{"id":23909,"date":"2021-06-04T13:12:43","date_gmt":"2021-06-04T07:42:43","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=23909"},"modified":"2021-06-04T13:12:43","modified_gmt":"2021-06-04T07:42:43","slug":"ncert-solutions-for-class-9-science-chapter-12","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-science-chapter-12\/","title":{"rendered":"NCERT Solutions for Class 9 Science Chapter 12 Sound"},"content":{"rendered":"

These NCERT Solutions for Class 9 Science<\/a> Chapter 12 Sound Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.<\/p>\n

Sound NCERT Solutions for Class 9 Science Chapter 12<\/h2>\n

Class 9 Scien<\/span>ce Chapter 12 Sound InText Questions and Answers<\/span><\/h3>\n

Question 1.
\nHow does the sound produced by a vibrating object in a medium reach your ear?
\nAnswer:
\nWhen an object vibrates, it forces the neighbouring particles of the medium to vibrate. These vibrating particles then force the particles adjacent to them to vibrate. In this way, vibrations produced by an object are transferred from one particle to another till it reaches the ear.<\/p>\n

Question 2.
\nExplain how sound is produced by your school bell.
\nAnswer:
\nWhen the school bell vibrates, it forces the adjacent particles in air to vibrate. This disturbance gives rise to a wave and when the bell moves forward, it pushes the air in front of it. This creates a region of high pressures known as compression. When the bell moves backwards, it creates a region of low pressure know as rarefaction. As the bell continues to move forward and backward, it produces a series of compressions and rarefactions. This makes the sound of a bell propagate through air.<\/p>\n

Question 3.
\nWhy are sound waves called mechanical waves?
\nAnswer:
\nSound waves force the medium particles to vibrate. Hence, these waves are known as mechanical waves. Sound waves propagate through a medium because of the interaction of the particles present in that medium.<\/p>\n

Question 4.
\nSuppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
\nAnswer:
\nSound needs a medium to propagate. Since the moon is devoid of any atmosphere, you cannot hear any sound on the moon.<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nDistinguish between loudness and intensity of sound.
\nAnswer:
\nIntensity of a sound wave is defined as the amount of sound energy passing through a unit area per second. Loudness is a measure of the response of the ear to the sound. The loudness of a sound is defined by its amplitude. The amplitude of a sound decides its intensity, which in turn is perceived by the ear as loudness.<\/p>\n

Question 6.
\nHow are the wavelength and frequency of a sound wave related to its speed?
\nAnswer:
\nSpeed, wavelength, and frequency of a sound wave are related by the following equation:
\nSpeed (v) = Wavelength (\u03bb) \u00d7 Frequency (u)
\nv = \u03bb \u00d7 u<\/p>\n

Question 7.
\nCalculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m\/s in a given medium.
\nAns. Frequency of the sound wave, \u03c5 = 220 Hz
\nSpeed of the sound wave, v = 440 m s-1<\/sup>
\nFor a sound wave,
\nSpeed = Wavelength \u00d7 Frequency
\nv = \u03bb \u00d7 u
\n\u2234 \u03bb = \\(\\frac{v}{u}=\\frac{440}{220}\\) = 2 m
\nHence, the wavelength of the sound wave is 2m.<\/p>\n

Question 8.
\nA person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?
\nAnswer:
\nThe time interval between two successive compressions is equal to the time period of the wave. This time period is reciprocal of the frequency of the wave and is given by the relation:
\nT = \\(\\frac{1}{\\text { Frequency }}=\\frac{1}{500}\\) = 0.002s<\/p>\n

Question 9.
\nIn which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
\nAnswer:
\nThe speed of sound depends on the nature of the medium. Sound travels the fastest in solids. Its speed decreases in liquids and it is the slowest in gases.
\nTherefore, for a given temperature, sound travels fastest in iron.<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nAn echo returned in 3s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 ms-1<\/sup>?
\nAnswer:
\nSpeed of sound, v = 342 ms-1<\/sup>
\nEcho returns in time, t = 3s
\nDistance travelled by sound = v \u00d7 t = 342 \u00d7 3 = 1026 m
\nIn the given time interval, sound has to travel a distance that is twice the distance of the reflecting surface and the source.
\nHence, the distance of the reflecting surface from the source = \\(\\frac{1026}{2}\\) m = 513 m<\/p>\n

Question 11.
\nWhy are the ceilings of concert halls curved?
\nAnswer:
\nCeilings of concert halls are curved so that sound after reflection (from the walls) spreads uniformly in all directions.<\/p>\n

Question 12.
\nWhat is the audible range of the average human ear?
\nAnswer:
\nThe audible range of an average human ear lies between 20 Hz to 20,000 Hz. Humans cannot hear sounds having frequencies less than 20 Hz and greater than 20,000 Hz.<\/p>\n

Question 13.
\nWhat is the range of frequencies associated with
\n(a) Infrasound?
\n(b) Ultrasound?
\nAnswer:
\n(a) Infrasound has frequencies less than 20 Hz.
\n(b) Ultrasound has frequencies more than 20,000 Hz.<\/p>\n

Question 14.
\nA submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in saltwater is 1531 m\/s, how far away is the cliff?
\nAnswer:
\nTime taken by the sonar pulse to return, t = 1.02 s
\nSpeed of sound in saltwater, v = 1531 ms-1<\/sup>
\nTotal distance covered by the sonar pulse = Speed of sound \u00d7 Time taken
\nTotal distance covered by the sonar pulse
\n= 1.02 \u00d7 1531 = 1561.62 …..(i)
\nLet d be the distance of the cliff from the submarine.
\nTotal distance covered by the sonar pulse = 2d
\n\u21d2 2d = 1561.62 [From(i)]
\n\u21d2 d = 780.81 m<\/p>\n

