's voice is rendered parallel following reflection from a concave or parabolic seer face when he stands at or near the focal of a curved surface. As a result, the voice can be heard from afar.

15. Show that Doppler effect in sound is asymmetric?

Ans: It is seen that apparent frequency of sound when source is approaching the stationary listener (with velocity ${{\text{v}}^1}$ ) is not the same as the apparent frequency of sound when the listener is approaching the stationary source with a velocity ${{\text{v}}^1}$ . This shows that Doppler Effect in sound is asymmetric.

Apparent frequency $ = {{\text{f}}^1} = \dfrac{{\text{V}}}{{{\text{V}} - {\text{us}}}} \times {\text{f}}$

This is when source approaches station any listener

${\text{V}} = $ Velocity of sound in air

${{\text{u}}_{\text{n}}} = $ Velocity of sound source $ = {{\text{v}}_1}$

${{\text{f}}^1} = $ Apparent frequency

${\text{f}} = $ Original frequency of sound

Apparent frequency ${f^{\prime \prime }} = \dfrac{{V + w0}}{V} \times f$

since ${f^{\prime \prime }} = {f^{\prime \prime }}$ (Hence Doppler effect is asymmetric)

16. An organ pipe ${{\text{P}}_1}$ closed at one end vibrating in its first overtone and another pipe ${{\text{P}}_2}$open at both the ends vibrating in its third overtone are in resonance with a given tuning fork. Find the ratio of length of ${{\text{P}}_1}$ and ${{\text{P}}_2}$ ?

Ans: Given data,

Length of pipe closed at one end for first overtone, ${l_1} = \dfrac{{3\pi }}{4}$

Length of pipe closed at both ends for third overtone; ${1_2} = \dfrac{{3\pi }}{4}$

$\pi = $ wavelength

We know that,

${l_2} = \dfrac{{4\pi }}{2} = 2\pi $

$\therefore \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{\dfrac{{3\pi }}{4}}}{{2\pi }} = \dfrac{{3\pi }}{{4 \times 2\pi }} = \dfrac{3}{8}$

${l_1};{l_2} = 3:8$

17. A simple Romanic wave has the equation

${\text{Y}} = 0.30{\text{Sin}}(314{\text{t}} - 1.57{\text{x}})$

${\text{t}} = \sec ,{\text{x}} = $ meters, ${\text{y}} = {\text{cm}}$. Find the frequency and wavelength of this wave.

Another wave has the equation.

${{\text{Y}}_1} = 0.1\sin (314{\text{t}} - 1.57{\text{x}} + 1.57)$

Deduce the phase difference and ratio of intensities of the above two waves?

If y is in meters, then equation becomes: $ \to $

$y = \dfrac{{0.30}}{{100}}\sin (314{\text{t}} - 1.57{\text{x}}) \to (1)$

The standard equation of plane progressive wave is

$y = a\operatorname{Sin} (wt - kx) \Rightarrow (2)$

Comparing equation 1) & 2)

$w = 314,k1.57;{\text{a}} = \dfrac{{0.30}}{{100}} = 3 \times {10^{ - 3}}\;{\text{m}}$

frequency $ = {\text{f}} = \dfrac{w}{{2\pi }} = \dfrac{{314}}{{2\pi }} = 50HZ$

wave velocity, ${\text{v}} = \dfrac{w}{k} = \dfrac{{314}}{{1.57}} = 200\;{\text{m}}/{\text{s}}$

wavelength $ = \pi - {\text{r}}\dfrac{{\text{v}}}{f} = \dfrac{{200}}{{50}} = 4\;{\text{m}}$

On inspection of the equations of the given two waves,

Phase difference,

$\Delta \phi = 1.57{\text{rad}}$

$ = 1.57 \times \left( {\dfrac{{180}}{\pi }} \right) = {90^2}$

Ratio of amplitudes of two waves $ = \dfrac{{{a_1}}}{{{a_1}}} = \dfrac{{0.3}}{{0.1}} = 3$

Ratio of intensities $ = \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{a_i^2}}{{a_2^2}} = {\left( {\dfrac{3}{1}} \right)^2} = \dfrac{9}{1}$

18. The component waves producing a stationary wave have amplitude, Frequency and velocity of $8\;{\text{cm}},30{\text{HZ}}$ and $180\;{\text{cm}}/{\text{s}}$ respected. Write the equation of the stationary wave?

Ans: Since the ware equation of a travelling wave

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

${\text{a}} = $ Amplitude

$t = $ time

${\text{T}} = $ Time Period

${\mathbf{x}} = $ Path difference

Let ${y_1} = a\sin 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\pi }} \right)$

${{\text{Y}}_2} = {\text{a}}\sin 2\pi \left( {\dfrac{t}{T} + \dfrac{x}{\pi }} \right)$

$(\because ){\text{t}}$ is travelling in opposite direction)

By principle of superposition, wave equation for the resultant wave $ = y = {y_1} + {y_2}$

$y = a\operatorname{Sin} 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\pi }} \right) + a\sin 2\pi \left( {\dfrac{t}{T} + \dfrac{x}{\pi }} \right)$

$ = a\left( {\operatorname{Sin} 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\pi }} \right) + a\operatorname{Sin} 2\pi \left( {\dfrac{t}{T} + \dfrac{x}{\pi }} \right)} \right)$

Using $\operatorname{Sin} C + \sin D = a\cos \dfrac{{C - D}}{z} \cdot \dfrac{{\operatorname{Sin} C + D}}{z}$

$y = 2a\operatorname{Cos} \dfrac{{2\pi x}}{\pi } \cdot \dfrac{{\sin 2\pi t}}{T}$

Here ${\text{a}} = 8\;{\text{cm}};{\text{f}} = 30\;{\text{Hz}},\;{\text{V}} = 180\;{\text{cm}}\;{\text{s}}$

$T = \dfrac{1}{{30}}$ sec, $\pi = $ wavelength $ = \dfrac{v}{f} = VT$

$\pi = 180 \times \dfrac{1}{{30}} = 6\;{\text{cm}}$

$y = 2a\operatorname{Cos} \dfrac{{2\pi x}}{\pi }\operatorname{Sin} \dfrac{{2\pi t}}{T}$

$y = 16\dfrac{{{\text{ Cos }}\pi x}}{3} \cdot \dfrac{{Sin60\pi t}}{{15}}$

19. A wine of density a $g$/cm $^3$ is stretched between two clamps $1\;{\text{m}}$ apart while subjected to an extension of $0.06\;{\text{cm}}$. What is the lowest frequency of transverse vibration in the wire? Let young's Modulus $ = {\text{y}} = 9 \times {10^{10}}\;{\text{N}}\;{{\text{m}}^2}$ ?

Ans: The lowest frequency of transverse vibrations is given by: -

Area $ = A$

Density $ = {\text{P}}$

$f = \dfrac{1}{{21}}\sqrt {\dfrac{T}{m}} $

Here ${\text{m}} = $ mass Per unit length = area $ \times $ Density

because Density $ = \dfrac{{{\text{ Mass }}}}{{{\text{ Volwme }}}}$

${\text{m}} = A \times P$

F=35.3 v/sec

20. Given two cases in which there is no Doppler effect in sound?

Ans: The two circumstances in which there is no Doppler effect in sound (i.e. no change in frequency) are as follows: -

1) When both the sound source and the listener are moving in the same direction and at the same speed.

2) When one of the source listeners is in the circle's center and the other is travelling around it at a constant speed.

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

Mass of the string, $M = 2.50\;{\text{kg}}$

Tension in the string, ${\text{T}} = 200\;{\text{N}}$

Length of the string, $l = 20.0\;{\text{m}}$

Mass per unit length, $\mu = \dfrac{M}{l} = \dfrac{{2.50}}{{20}} = 0.125\;{\text{kg}}{{\text{w}}^{ - 1}}$

The velocity (v) of the transverse wave in the string is given by the relation:

$v = \sqrt {\dfrac{T}{\mu }} $

$ = \sqrt {\dfrac{{200}}{{0.125}}} = \sqrt {1600} = 40\;{\text{m}}/{\text{s}}$

Time taken by the disturbance to reach the other end, $t = \dfrac{l}{v} = \dfrac{{20}}{{40}} = 0.50\;{\text{s}}$

22.$A$ steel wire has a length of $12.0\;{\text{m}}$ and a mass of $2.10\;{\text{kg}}$. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at ${20^{ - 1}}{\text{C}} = 343\;{\text{m}}\;{{\text{s}}^{ - 1}}$ ?

Length of the steel wire, $I = 12\;{\text{m}}$

Mass of the steel wire, $m = 2.10\;{\text{kg}}$

Velocity of the transverse wave, $v = 343\;{\text{m}}/{\text{s}}$

Mass per unit length, $\mu = \dfrac{m}{l} = \dfrac{{2.10}}{{12}} = 0.175\;{\text{kg}}\;{{\text{m}}^{ - 1}}$

For tension $T$ , velocity of the transverse wave can be obtained using the relation:

$v = \sqrt {\dfrac{{\bar I}}{H}} $

$\therefore T = {v^2}\mu $

$ = {(343)^2} \times 0.175 = 20588.575 \approx 206 \times {10^4}\;{\text{N}}$

23. A bat emits ultrasonic sound of frequency $1000{\text{kHz}}$ in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is $340\;{\text{m}}\;{\text{s}} - 1$ and in water $1486\;{\text{m}}_5^{ - 1}$.

Ans: (a) Frequency of the ultrasonic sound, ${\text{v}} = 1000{\text{kHz}} = 106{\text{H}}$ ?

Speed of sound in air, ${v_a} = 340\;{\text{m}}/{\text{s}}$

The wavelength $({h_r})$ , of the reflected sound is given by the relation:

${\lambda _r} = \dfrac{V}{V}$

$ = \dfrac{{340}}{{{{10}^5}}} = 3.4 \times {10^{ - 6}}n$

(b) Frequency of the ultrasonic sound ${\text{v}} = 1000{\text{kHz}} = 106\;{\text{Hz}}$

Speed of sound in water, $vw = 1486\;{\text{m}}/{\text{s}}$

The wavelength of the transmitted sound is given as:

${\lambda _{\text{r}}} = \dfrac{{1486}}{{{{10}^8}}} = 1.49 \times {10^{ - 3}}\;{\text{m}}$

24. (1) For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (i) What is the amplitude of a point 0.375 m away from one end?

(a) Yes, except at the nodes

(b) Yes, except at the nodes

(ii) 0.042 m

(a) All the points on the string oscillate with the same frequency, except at the nodes which have zero frequency.

(b) All the points in any vibrating loop have the same phase, except at the nodes.

(ii) The given equation is:

$y(x,t) = 0.06\sin \left( {\dfrac{{2\pi }}{3}x} \right)\cos (120\pi t)$

For $x = 0.375\;{\text{m}}$ and $t = 0$

Amplitude = Displacement $ = 0.06\sin \left( {\dfrac{{2\pi }}{3}x} \right)\cos 0$

$ = 0.06\sin \left( {\dfrac{{2\pi }}{3}0.375} \right) \times 1$

$ = 0.06\sin (0.25\pi ) = 0.06\sin \left( {\dfrac{\pi }{4}} \right)$

$ = 0.06 \times \dfrac{1}{{\sqrt 2 }} = 0.042\;{\text{m}}$

25. A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $45\;{\text{Hz}}$ The mass of the wire is $3.5 \times {10^{ - 2}}\;{\text{kg}}$ and its linear mass density is $4.0 \times {10^{ - 2}}\;{\text{kg}}\;{\text{m}} - 1.$ What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

(a) Mass of the wire, $m = 3.5 \times {10^{ - 2}}\;{\text{kg}}$

Linear mass density, $\mu = \dfrac{m}{l} = 4.0 \times {10^{ - 1}}\;{\text{kg}}\;{{\text{m}}^{ - 1}}$

: Length of the wire, $l = \dfrac{m}{\mu } = \dfrac{{3.5 \times {{10}^{ - 1}}}}{{4.0 \times {{10}^{ - 2}}}} = 0.875\;{\text{m}}$

The wavelength of the stationary wave $(2)$ is related to the length of the wire by the relation:

$\lambda = \dfrac{2}{n}$

Where, ${\text{n}} = $ Number of nodes in the wire

For fundamental node, $n = 1$ :

$\lambda = 2l$

$\lambda = 2 \times 0.875 = 1.75\;{\text{m}}$

The speed of the transverse wave in the string is given as:

$v = v\lambda 45 \times 1.75 = 78.75\;{\text{m}}/{\text{s}}$

(b) The tension produced In the string is given by the relation:

$T = {V_1}\mu $

$ = {(78.75)^2} \times 4.0 \times {10^{ - 1}} = 248.06\;{\text{N}}$

26. Two sitar strings $A$ and B playing the note "Ga' are slightly out of tune and produce beats of frequency $6{\text{H}}$. The tension in the string ${\text{A}}$ is slightly reduced and the beat frequency is found to reduce to $3\;{\text{Hz}}$ if the original frequency of ${\text{A}}$ is $324\;{\text{Hz}}$, what is the frequency of $\mathbb{B}$ ?

Frequency of string A, ${f_A} = 324\;{\text{Hz}}$

Frequency of string ${\text{B}} = {f_{\text{B}}}$

Beat's frequency, $n = 6\;{\text{Hz}}$

Beat's frequency is given as:

$n = {f_A} \pm {f_B}$

$6 = 324 \pm {f_B}$

\[{f_B} = 330{\text{Hz or }}318\;{\text{Hz}}\]

Frequency decreases with a decrease in the tension in a string. This is because frequency is directly proportional to the square root of tension. It is given as:

$v \propto \sqrt T $

Hence, the beat frequency cannot be $330\;{\text{Hz}}$

$\therefore {f_S} = 318\;{\text{Hz}}$

3 Marks Questions

1. Explain briefly the analytical method of formation of heats?

Let us consider two wares’ trains at equal amplitude 'a' and with different frequencies ${v_1}$ 4 and ${v_2}$ in same direction.

Let displacements are ${{\text{y}}_1}$ and ${{\text{y}}_2}$ in time ${\text{t}}$ .

${w_1} = $ Angular vel. of first ware

${w_1} = 2\pi {v_1}$

${y_1} = a\sin {w_1}t$

${y_1} = a\operatorname{Sin} 2\pi {v_1}t$

${y_2} = a\operatorname{Sin} 2{v_2}t$

According to superposition principle, the resultant displacement ${y^\prime }$ at the same time t' is-

$y = y1 + yz$

$y = a\left[ {\operatorname{Sin} 2\pi {u_1}t + \sin 2\pi {v_1}t} \right]$

Using $\sin C + \sin D = 2\cos \dfrac{{C - D}}{2}\sin \dfrac{{C + D}}{2}$

$y = 2a\cos \dfrac{{2\pi \left( {{u_1} - {v_i}} \right)t}}{2}\sin \dfrac{{2\pi \left( {{v_1} + {v_2}} \right)t}}{2}$

$y = 2a\operatorname{Cos} \pi \left( {{v_1} - {v_2}} \right)t\sin \pi \left( {{v_2} + {v_2}} \right)t$

$y = A\operatorname{Sin} \pi \left( {{v_1} + {v_2}} \right)t$

Where, ${\text{A}} = 2{\text{a}}\cos \pi \left( {{v_1} - {v_2}} \right)t$ .

Now, amplitude A is maximum when,

$\operatorname{Cos} \pi \left( {{u_1} - {v_2}} \right) + = \max + \pm 1 = \operatorname{Cos} {\text{k}}\pi $

$\pi \left( {{v_1} - {v_2}} \right)t = k\pi ,k = 0,1,2 - $

${\text{t}} = \dfrac{k}{{\left( {{v_1} - {v_1}} \right)}}$

i.e. resultant intensity of sound will be maximum at times, $t = 0,\dfrac{1}{{\left( {{u_1} - {v_2}} \right)}} \cdot \dfrac{3}{{\left( {{u_1} - {v_2}} \right)}} - - - - $

Time interval between 2 successive Maximax’s $ = \dfrac{1}{{\left( {{v_1} - {v_2}} \right)}} - 0 = \dfrac{1}{{\left( {{v_1} - {v_2}} \right)}} \to $ (1)

similarly, a will be minimum,

$\operatorname{Cos} \pi \left( {{v_1} - {v_2}} \right)t = 0$

$\operatorname{Cos} \pi \left( {{u_1} - {v_1}} \right)t = \operatorname{Cos} (2k + 1)\dfrac{\pi }{2},k = 0,1,2,3 - $

$\pi \left( {{v_1} - {v_2}} \right)t = (2k + 1)\dfrac{\pi }{2}$

$t = \dfrac{{(2k + 1)}}{{2\left( {{v_1} - {v_2}} \right)}}$

i.e. resultant intensity of sound will be minimum at times

$t = \dfrac{1}{{2\left( {{v_1} - {v_2}} \right)}},\dfrac{3}{{2\left( {{v_1} - {v_2}} \right)}},\dfrac{5}{{2\left( {{u_1} - {v_2}} \right)}}$

Hence time interval between 2 successive minim as are

$ = \dfrac{3}{{2\left( {{v_1} - {v_2}} \right)}} - \dfrac{1}{{2\left( {{v_1} - {u_2}} \right)}} = \dfrac{1}{{\left( {{q_1} - {v_2}} \right)}} \to (2)$

Combining 1) &: 2) frequency of beats $ = \left( {{v_1} - {v_1}} \right)$

$\therefore \dfrac{{{\text{ No}}{\text{. of beats }}}}{{{\text{ seconds }}}} = $ Difference in frequencies of two sources of sound.

2. Show that the frequency of nth harmonic mode in a vibrating string which is closed at both the end is 'n" times the frequency of the first harmonic mode?

Ans: When a sting under tension is placed into vibration, transverse harmonic waves propagate along its length, and reflected waves exist when the length of the sting is fixed. In sting, the incident and reflected waves will superimpose on one other, resulting in transverse stationary waves.

Let a harmonic wave be set up in a sting of length $ = \lambda $ which is fixed at 2 ends: $ \to {\text{x}} = 0$ and $x = L$

Let the incident wave travels from left to right direction, the wave equation is:

$\left. {{y_1} = r\operatorname{Sin} \dfrac{{2\pi }}{\pi }(Vt + x) \to 1} \right)$

The reflected wave equation will have the same amplitude, wavelength, velocity, and time as the incident wave equation, but the only difference between the incident and reflected waves will be in their propagation direction.