Class 9 Science Chapter 12 Sound Textbook Questions and Answers<\/span><\/h3>\n

Question 1.
\nWhat is sound and how is it produced?
\nAnswer:
\nSound is produced by vibration. When a body vibrates, it forces the neighbouring particles of the medium to vibrate. This creates a disturbance in the medium, which travels in the form of waves. This disturbance, when reaches the ear, produces sound.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nDescribe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
\nAnswer:
\nWhen a vibrating body moves forward, it creates a region of high pressure in its vicinity. This region of high pressure is known as compressions. When it moves backward, it creates a region of low pressure in its vicinity. This region is known as a rarefaction. As the body continues to move forward and backwards, it produces a series of compressions and rarefactions (as shown in the following figure).
\n\"NCERT<\/p>\n

Question 3.
\nCite an experiment to show that sound needs a material medium for its propagation.
\nAnswer:
\nTake an electric bell and hang this bell inside an empty bell-jar fitted with a vacuum pump (as shown in the following figure).
\n\"NCERT
\nInitially, one can hear the sound of the ringing bell. Now, pump out some air from the bell-jar using the vacuum pump. It will be observed that the sound of the ringing bell decreases. If one keeps on pumping the air out of the bell-jar, then at one point, the glass-jar will be devoid of any air. At this moment, no sound can be heard from the ringing bell although one can see that the prong of the bell is still vibrating. When there is no air present inside, we can say that a vacuum is produced. Sound cannot travel through vacuum. This shows that sound needs a material medium for its propagation.<\/p>\n

Question 4.
\nWhy is sound wave called a longitudinal wave?
\nAnswer:
\nThe vibration of the medium that travels along or parallel to the direction of the wave is called a longitudinal wave. In a sound wave, the particles of the medium vibrate in the direction parallel to the direction of the propagation of disturbance. Hence, a sound wave is called a longitudinal wave.<\/p>\n

Question 5.
\nWhich characteristics of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
\nAnswer:
\nQuality of sound is that characteristic which helps us identify a particular person. Sound produced by two persons may have the same pitch and loudness, but the quality of the two sounds will be different.<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nFlash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
\nAnswer:
\nThe speed of sound (344 m\/s) is less than the speed of light (3 \u00d7 108<\/sup> m\/s). Sound of thunder takes more time to reach the Earth as compared to light. Hence, a flash is seen before we hear a thunder.<\/p>\n

Question 7.
\nA person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 ms-1<\/sup>.
\nAnswer:
\nFor a sound wave,
\nSpeed = Wavelength \u00d7 Frequency
\nv = \u03bb \u00d7 \u03c5
\nGiven that the speed of sound in air = 344 m\/s
\n(j) For, \u03c5 = 20.Hz
\n\u03bb1 = \\(\\frac{v}{v_{1}}=\\frac{344}{20}\\) = 17.2m
\n(ii) For, \u03c52<\/sub> = 20,000 Hz
\n\u03bb2<\/sub> = \\(\\frac{v}{v_{2}}=\\frac{344}{20,000}\\) = 0.0172 m
\nHence, for humans, the wavelength range for hearing is 0.0172 m to 17.2 m.<\/p>\n

Question 8.
\nTwo children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.
\nAnswer:
\nLet the length of the aluminium rod be d.
\nSpeed of sound wave in aluminium at 25\u00b0C,
\nvAl<\/sub> = 6420 ms-1<\/sup>
\nTherefore, time taken by the sound wave to reach the other end,
\n\\(t_{\\mathrm{Al}}=\\frac{d}{v_{\\mathrm{Al}}}=\\frac{d}{6420}\\)
\nSpeed of sound wave in air at 25\u00b0C, vAir = 346 ms-1<\/sup> .
\nTherefore, time taken by sound wave to reach the other end,
\n\\(t_{\\text {Air }}=\\frac{d}{v_{\\text {Air }}}=\\frac{d}{346}\\)
\nThe ratio of time taken by the sound wave in air and aluminium: \\(\\frac{t_{\\text {Air }}}{t_{\\mathrm{Al}}}=\\frac{\\frac{d}{346}}{\\frac{d}{6420}}=\\frac{6420}{346}=18.55\\)<\/p>\n

Question 9.
\nThe frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
\nAnswer:
\nFrequency is defined as the number of oscillations per second. It is given by the relation:
\nFrequency = \\(\\frac{\\text { Number of oscillations }}{\\text { Total Time }}\\)
\nNumber of oscillations = Frequency \u00d7 Total time
\nGiven, Frequency of sound = 100 Hz
\nTotal time = 1 min = 60 s
\nNumber of oscillations \/ Vibrations = 100 \u00d7 60 = 6000
\nHence, the source vibrates 6000 times in a minute, producing a frequency of 100 Hz.<\/p>\n

Question 10.
\nDoes sound follow the same laws of reflection as light does? Explain.
\nAnswer:
\nSound follows the same laws of reflection as light does. The incident sound wave and the reflected sound wave make the same angle with the normal to the surface at the point of incidence. Also, the incident sound wave, the reflected sound wave, and the normal to the point of incidence all lie in the same plane.<\/p>\n

\"NCERT<\/p>\n

Question 11.
\nWhen a sound is reflected from a distant object, an echo is produced, Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day?
\nAnswer:
\nAn echo is heard when the time interval between the original sound and the reflected sound is at least 0.1 s. The speed of sound in a medium increases with an increase in temperature. Hence, on a hotter day, the time interval between the original sound and the reflected sound will decrease. Therefore, an echo can be heard only if the time interval between the original sound and the reflected sound is greater than 0.1 s.<\/p>\n

Question 12.
\nGive two practical applications of reflection of sound waves.
\nAnswer:<\/p>\n