${\text{So}},: \to $

${y_2} = r\operatorname{Sin} \dfrac{{2\pi }}{\pi }(v - x)$

$\because $ Reflected wave travels from right to left

Reflection, the wave will suffer a phase change of $\pi .S0$ ,

$y2 = r\operatorname{Sin} \dfrac{{2\pi }}{\pi }[vt - x + \pi ]$

$y2 = - r\operatorname{Sin} \dfrac{{2\pi }}{\pi }[{\text{vt}} - x] \to (2)$

According to the principle of superposition, the wave equation of resultant stationary wave will be-

$y = {y_1} + {y_2}$

Using equation 1) & 2)

$y = r\operatorname{Sin} \dfrac{{2\pi }}{\lambda }[vt + x] - r\sin \dfrac{{2\pi }}{\lambda }(v - x)$

$ = r\left[ {\sin \dfrac{{2\pi }}{\lambda }(vt + x) - \sin \dfrac{{2\pi }}{\lambda }(vt - x)} \right]$

Using $\sin C \cdot \sin D = 2\operatorname{Cos} \dfrac{{C + D}}{Z}\sin \dfrac{{C - D}}{Z}$

$y = r \times 2\operatorname{Cos} \left[ {\dfrac{{\dfrac{{2\pi }}{2}(v + x) + \dfrac{{2\pi }}{\lambda }(v - x)}}{2}} \right]$

$\operatorname{Sin} \left[ {\dfrac{{2\pi }}{\lambda }(v + x) - \dfrac{{2\pi }}{\lambda }(v - x)} \right]$

$y = 2r\cos \left( {\dfrac{{\dfrac{{2\pi }}{2}vt + \dfrac{{2\pi }}{7}x + \dfrac{{2\pi }}{2}vt - \dfrac{{2\pi }}{7}x}}{2}} \right)$

$y = 2r\operatorname{Cos} \dfrac{{2\pi }}{\lambda }\left( {\dfrac{{{2^\prime }vt}}{{{7^\prime }}}} \right) \cdot \sin \dfrac{{2\pi }}{\lambda }\left( {\dfrac{{2x}}{2}} \right)$

$y = 2r\operatorname{Cos} \dfrac{{2\pi }}{2}vt.\sin \dfrac{{2\pi x}}{\lambda }$

Now, at $x = 0{\text{ & }}\,x = L;y = 0$

1) At $x = 0,y = 0$ is satisfied

2) $Atx = {L_n}y = 0$

$y = 2{\text{r}}\operatorname{Cos} \left( {\dfrac{{2\pi }}{\lambda }vt} \right) \cdot \sin \left( {\dfrac{{2\pi }}{\lambda } \times L} \right)$

$0 = 2{\text{t}}\operatorname{Cos} \left( {\dfrac{{2\pi }}{\lambda }vt} \right)\operatorname{Sin} \left( {\dfrac{{2\pi }}{\lambda } \times L} \right)$

Now, $r \ne 0,\dfrac{{2\pi }}{\lambda }vt \ne 0$ so

$o = \operatorname{Sin} \left( {\dfrac{{2\pi L}}{\lambda }} \right)$

Sin ${\text{n}}{z^\prime } = \operatorname{stn} \dfrac{{2\pi 2}}{\lambda }$

i.e ${\text{L}} = {\text{n}}\left( {\dfrac{x}{2}} \right)$

1) Let ${\text{n}} = 1$ (Le first harmonic mode or fundamental frequency

$\lambda = {\lambda _1}$

$L = \dfrac{{{\lambda _1}}}{2}$

${\lambda _1} = 2L$

Let ${\text{v}} = $ velocity, ${\gamma _1} = $ frequency of first harmonic Mode ${\text{v}} = {\gamma _1}{\lambda _1}$

$v = {\gamma _1} \times 2L$

$\dfrac{v}{{2L}} = {\gamma _1}$

If ${\text{n}} = 2$ (second harmonic Mode or first overtone)

$L = \lambda 2$

$v = $ velocity, ${\gamma _2} = $ ferq4 of second harmonic Mode

${\text{v}} = {\gamma _1}{\hat h_2}$

${\gamma _2} = \dfrac{v}{{{\lambda _2}}} = \dfrac{v}{2}$

$\gamma 2 = \dfrac{{v \times 2}}{{L \times 2}}\quad \left( {\because \dfrac{v}{{2L}} = {\gamma _1}} \right)$

${\gamma _2} = 2{\gamma _1}$

i.e. frequency of second harmonic Mode is twice the frequency of first harmonic Mode Similarly, $m = n{\% _1}$ and frequency of ${{\text{n}}^{{\text{th }}}}$ harmonic Mode is n times the frequency of first harmonic Mode.

3. Differentiate between the types of vibration in closed and open organ pipes?

Ans: 1) In closed pipe, the wavelength of nth mode $ = \pi n = \dfrac{{4l}}{n}$

where ${\text{n}} = $ odd integer

where as in open pipe, $\pi N = \dfrac{{2l}}{n}$ and $n = $ all integer

2) The fundamental frequency of open pipe is twice that of closed pipe of same length.

3) A closed pipe of length $\dfrac{2}{2}$ produces the same fundamental frequency as an open pipe of length L .

4) For an open pipe, harmonics are present for all integers and for a closed pipe, harmonics are present for only odd integers hence, open pipe gives richer note.

Put the value of ${\text{n}}$ in equation for frequency

$f = \dfrac{1}{{21}}\sqrt {\dfrac{T}{{AP}}} $

$f = \dfrac{1}{{2!}}\sqrt {\dfrac{{T/A}}{P}} $

Now, Force = Tension and $\dfrac{{{\text{ Farce }}}}{{{\text{ Area }}}} = $ Stress

$\left. {{\text{f}} = \dfrac{1}{{2\;{\text{J}}}}\sqrt {\dfrac{{s\operatorname{tress} }}{{\text{P}}}} \to 2} \right)$

Now, young's Modulus $ = {\text{y}} = \dfrac{{{\text{ Stress }}}}{{{\text{ Strairt }}}}$

Or Stress $ = yx$ strain

Stress $ = y \times \dfrac{{\Delta L}}{L}$

Put the value of Stress in equation

$f = \dfrac{1}{{2l}}\sqrt {\dfrac{{y \times \Delta L}}{{Z \times P}}} $

Put $y = 9 \times {10^{10}}N/m,\vartriangle L = 0.05 \times {10^{ - 2}}ms$

$L= 1.0 m , P = 9 \times 10^3 \frac{kg}{m^3} f =\frac{1}{2 \times 1}\sqrt{\frac{9 \times 10^9 \times 0.05 \times 10^{-2}}{1 \times 9 \times 10^3}}$

Since fundamental frequency of a stretched spring $\dfrac{{\alpha 1}}{l}$

$\frac{f_2}{f_1} = \frac{l_2}{l_1} = \frac{50.1}{49.9}\\f_2 = \frac{50.1}{49.9}f_1 \\\text{now} ,f_2-f_1 = 1 \text{or} \frac{50.1}{49.9}f_1 -f_1\\l =\frac{50.1 f_1 - 49.9 f_2}{49.9}\\l = \frac{0.2 f_1}{49.9} \\l \times 49.9 = 0.2 f_1\\ f_1 = \frac{49.9}{0.2}\\f_1 = 249.5 \text{HZ} \\f_2 - f_1 =1 \\f_2 = 1+f_1 \\=1+249.5$

${{\text{f}}_2} = 250.5{\text{HZ}}$

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

Ans: Ultrasonic beep frequency emitted by the bat, ${\text{V}} = 40{\text{k}}{{\text{H}}_z}$

Velocity of the bat, ${v_2} = 0.03v$

Where, $v$ = velocity of sound in air

The apparent frequency of the sound striking the wall is given as:

${v^\prime } = \left( {\dfrac{v}{{v - {v_b}}}} \right)v$

$ = \left( {\dfrac{v}{{v - 0.03v}}} \right)40$

$ = \dfrac{{40}}{{0.97}}{\text{kHz}}$

This frequency is reflected by the stationary wall $\left( {{v_4} = 0} \right)$ toward the bat.

The frequency ( $V''$ ) of the received sound is given by the relation:

${v^{\prime \prime }} = \left( {\dfrac{v}{{v + {v_z}}}} \right){v^\prime }$

$ = \left( {\dfrac{{v + 0.3v}}{v}} \right) \times \dfrac{{40}}{{0.97}}$

$ = \dfrac{{1.03 \times 40}}{{0.97}} = 42.47{\text{kHz}}$

5. A stone dropped from the top of a tower of height $300\;{\text{m}}$ high 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 $340\;{\text{m}}\;{{\text{s}}^{ - 1}}?\left( {\;{\text{g}} = 9.8\;{\text{m}},{5^{ - 1}}} \right)$

Height of the tower, $s = 300\;{\text{m}}$

Initial velocity of the stone, $u = 0$

Acceleration, $a = g = 9.8\;{\text{m}}{5^{ - 1}}$

Speed of sound in air $ = 340\;{\text{m}}/{\text{s}}$

The time $\left( {{t_1}} \right)$ taken by the stone to strike the water in the pond can be calculated using the

second equation of motion, as:

$s = s{t_1} + \dfrac{1}{2}gt_1^2$

$300 = 0 + \dfrac{1}{2} \times 9.8xt_1^2$

$\therefore {t_1} = \sqrt {\dfrac{{300 \times 2}}{{9.8}}} = 7,82s$

Time taken by the sound to reach the top of the tower, ${t_1} = \dfrac{{300}}{{340}} = 0.88s$

Therefore, the time after which the splash is heard, $t = {t_1} + {t_2}$

$ = 7.82 + 0.88 = 8.7\;{\text{s}}$

6. You have learnt that a traveling wave in one dimension is represented by a function $y = f(x$, there $x$ and $t$ must appear in the combination $x - vt$ or $x + vt$, 1.e. $y = f$ $(x \pm vt).$ Is the converse true? Examine if the following functions for y can possibly represent a travelling wave:

(a) ${(x - vt)^2}({\mathbf{b}})\log \left[ {\dfrac{{x + vt}}{{{x_0}}}} \right]$ (c) $\dfrac{1}{{(x + vt)}}$

(a) Does not represent a wave

(b) Represents a wave

The converse of the given statement is not true. The essential requirement for a function to represent a travelling wave ${\text{k}}$ : that it should remain finite for all values of $x$ and $t$ .

(a) Far $x = 0$ and $t = 0$ , the function ${(x - v)^2}$ becomes 0 .

Hence, for $x = 0$ and $t = 0$ , the function represents a point and not a wave.

(b) For $x = 0$ and $t = 0$ , the function

$\log \left( {\dfrac{{x + vt}}{{{x_0}}}} \right) = \log 0 = x$

Since the function does not converge to a finite value for $x = 0$ and $t = 0$ , it represents a travelling wave.

$\dfrac{1}{{x + vt}} = \dfrac{1}{0} = x$

Since the function does not converge to a finite value for $x = 0$ and $t = 0$ , it does not represent a travelling wave.

7: For the travelling harmonic wave

$y(x,t) = 2.0\cos 2\pi (10t - 0.0080x + 0.35)$

Where $x$ and $y$ are in cm and $t$ in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of

(a) $4\;{\text{m}}$, (b) $0.5\;{\text{m}},({\text{c}}),\dfrac{\lambda }{2}$ (d) $\dfrac{{3.\lambda }}{4}$

Ans: Equation for a travelling harmonic wave is given as:

$y(x,t) = 20\cos 2\pi (10t - 0.0050x + 0.35)$

$ = 2.0\cos (20\pi t - 0.016 - x + 0.70\pi )$

Propagation constant, $k = 0.0160\pi $

Amplitude, $a = 2\;{\text{cm}}$

Angular frequency, $\emptyset = 20\pi $ rad/s

Phase difference is given by the relation:

$\phi = kx = \dfrac{{2\pi }}{\lambda }$

(a) For $x = 4\;{\text{m}} = 400\;{\text{cm}}$

$\Phi = 0.016\pi \times 400 = 6.4\pi {\text{rad}}$

(b) For $0.5\;{\text{m}} = 50\;{\text{cm}}$

$\Phi = 0.016\pi \times 50 = 0.8{\text{mrad}}$

$(c){\text{For}}x = \dfrac{\lambda }{2}$

$\phi = \dfrac{{2\pi }}{\lambda } \times \dfrac{\lambda }{2} = \pi rad$

(d) For $x = \dfrac{{3\lambda }}{4}$

$\phi = \dfrac{{2\pi }}{2} \times \dfrac{{3\lambda }}{4} = 1.5\pi rad$

8. A meter-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency $340\;{\text{Hz}}$ ) when the tube length is $25.5\;{\text{cm}}$ or $79.3\;{\text{cm}}$. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.

Ans: Given data,

Frequency of the turning fork, ${\text{v}} = 340\;{\text{Hz}}$

Because one end of the supplied pipe is connected to a piston, it will behave as a pipe with one end closed and the other open, as illustrated in the diagram.

Such a system produces odd harmonics. The fundamental note in a closed pipe is given by the relation:

${l_1} = \dfrac{\lambda }{4}$

Length at the pipe, ${l_1} = 25.5\;{\text{cm}} = 0.255\;{\text{m}}$

$\therefore \lambda = 4{l_1} = 4 \times 0.255\;{\text{m}} = 1.02\;{\text{m}}$

The speed of sound is given by the relation:

$v = v\lambda = 340 \times 1.02 = 346.8\;{\text{m}}/{\text{s}}$

9. A steel rod $100\;{\text{cm}}$ long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be $2.53{\text{kHz}}$. What is the speed of sound in steel?

Length of the steel rod, $l = 100\;{\text{cm}} = 1\;{\text{m}}$

Fundamental frequency of vibration, $v = 2.53{\text{kHz}} = 2.53 \times {10^{{\text{iHz}}}}$

When the rod is plucked at its middle, an antinode (A) is formed at its center, and nodes (N) are formed at its two ends, as shown in the given figure.

The distance between two successive nodes is $\dfrac{\lambda }{2}$ .

$\therefore l = \dfrac{\lambda }{2}$

$\lambda = 2l = 2 \times 1 = 2m$

The speed of sound in steel is given by the relation:

$v = {v_\lambda }$

$ = 2.53 \times {10^3} \times 2$

$ = 5.06 \times {10^3}\;{\text{m}}/{\text{s}}$

$ = 5.06\;{\text{km}}/{\text{s}}$

(e) A pulse is made up of many waves with varying wavelengths. Depending on the nature of the medium, these waves travel at varying speeds in a dispersive medium. The shape of a wave pulse is distorted as a result of this.

10. A train, standing at the outer signal of a railway station blows a whistle of frequency $400\;{\text{Hz}}$ 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 $10\;{\text{m}}_5^{ - 1},(\;{\text{b}})$ recedes from the platform with a speed of $10\;{\text{m}} = 1?$ (ii) What is the speed of sound in each ease? The speed of sound in still air can be taken as $340\;{\text{m}}\;{\text{s}} - 1$.

Ans: (i) Given that,

(a) Frequency of the whistle, ${\text{v}} = 400\;{\text{Hz}}$

Speed of the train, ${v_T} = 10\;{\text{m}}/{\text{s}}$

Speed of sound, $v = 340\;{\text{m}}/{\text{s}}$

The apparent frequency $\left( {{v^7}} \right)$ of the whistle as the train approaches the platform is given by

the relation:

${v^\prime } = \left( {\dfrac{v}{{v - {v_r}}}} \right)v$

$ = \left( {\dfrac{{340}}{{340 - 10}}} \right) \times 400 = 412.12\;{\text{Hz}}$

(b) The apparent frequency $\left( {{v^{\prime \prime }}} \right)$ ot the whistle as the train recedes from the platform is given by the relation:

${v^{\prime \prime }} = \left( {\dfrac{v}{{v - {v_r}}}} \right)v$

$ = \left( {\dfrac{{340}}{{340 + 10}}} \right) \times 400 = 388.57\;{\text{Hz}}$

(ii) The apparent shift in sound frequency is produced by the source and observer's relative motions. The speed of sound is unaffected by these relative motions. As a result, the speed of sound in air is constant in both circumstances. i.e, $340\;{\text{m}}$ /s

11. A SONAR system fixed in a submarine operates at a frequency $40.0{\text{kH}}$. An enemy submarine moves towards the SONAR with a speed of $360\;{\text{km}}\;{{\text{h}}^{ - 1}}$ - What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be $1450\;{\text{m}}\;{{\text{s}}^{ - 1}}$

Ans: Given that,

Operating frequency of the $50{\text{NAR}}5{\text{ssem}},{\text{V}} = 40{\text{kHz}}$

Speed of the enemy submarine, ${v_4} = 360\;{\text{km}}/{\text{h}} = 100\;{\text{m}}/{\text{s}}$

Speed of sound in water, $v = 1450\;{\text{m}}/{\text{s}}$

The source is at rest and the observer (enemy submarine) is moving toward it. Hence, the apparent frequency $\left( {{v^\prime }} \right)$ received and reflected by the submarine is given by the relation:

${v^\prime } = \left( {\dfrac{{v + {v_A}}}{v}} \right)v$

$ = \left( {\dfrac{{1450 + 100}}{{1450}}} \right) \times 40 = 42.76{\text{ldH}}$

The frequency (v") received by the enemy submarine is given by the relation:

${{\text{v}}^{\prime \prime }} = \left( {\dfrac{v}{{v + {v_2}}}} \right){v^\prime }$

where, ${v_2} = 100\;{\text{m}}/{\text{s}}$

$\therefore {V^{\prime \prime }} = \left( {\dfrac{{1450}}{{1450 - 100}}} \right) \times 42.76 = 45.93{\text{kHz}}$

4 Marks Questions

1. A pipe $20\;{\text{cm}}$ long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a $430\;{\text{Hz}}$ source? Will the same source be in resonance with the pipe if both ends are open? (Speed of sound in air is $340\;{\text{m}}_5^{ - 1}$ )

First (Fundamental); No

Length of the pipe, ${\text{I}} = 20\;{\text{cm}} = 0.2\;{\text{m}}$

Source frequency $ = n$ the normal mode of frequency, ${v_n} = 430\;{\text{Hz}}$

Speed of sound, $v = 340\;{\text{m}}\;{\text{s}}$

In a closed pipe, the $n$ the normal mode of frequency is given by the relation:

${v_n} = (2n - 1)\dfrac{v}{4};{\text{n}}$ is an integer $ = 0,1,23 \ldots \ldots $

$430 = (2n - 1)\dfrac{{340}}{{4 \times 0.2}}$

$2n - 1 = \dfrac{{430 \times 4 \times 0.2}}{{340}} = 1.01$

$2n = 2.01$

Hence, the first mode of vibration frequency is resonantly excited by the given source.

In a pipe open at both ends, the nth mode of vibration frequency is given by the relation:

${v_n} = \dfrac{{Nv}}{{2l}}$

${v_n} = \dfrac{{2{{\sqrt V }_A}}}{v}$

$ = \dfrac{{2 \times 0.2 \times 430}}{{340}} = 0.5$

Since the number of the mode of vibration $(n)$ has to be an integer, the given source does not produce a resonant vibration in an open pipe.

2. Explain why (or how):

(a) In a sound wave, a displacement node is a pressure antinode and vice versa,

(b) Bats can ascertain distances, directions, nature, and sizes of the obstacles without any "eyes",

(e) A violin note and sitar note may have the same frequency. yet we can distinguish between the two notes.

(d) Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and

(e) The shape of a pulse gets distorted during propagation in a dispersive medium.

Ans: (a) A node is a position where the vibration amplitude is the smallest and the pressure is the highest. An antinode, on the other hand, is a place where the vibration amplitude is greatest and the pressure is lowest.

As a result, a pressure antinode is nothing more than a displacement node, and vice versa.

(b) Bats produce ultrasonic sound waves at a very high frequency. Obstacles re-direct these waves back toward them. With the help of its cerebral sensors, a bat receives a reflected wave (frequency) and calculates the distance, direction, nature, and size of an obstacle.

(c) A sitar's and a violin's overtones, as well as the strength of these overtones, are different. Hence, one can distinguish between the notes produced by a sitar and a Violin even if they have the same frequency of vibration.

(d) Shear modulus is a property of solids. They are able to withstand shearing force. Fluids yield to shearing stress because they have no defined shape. A transverse wave propagates in such a way that it causes shearing stress in a medium. A wave of this nature can only propagate in solids, not in gases.

The bulk moduli of solids and fluids are different. They have the ability to withstand compressive stress.

3. One end of a long string of linear mass density $8.0 \times {10^{ - 3}}\;{\text{kg}}\;{{\text{m}}^{ - 1}}$ is connected to an electrically driven tuning fork of frequency $256{\text{H}}$. The other end passes over a pulley and is tied to a pan containing a mass of $90\;{\text{kg}}$. 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 displacement $(y = 0)$ and is moving along positive $y$ -direction. The amplitude of the wave is $5.0{\text{em}}$. Write down the transverse displacement $y$ as a function of $x$ and $t$ that describes the wave on the string.

Ans: The equation of a travelling wave propagating along the positive $y$ -direction is given by the displacement equation:

$y(x,t) = a\sin (wt - kx) - (0)$

Linear mass density, $\lambda = 8.0 \times {10^{ - 3}}$ kg ${{\text{m}}^{ - 1}}$

Frequency of the tuning fork, ${\text{v}} = 256{\text{Hz}}$

Amplitude of the wave, $a = 5.0\;{\text{cm}} = 0.05\;{\text{m}} \ldots ({\text{m}}$

Mass of the pan, $m = 90\;{\text{kg}}$

Tension in the string. ${\text{T}} = {\text{mg}} = 90 \times 9.8 = {\text{BRZN}}$

The velocity of the transverse wave $v$ , is given by the relation:

$ = \sqrt {\dfrac{{882}}{{8.0 \times {{10}^{ - 1}}}}} = 332\;{\text{m}}/{\text{s}}$

Angular frequency $ \cdot \omega = 2\pi v$

$ = 2 \times 3.14 \times 256$

$ = 1608.5 = 1.6 \times {10^3}{\text{rad}}/{\text{s}} - ({\text{ui}})$

Wavelength, $\lambda = \dfrac{v}{v} = \dfrac{{332}}{{256}}m$

$\therefore $ Propagation constant, $k = \dfrac{{2\pi }}{\lambda }$

$ = \dfrac{{2 \times 3.14}}{{\dfrac{{332}}{{256}}}} = 4.84{w^{ - 1}}\quad \ldots \ldots ..0$

displacement equation:

\[y(x,t) = 0.05\sin \left( {1.6 \times {{10}^3}t - 4.84x} \right)m\]

4. Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse $(5)$ and longitudinal $(P)$ sound waves. Typically the speed of S wave is about $4.0\;{\text{km}}\;{\text{s}} - 1$, and that of ${\text{P}}$ wave is $80\;{\text{l}}{{\text{m}}_5}\;{{\text{s}}^{ - 1}}$ - 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 straight line, at what distance does the earthquake occur?

Ans: Assume ,

Let ${v_s}$ and ${v_F}$ be the velocities of $S$ and $P$ waves respectively.

Let $L$ be the distance between the epicenter and the seismograph.

L. $ = {v_g}{t_3}(a)$

$1. = {v_P}{l_P}(i)$

${t_5}$ and ${t_p}$ are the respective times taken by the S and P waves to reach the seismograph from the epicenter

It is given that:

${v_p} = 8\;{\text{km}}/{\text{s}}$

${v_s} = 4\;{\text{km}}/{\text{s}}$

From equations (i) and (ii), we have:

${v_3}{t_S} = {v_k}{t_p}4{t_3} = 8{t_p}$

${t_s} = 2{t_P}(iii)$

It is also given that

${t_y} - {t_P} = 4\;{\text{min}} = 240s$

$2t - {t_p} = 240$

${t_P} = 240$

And ${t_3} = 2 \times 240 = 480\;{\text{s}}$

From equation (ii), we get

$I = 8 \times 240$

$ = 1920\;{\text{km}}$

Hence, the earthquake occurs at a distance of $1920\;{\text{km}}$ from the seismograph.

5. A train, standing in a station-yard, blows a whistle ot frequency $400\;{\text{Hz}}$ in still air. The wind starts blowing in the direction from the yard to the station with at a speed of $10\;{\text{m}}$ ${5^{ - 1}}$. 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 ease when the air is still and the observer runs towards the yard at a speed of $10\;{\text{m}}\;{{\text{s}}^{ - 1}}$ ? The speed of sound in still air can be taken as $340\;{\text{m}}\;{{\text{s}}^{ - 1}}$ -

For the stationary observer: $400\;{\text{Hz}},0.875\;{\text{m}};350\;{\text{m}}/{\text{s}}$

For the running observer Not exactly identical

For the stationary observer:

Frequency of the sound produced by the whistle, ${\text{v}} = 400\;{\text{Hz}}$

Speed of sound $ = 340\;{\text{m}}/{\text{s}}$

Velocity of the wind, $v = 10\;{\text{m}}/{\text{s}}$

As there is no relative motion between the source and the observer, the frequency of the sound heard by the observer wit be the same as that produced by the source, Le. $400\;{\text{Hz}}$ .

The wind is blowing toward the observer. Hence, the effective speed of the sound increases by 10 units, i.e.,

Effective speed of the sound, ${v_4} = 340 + 10 = 350\;{\text{m}}/{\text{s}}$

The wavelength ( $\lambda $ ) of the sound heard by the observer is given by the relation:

$\lambda = \dfrac{{{v_4}}}{v} - = \dfrac{{350}}{{400}} = 0.875\;{\text{m}}$

For the running observer:

Velocity of the observer, ${v_s} = 10\;{\text{m}}/{\text{s}}$

The observer is moving toward the source. As a result of the relative motions of the source and the observer, there is a change in frequency $\left( {{v^n}} \right)$ .

This is given by the relation:

$ = \left( {\dfrac{{340 + 10}}{{340}}} \right) \times 400 = 411.76Hz$

since the air is still, the effective speed of sound $ = 340 + 0 = 340\;{\text{m}}/{\text{s}}$

The source is at rest. Hence, the wavelength of the sound will not change, Le., $\lambda $ remains

$0.875\;{\text{m}}$ .

Hence, the given two situations are not exactly identical

5 Marks Questions

1. Use the formula $v = \sqrt {\dfrac{{yP}}{\rho }} $ to explain why the speed of sound in air

(a) is independent of pressure,

(b) increases with temperature,

(e) Increases with humidity-

(a) Take the relation:

$v = \sqrt {\dfrac{{{y^P}}}{\rho }} ........(i)$

Density, $\rho = \dfrac{{{\text{ Moss }}}}{{{\text{ Volume }}}} = \dfrac{N}{V}$

${\text{M}} = $ molecular weight of the ${\text{gas}}$

${\text{V}} = $ Volume of the ${\text{gas}}$

Hence, equation(i) reduces to

${\text{v}} = \sqrt {\dfrac{{yP\;{\text{V}}}}{{\text{M}}}} \cdots (ii)$

Now from the ideal gas equation for $n = 1:$

${\text{PV}} = {\text{m}}T$

For constant $T,PV = $

Since both $M$ and ${\text{Y}}$ are constants, $v$ = Constant

Hence, at a constant temperature, the speed of sound in a gaseous medium is independent at the change in the pressure of the gas.

Tale the relation:

$v = \sqrt {\dfrac{{yP}}{\rho }} \ldots \ldots (i)$

For one mole at an ideal gas, the gas equation can be written as:

${\text{PV}} = {\text{FT}}$

$P = \dfrac{{RT}}{V} \ldots (ii)$

Substituting equation (ii) in equation (i), we get:

$v = \sqrt {\dfrac{{yRT}}{{V\rho }}} = \sqrt {\dfrac{{yRT}}{M}} ........(iv)$

$M = pv$ is a constant

${\text{Y}}$ and ${\text{F}}$ are also constants

We conclude from equation (iv) that $v\alpha \sqrt T $

As a result, the speed of sound in a gas is proportional to the square root of the temperature of the gaseous medium, Le. In other words, the speed of sound increases as the temperature of the gaseous medium rises and vice versa.

(e) Now assume that,

${v_m}$ and ${V_d}$ be the speeds of sound in moist air and dry air respectively.

And ${\rho _n}$ and ${\rho _A}$ be the densities of moist air and dry air respectively.

Take the relation:

$v = \sqrt {\dfrac{{yP}}{\rho }} $

Hence, the speed of sound in moist air is:

${v_n} = \sqrt {\dfrac{{yP}}{{{\rho _m}}}} $

And the speed of sound in dry air is:

${v_f} = \sqrt {\dfrac{{yP}}{{{\rho _d}}}} $

On dividing equations $(1)$ and (ii), we get:

$\dfrac{{{v_m}}}{{{v_A}}} = \sqrt {\dfrac{{1P}}{{{\rho _x}}} \times \dfrac{\rho }{{y\rho }}} = \sqrt {\dfrac{{{\rho _A}}}{{{P_m}}}} $

However, the presence of water vapor reduces the density of air, 1e.,

${\rho _d} < {\rho _m}$

$\therefore {V_k} > {v_6}$

As a result, sound travels faster in moist air than in dry air. As a result, the speed of sound increases with humidity in a gaseous medium.

2. A transverse harmonic wave on a string is described by

$y(x,t) = 3.0$ sin $\left( {36t + 0.018x + \dfrac{\pi }{4}} \right) \ldots .......(i)$

Where $x$ and $y$ are in $c m$ and $t$ in s. The positive direction of $x$ is from left to right.

(a) Is this a travelling wave or a stationary wave?

If it is travelling, what are the speed and direction of its propagation?

(b) What are its amplitude and frequency?

(d) What is the least distance between two successive crests in the wave?

Ans: (a) Yes; Speed = $20\;{\text{m}}4$ , Direction = Right to left

(b) $3\;{\text{cm}},5.73{\text{Hz}}$

(d) $3.49\;{\text{m}}$

(a) The equation of a progressive wave travelling from right to Left is given by the displacement function:

$y(x,t) = a\sin (\omega t + kx + \theta ) \ldots (2)$

The given equation is:

$y(x,t) = 3.0$ sin $\left( {36t + 0.018x + \dfrac{\pi }{4}} \right).... \ldots (ii)$

On comparing both the equations, we find that equation $(i)$ represents a travelling wave, propagating from right to left.

Now, using equations $(i)$ and (ii), we can write:

$\phi = 36{\text{rad}}/{\text{s}}$ and $k = 0.018\;{{\text{m}}^{ - 1}}$

We know that:

$v = \dfrac{\omega }{{2\pi }}$ and $\lambda = \dfrac{{2\pi }}{k}$

$\therefore v = \left( {\dfrac{\omega }{{2\pi }}} \right) \times \left( {\dfrac{{2\pi }}{k}} \right) = \dfrac{\omega }{k}$

$ = \dfrac{{36}}{{0.018}} = 2000\;{\text{cm}}/{\text{s}} = 20\;{\text{m}}/{\text{s}}$

Hence, the speed of the given travelling wave is $20\;{\text{m}}$ k.

(b) Amplitude of the given wave, $a = 3\;{\text{cm}}$

Frequency of the given wave:

$v = \dfrac{\omega }{{2\pi }} = \dfrac{{36}}{{2x3.14}} = 5.73\;{\text{Hz}}$

(d) The distance between two successive crests ơ troughs is equal to the wavelength of the wave.

Wavelength is given by the relation:

$k = \dfrac{{2\pi }}{\lambda }$

$\therefore k = \dfrac{{2\pi }}{\lambda } = \dfrac{{2 \times 3.14}}{{0.018}} = 348.89\;{\text{cm}} = 3.49\;{\text{m}}$

3 . For the wave described in Exercise 15.8, plot the displacement $(y)$ versus $(t)$ graphs for $x = 0,2$and $4\;{\text{cm}}$. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?

Ans: All the waves have different phases.

The given transverse harmonic wave is

$y(x,t) = 3.0\sin \left( {36t + 0.018x + \dfrac{\pi }{4}} \right)(1)$

For $x = 0$ , the equation reduces to:

$y(0,t) = 3.0\sin \left( {36t + \dfrac{\pi }{4}} \right)$

Also, $\omega = \dfrac{{2\pi }}{T} = 36{\text{rad}}/{{\text{s}}^{ - 1}}$

$\therefore T = \dfrac{\pi }{8}s$

Now, plotting $y$ vs. t graphs using the different values of $t$ , as listed in the given table.

For $x = 0,x = 2$ , and $x = 4$ , the phases of the three waves wit get changed. This is because amplitude and frequency are invariant for any change in $x$ . The $y$ -t plots of the three waves are shown in the given figure.

4. The transverse displacement of a string (clamped at its both ends) is given by

$y(x,t) = 0.06\sin \dfrac{2}{3}x\cos (120\pi t)$

Where $x$ and $y$ are in $m$ and $t$ in $s$. The length of the string is $1.5\;{\text{m}}$ and its mass is $3.0 \times {10^{ - 2}}\;{\text{kg}}$

Answer the following:

(a) Does the function represent a travelling wave or a stationary wave?

(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?

Ans: (a)The general equation representing a stationary wave is given by the displacement function:

$y(x,t) = 2a\sin kxdb{\cos ^n}{\text{r}}$

This equation is similar to the given equation:

$y(x,t) = 0.06\sin \left( {\dfrac{2}{3}x} \right)\cos (120\pi t)$

Hence, the given function represents a stationary wave.

(b) A wave travelling along the positive $x$ -direction is given as:

${y_1} = a\sin \left( {{\alpha ^2} - kx} \right)$

The wave travelling along the negative $x$ -direction is given as:

${y_2} = a\sin (ct - kx)$

The superposition of these two waves yields:

$y = {y_1} + {y_2} = a\sin (\omega t - kx) = a\sin (\omega t + kx)$

$ = a\sin (\omega r)\cos (kx) = a\sin (kx)\cos (ar) = a\sin (\omega t)\cos (kx) = a\sin (kx)\cos (\omega t)$

$ = - 2a\sin (kx)\cos (at)$

$ = - 2a\sin \left( {\dfrac{{2\pi }}{\lambda }x} \right)\cos (2\pi v) \ldots \ldots \ldots (1)$

The transverse displacement of the string is given as:

$y(x,t) = 0.06\sin \left( {\dfrac{{2\pi }}{3}x} \right)\cos (120\pi )$

Comparing equations $(1)$ and $(10)$ , we have:

$\dfrac{{2\pi }}{\lambda } = \dfrac{{2\pi }}{3}$

Wavelength, ${\text{h}} = 3\;{\text{m}}$

$120\pi = 2\pi h$

Frequency, ${\text{v}} = 60{{\text{H}}_2}$

Wave speed, $v = $ vh

$ = 60 \times 3 = 180\;{\text{m}}/{\text{s}}$

(e) The velocity of a transverse wave travelling in a string is given by the relation:

$v = \sqrt {\dfrac{T}{\lambda }} \cdots \ldots \ldots (1)$

Velocity of the transverse wave, $v = 180\;{\text{m}}/{\text{s}}$

Mass of the string, $m = 3.0 \times {10^{ - 1}}\;{\text{kg}}$

Length of the string, $l = 1.5\;{\text{m}}$

Mass per unit length of the string $\lambda = \dfrac{m}{l}$

$ = \dfrac{{3.0}}{{1.5}} \times {10^{ - 2}}$

$ = 2 \times {10^{ - 1}}\;{\text{kg}}\;{{\text{m}}^{ - 1}}$

Tension in the string $ = T$

From equation (i), tension can be obtained as:

${\text{T}} = {{\text{v}}^2}\mu $

$ = {(180)^2} \times 2 \times {10^{ - 2}}$

$ = 648\;{\text{N}}$

5. A travelling harmonic wave on a string is described by

$y(x,t) = 7.5{\text{sin}}\left( {0.0050x + 12t + \dfrac{\pi }{4}} \right)$

(a) What are the displacement and velocity of oscillation of a point at $x = 1\;{\text{cm}}$, 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 $t = 2s,5s$ and 11 s.

Ans: (a) The given harmonic wave is:

$y\left( {{x_1}t} \right) = 7,5\sin \left( {0.0050x + 12t + \dfrac{\pi }{4}} \right)$

Far $x = 1\;{\text{cm}}$ and $t = 15$ ,

$y(1,1) = 7.5\sin \left( {0.0050x + 12t + \dfrac{\pi }{4}} \right)$

$ = 7.5\sin \left( {12.0050 + \dfrac{\pi }{4}} \right)$

Where, $\theta = 12.0050 + \dfrac{\pi }{4} = 12.0050 + \dfrac{{3.14}}{4} = 12.79{\text{rad}}$

$ = \dfrac{{180}}{{3.14}} \times 12.79 = {732.81^{\prime \prime }}$

$\therefore y = (1,1) = 7.5\sin \left( {{{732.81}^*}} \right)$

$ = 7.5\sin (90 \times 8 + 12.81) = 7.5\sin {{12.81}^{0}}$

$ = 7.5 \times 0.2217$

$ = 1.6629 \times 1.663\;{\text{cm}}$

The velocity of the oscillation at a given point and time is given as

$v = \dfrac{d}{{dt}}y\left( {{x_1}t} \right) = \dfrac{d}{{dt}}\left[ {7.5\sin \left( {0.0050x + 12t + \dfrac{\pi }{4}} \right)} \right]$

$7.5 \times 12\cos \left( {0.0050x + 12t + \dfrac{\pi }{4}} \right)$

A ${t_x} = 1$ cm and t=1 1 :

$v = y(1,1) = 90\cos \left( {12.005 + \dfrac{\pi }{4}} \right)$

\[=90\cos 73281{}^\circ =90\cos \left( 90\cos \text{ }\!\!~\!\!\text{ }\times 8+{{12.81}^{*}} \right)\]

$ = 90\cos \left( {{{12.81}^{\prime \prime }}} \right)$

$ = 90 \times 0.975 = 87.75\;{\text{cm}}/{\text{s}}$

Now, the equation of a propagating wave ss given by:

$y\left( {{x_n}t} \right) = a\sin (kx + wt + \phi )$

$\therefore k = \dfrac{{2\pi }}{\lambda }$

And $\theta = 2\pi v$

$\therefore v = \dfrac{\omega }{{2\pi }}$

speed, $v = v\lambda = \dfrac{\omega }{k}$

$\omega = 12{\text{rad}}/{\text{s}}$

${\text{K}} = 0.0050\;{{\text{m}}^{ - 1}}$

$\therefore v = \dfrac{{12}}{{0.0050}} = 2400\;{\text{cm}}/{\text{s}}$

Hence, the velocity of the wave oscillation at $x = 1\;{\text{cm}}$ and $t = 1\;{\text{s}}$ is not equal to the velocity of the wave propagation.

(b) Propagation constant is related to wavelength as:

$\therefore A = \dfrac{{2\pi }}{k} = \dfrac{{2 \times 3.14}}{{0.0050}}$

$ = 1253\;{\text{cm}} = 12.56\;{\text{m}}$

Therefore, all the points at distances $\lambda .(n = \pm 1, \pm 2, \ldots $ and $s00n)$ , Le. $ \pm 12.56\;{\text{m}}, \pm 25.12$ ${\text{m}}$ , . and $s$ on for $x = 1\;{\text{cm}}$ , will have the same displacement as the $x = 1\;{\text{cm}}$ points at $t = 2\;{\text{s}}$ , 5 s, and $11\;{\text{s}}$ .

Important Related Links for CBSE Class 11

FAQs on Important Questions for CBSE Class 11 Physics Chapter 15 - Waves

1. Is Class 11 Physics Chapter 15 difficult?

All the chapters of Class 11 Physics require serious effort and striving hard on the part of the students. The students may find some topics difficult. But it is also an important chapter. Thus, you need to devote time, effort, and energy to understanding, reading and practising this chapter with all that you got. In fact, without sheer hard work, all the chapters of Class 11 Physics will seem challenging.

2. Why are waves important?

The area of "Waves" forms an important aspect of Physics. If you understand the workings of a wave, you will be able to explain various physical phenomena like the wave properties of matter. We encounter several waves in our daily lives including light waves, radio waves, microwaves, X-rays, etc. If you aspire to have a career in Engineering or a field of expertise related to Physics, then you need to have a good understanding of the working of waves.

3. What are some important questions from Chapter 15 “waves”?

As we have mentioned previously, Chapter 15 "Waves'' is an important chapter that carries significant marks in exams. Hence students should practice the important questions from this chapter, such as;

Why are all stringed instruments provided with hollow boxes?

Show that the Doppler effect in sound is asymmetric?

The chapter has many more important questions, which can be accessed from the page Important questions for Class 11 Physics and can also be downloaded at absolutely no cost.

4. Is “Waves” important for NEET?

"Waves" and "Oscillations" form an important part of the NEET syllabus. Wave mechanics usually carry around 4% of the weightage in NEET exams. This chapter is important also for CBSE exams and also for understanding basic concepts of Physics for Class 12th. So even NEET aspirants need to practice this chapter well. You can always refer to Vedantu's app to access the resources for help in exam preparation; be it for CBSE or NEET exams. You can find all the important questions from this chapter by visiting the page Important questions for Class 11 Physics.

5. What are beats?

When two sound waves of similar but different frequencies and amplitudes travel in the same direction, they form beats. Two slightly different frequencies having comparable amplitudes, superpose forming beats. The concept might appear complicated. But interestingly we experience beats around us in our lives very frequently. However, the fact that these beats are not physically visible makes them hard to comprehend. We experience beats while striking a fork or playing the piano. Beats are necessary for musicians.

CBSE Class 11 Physics Important Questions

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Class 11 Physics Practicals Viva Questions and Answers

Physics Class 11 Viva Questions and Answers

Physics viva questions are questions intended to examine a student’s understanding of the core concepts behind each physics experiment.

Concrete knowledge about each fundamental topic behind the experiments is the best way to prepare for practical lab exams. If students are able to answer questions regarding the fundamentals of the experiment, then they will be able to attend the practical test without any serious hassles. Viva questions will extensively motivate students to study the fundamental ideas behind each experiment.

Viva questions have a different approach compared to numerical and analytical questions. Such questions do not attempt to ask the fundamental questions related to the core concepts behind the experiments. In order to answer the complex and application-level questions, one should get a hold of all the basic concepts. This is what viva questions bring. It challenges the basic awareness of students regarding the inherent concepts behind the lab experiments.

Viva questions demand more than just simple recall. Such questions force students to cover all the core topics that drive each practical experiment.

Here is the list of questions and answers on various core physics concepts to help you understand the fundamentals in detail.

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Class 11 Physics Case Study Questions Chapter 8 Gravitation

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In Class 11 Final Exams there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Downloads of CBSE Class 11 Physics Chapter 8 Case Study and Passage-Based Questions with Answers were Prepared Based on the Latest Exam Pattern. Students can solve Class 11 Physics Case Study Questions Gravitation to know their preparation level.

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

In CBSE Class 11 Physics Paper, There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Gravitation Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 11 Physics Chapter 8 Gravitation

Case Study/Passage-Based Questions

Case Study 1: Satellites in a circular orbit around the earth in the equatorial plane with T = 24 hours are called Geostationary Satellites. Clearly, since the earth rotates within the same period, the satellite would appear fixed from any point on Earth. It takes very powerful rockets to throw up a satellite to such large heights above the earth but this has been done in view of the several benefits of much practical application. The weight of an object is the force with which the earth attracts it. We are conscious of our own weight when we stand on a surface since the surface exerts a force opposite to our weight to keep us at rest. The same principle holds good when we measure the weight of an object by a Spring balance hung from a fixed point e.g. the ceiling. The object would fall down unless it is subject to a force opposite to gravity. This is exactly what the spring exerts on the object. This is because the spring is pulled down a little by the gravitational pull of the object and in turn, the spring exerts a force on the object vertically upwards. Now, imagine that the top end of the balance is no longer held fixed to the top ceiling of the room. Both ends of the spring as well as the object move with identical acceleration g. The spring is not stretched and does not exert any upward force on the object which is moving down with acceleration g due to gravity. The reading recorded in the spring balance is zero since the spring is not stretched at all. If the object were a human being, he or she will not feel his weight since there is no upward force on him. Thus, when an object is in free fall, it is weightless and this phenomenon is usually called the phenomenon of weightlessness. In a satellite around the earth, every part and parcel of the satellite has acceleration towards the center of the earth which is exactly the value of the earth’s acceleration due to gravity at that position. Thus in the satellite, everything inside it is in a state of free fall. This is just as if we were falling towards the earth from a height. Thus, in a manned satellite, people inside experience no gravity. Gravity for us defines the vertical direction and thus for them there are no horizontal or vertical directions, all directions are the same.

1) Astronaut experiences weightlessness in space because

a) Acceleration due to gravity is zero
b) Actual weight of the astronaut is zero
c) They are going with the same acceleration due to gravity
d) None of these

Answer: c) They are going with same acceleration due to gravity

2) Weighing machine measures

a) Mass of the person
b) Normal reaction exerted by machine on person
c) Both a and b

Answer:b) Normal reaction exerted by machine on person

What are Geostationary Satellites? A) Satellites rotating faster than the Earth. B) Satellites in a circular orbit with T = 24 hours, appearing fixed from any point on Earth. C) Satellites that are closer to the poles. D) Satellites that rotate opposite to the Earth’s rotation.

Why do we feel our weight when standing on a surface? A) Because gravity pushes us upwards. B) Because the surface exerts a force opposite to our weight, keeping us at rest. C) Because our weight is always directed sideways. D) Because of the magnetic field of the Earth.

What is recorded in a spring balance when an object is in free fall? A) The weight of the object. B) Double the weight of the object. C) Zero, since the spring is not stretched. D) An erratic value due to gravitational forces.

Why do objects inside a satellite experience weightlessness? A) Because gravity doesn’t exist in space. B) Because everything inside the satellite is in a state of free fall. C) Because the satellite moves at the speed of light. D) Because objects inside are not affected by gravity.

What is the consequence of weightlessness for people inside a manned satellite? A) They feel heavier than on Earth. B) They can distinguish between horizontal and vertical directions. C) They feel no gravity and there are no distinct horizontal or vertical directions. D) They are attracted to the walls of the satellite.

How is weight measured using a spring balance? A) The spring exerts a force on the object vertically downwards. B) The object pulls the spring down by gravitational pull, and the spring exerts an upward force. C) The spring is compressed by the object’s weight. D) The object’s weight causes the spring to expand horizontally.

Hope the information shed above regarding Case Study and Passage Based Questions for Class 11 Physics Chapter 8 Gravitation with Answers Pdf free download has been useful to an extent. If you have any other queries about the CBSE Class 11 Physics Gravitation Case Study and Passage-Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible. By Team Study Rate

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Case Study Questions for Class 9 Science Chapter 1 Matter in Our Surroundings

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Here we are providing case study questions for class 9 science chapter 12 sound. Students are suggested to go through each and every case study questions for better understanding of the chapter.

Case Study/Passage Based Questions:

Question 1:

Read the following passage and answer the questions given below.

Every matter is made up of tiny particles. These particles are so tiny that they can’t be seen with naked eyes.

The three characteristics shown by particles of matter are as follows:

(i) There are small voids between particles in a matter. This characteristic is the concept behind the solubility of a substance in other substances.

(ii) Particles of matter show continuous random movements, that is they possess kinetic energy. The spreading of ink in a beaker of glass, smell of agarbattis, etc. are few illustrations that show the movement of particles of a substance.

(iii) The particles of matter attract each other with a force called interparticle force of attraction. Read the given passage carefully and give the answer of the following questions:

Q 1. Spreading of fragrance of a burning incense stick in a room shows that:

a. particles of matter have spaces between them.

b. particles of matter attract each other.

c. particles of matter are constantly moving.

d. None of the above

Q 2. What happens when we add sugar to water?

a. Volume of water doubles.

b. Volume of water decreases

c. Volume of water remains the same.

Q 3. A stream of water cannot be cut by fingers. Which property of matter does this observation show?

a. Particles of matter attract each other.

b. Particles of matter have spaces between them.

c. Particles of matter are continuously moving.

Q 4. When we put some crystals of potassium permanganate in a beaker containing water, we observe that after some time, the whole water turns pink. This intermixing of particles of two different types of matter on their own is called:

a. Brownian motion

c. sublimation

d. diffusion

Q 5. Why is the rate of diffusion of liquids higher than that of solids?

a. In the liquid state, particles are tightly packed as compared to solids.

b. In the liquid state, particles move freely as compared to solids.

c. In solid state, particles have least force of attraction between the particles.

d. In solid state, particles cannot be compressed easily.

(c) particles of matter are constantly moving.
(c) Volume of water remains the same.
(a) Particles of matter attract each other.
(d) diffusion
(b) In the liquid state, particles move freely as compared to solids

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Class 11th Physics - Thermal Properties of Matter Case Study Questions and Answers 2022 - 2023

By QB365 on 09 Sep, 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 11 Physics Subject - Thermal Properties of Matter, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Thermal properties of matter case study questions with answer key.

11th Standard CBSE

Final Semester - June 2015

Heat is a form of energy which produces the sensation of warmth. It is total thermal energy of the body which is sum of kinetic energies of all the individual molecules of the body. It is transferred from one body to the other on account of temperature difference between two bodies. Joule found that when' mechanical work (W) is converted into heat (Q), then the ratio of Wand Q is always constant, represented by J i.e., Joule's mechanical equivalent of heat. $J=\frac{W}{Q} \text { or } W=J Q$ A measure of temperature is obtained using a thermometer that use Physical Properties which change uniformly with temperature, for example, in common liquid in glass thermometers, mercury, alcohol etc are used whose volume varies linearly with temperature over a wide range. (i) Define temperature and its significance. (ii) Define One calorie. (iii) How does a thermometer work? (iv) Name the two convenient fixed reference point for measuring temperature. (v) How celsius temperature and Fahrenheit scale are related to each other? (vi) What is absolute zero temperature? (vii) What is Kelvin scale? How it is related with celsius scale?

When solid is heated, the amplitude of vibration of atoms and molecules increases. Therefore effective interatomic separation increases and cause thermal expansion. Thermal expansion of solids are of three types: (i) Linear expansion (ii) Area expansion (iii) Volume expansion. Thermal expansion of liquid are of two kinds. Real expansion and apparent expansion that occurs in form of volume expansion. Thermal expansion of water is anomalous i.e., volume of given amount of water first decreases with increase in temperature from 0 °C to 4 °C and beyond 4 °C volume of water increases with rise of temperature. In thermal expansion of gases is more than solid and liquid but coefficient of volume expansion is dependent on temperature for gases. (i) On what factors does the coefficient of thermal expansion depend? Write its S.I. unit. (ii) Write the relation between the three coefficients of expansion $\alpha, \beta$ and $\gamma$ for solid. (iii) A body at higher temperature contains more heat. Comment. (iv) Why are clock pendulums usually made of invar? (v) Draw graphical variation of volume and temperature for water. (vi) How is coefficient of thermal expansion of gases related to temperature? (vii) State water equivalent.

States of matter viz: solid, liquid and gas are function of temperature and heat content. During the change of state of a substance, the exchange of heat takes place between the substance and surrounding. In this process temperature of substance remains constant. At certain temperature known as melting point. Both the solid and liquid states of the substance coexist in thermal equilibrium. Similarly, at boiling point both the liquid and vapour states of the substance co-exist in the thermal equilibrium. There are certain substance which on heating directly pass from solid to vapour state without passing through the liquid state. This is sublimation process in which solid changes to vapour state of the substance. Process of change of state depends on pressure and temperature. (i) Define triple point. (ii) Define latent heat of a substance. (iii) What is principle of calorimetry? (iv) What is effect of pressure on the melting point of a substance? (v) State phenomenon of relegation. (vi) What is boiling point? What is effect of pressure on the boiling point? (vii) What is sublimation?

Three cylindrical rods A, Band C of equal lengths and equal diameters are joined in series as shown in the figure. Their thermal conductivities are 2K, K and 0.5 K respectively. 100°C $\begin{array}{|l|l|l|} \hline \mathrm{A} & \mathrm{B} & \mathrm{C} \\ \hline \end{array}$ 0°C In the steady-state, the free ends of rods A and C are at 100 °C and a 0°C respectively. Neglecting loss of heat from the curved surfaces of rods. (i) Determine the temperature of the junction between rods A and B. (ii) Determine the temperature of the junction between rods Band C. (iii) Determine the equivalent thermal conductivity of the combination. (iv) Define coefficient of thermal conductivity of a solid. Write its S.I unit. Two rods A and B are of equal length. Each rod has the ends at temperature T 1 and T 2 What is the condition that will ensure equal rates of flow of heat through the rods A and B?

All bodies emit heat energy from their surface by virtue of their temperature. This heat energy is called radiant energy or thermal radiation. The heat that we receive from the sun is transferred to us by a process which, unlike conduction and convection, does not require the help of a medium in intervening space which is almost free of particles. Radiant energy travels in space as electromagnetic waves in the infra-red region of electromagnetic spectrum. They exhibit the phenomenon of interference, diffraction, and polarization as light does. The emission of radiation from a hot body is expressed in terms of that emitted from a reference body (called the black body) at the same temperature. A black body absorbs and emits radiations of all wavelengths. The total energy E emitted by a unit area of a black body per second is given by E = $\sigma T^{4}$ Where T - is absolute temperature of the body and a is Stefan's constant, if the body is not the perfect black body, then E = $\varepsilon \sigma T^{4}$ , where $\varepsilon$ is the emissivity of the body. (i) Determine the dimensions of Stefan's constant a from Stefan - Boltzman law. (ii) What is S.I unit of Stefan's constant? (iii) In which region of the electromagnetic spectrum do thermal radiation lie? (iv) Which device is used to detect thermal radiation? (v) When a body A at a higher temperature T 1 is surrounded by another body B at a lower temperature T 2 . Write the relation between the rate of loss of heat from body A and temperature. (vi) On what factor does the rate at which energy is radiated by a body depends? (vii) On which parameter does the colour of a star depends upon?

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Thermal properties of matter case study questions with answer key answer keys.

(i) Temperature of a body is the measure of degree of hotness or coldness. It determines the direction of flow of heat when this body is placed in contact with another body i.e., hotter end to colder end. (ii) One calorie is the amount of heat required at a pressure of I atm. to raise the temperature of 1 gram of water through 1 o C. 1 calorie = 4.186 Joules. (iii) Thermometer is based on change in property of matter with temperature. Some common properties are: (i) variation of volume of liquid with temperature. (ii) variation of pressure with temperature. (iii) variation of resistance of metal with temperature. (iv) variation of thermo emf with temperature of a junction in a thermocouple etc. (iv) Reference point for measuring temperature are steam point and ice point of water at standard pressure. (v) $\frac{T_{C}}{100}=\frac{T_{F}-32}{180}$ (vi) It is the temperature at which pressure P of gas would reduce to zero so also volume V of the gas. It is the lowest attainable temperature. 0 °K = -273.15 °C. (vii) Scale of temperature with its zero at -273.15 °C and size of each degree same as on the celsius scale, is called kelvin scale T K = t °C + 273.15

(i) It is a particular point on the P- T phase diagram representing a particular pressure and temperature at which the substance can co-exist in all the three states viz. Solid, liquid and vapour. (ii) It is the amount of heat energy required to change the state of unit mass of a substance from solid to liquid or from liquid to vapour without any change in temperature. (iii) Calorimeter works on the principle that heat lost by one body is equal to heat gained by other provided no heat escape from system to surrounding in any form. (iv) Melting point of a substance decreases with increase in pressure. (v) Phenomenon of refreezing of ice when pressure is reduced is known as relegation. (vi) The temperature at which the liquid and vapour states of substance are in thermal equilibrium with each other i.e., vapour pressure becomes equal to external pressure. Boiling point of a substance increases with increase in external pressure. (vii) On heating a substance, the change from solid state to vapour state without passing through the liquid state is called sublimation. e.g., camphor, iodine and solid carbon dioxide are sublime substance.

(i) In the steady-state, the rate of flow of heat is the same for all rods. If T 1 and T 2 are the temperatures at the junction points between A, Band C respectively. Then $\frac{Q}{t}=\frac{K_{A} A\left(100-T_{1}\right)}{d}$ = $\frac{K_{B} A\left(T_{1}-T_{2}\right)}{d}$ = $\frac{K_{C} A\left(T_{2}-0\right)}{d}$ Given K A =2K 1 K B = K and K c =0.5K. Hence 2(100 - T 1 ) = (T 1 - T 2 ) = 0.5(T 2 - 0) and T 1 - T 2 = 0.5T 2 T 1 = 85.7 °C. (ii) Since T 1 = 85.7 °C, therefore T 2 = 57.1 °C using 200 - 2T 1 = 0.5T 2 (iii) Since the rods have the same lengths and the same diameters, the equivalent thermal conductivity of the series combination is $\frac{1}{K_{e}}=\frac{1}{K_{A}}+\frac{1}{K_{B}}+\frac{1}{K_{C}}$ = $\frac{1}{2 K}+\frac{1}{K}+\frac{1}{0.5 K}$ K e = $\frac{2 K}{7}$ (iv) Coefficient of thermal conductivity of a solid is equal to the rate of flow of heat per unit area per unit temperature gradient across the solid. S.I unit of (Coefficient of thermal conductivity) K is Wm -1 K -1 . (v) As $\frac{\Delta Q_{1}}{\Delta t}=\frac{\Delta Q_{2}}{\Delta t}$ $\therefore \quad K_{1} A_{1} \frac{\left(T_{1}-T_{2}\right)}{L_{1}}=K_{2} A_{2} \frac{\left(T_{1}-T_{2}\right)}{L_{2}}$ $\Rightarrow \quad \frac{A_{1}}{A_{2}}=\frac{K_{2}}{K_{1}}$ i.e., the cross-sectional area of two rods must be in inverse ratio of their thermal conductivities.

(i) Stefan - Boltzman law states that E = $\sigma T^{4}$ where E - stands for the total energy emitted per unit area per second. Thus dimensions of E = dimensions of $\left(\frac{\text { Energy }}{\text { area } \times \text { time }}\right)$ i.e., $\frac{\mathrm{ML}^{2} \mathrm{~T}^{-2}}{\mathrm{~L}^{2} \mathrm{~T}}=\mathrm{MT}$ ,therefore, dimension of $\sigma=\frac{\mathrm{MT}^{-3}}{\mathrm{~K}^{4}}=\mathrm{MT}^{-3} \mathrm{~K}^{-4}$ (ii) S.I unit of Stefan's constant is Wm -2 K -4 (iii) Thermal radiations lie in the infrared region of electro magnetic spectrum. (iv) Thermopile is used to detect thermal radiation. (v) $E \propto\left(T_{1}^{4}-T_{2}^{4}\right)$ (vi) The rate at which energy is radiated by a body depends upon. (i) the surface area of the body (ii) the temperature of the body (iii) nature of the surface of the body (iv) the emissivity of the surface of the body. (vii) According to Wein's law $\lambda_{\mathrm{m}} \mathrm{T}$ = Constant. Colour of star depends upon its surface temperature.

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