NCERT Solutions for Class 9 Science Chapter 12

NCERT Solutions for Class 9 Science Chapter 12 Sound intext questions on Page 162 or Page 163 or Page 166 or Page 167 or Page 168 or Page 169 or Page 170 or Page 172 and Exercises in English Medium free to download in PDF updated for new academic session 2020-21. Download Hindi Medium solutions of Class 9 Science Chapter 12 given on पेज 182 के उत्तर or पेज 186 के उत्तर or पेज 187 के उत्तर or पेज 188 के उत्तर or पेज 189 के उत्तर or पेज 190 के उत्तर or पेज 191 के उत्तर or पेज 193 के उत्तर and अभ्यास के प्रश्न उत्तर in PDF format.

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NCERT Solutions for Class 9 Science Chapter 12

Class:	9
Subject:	Science – विज्ञान
Chapter 12:	Sound

9th Science Chapter 12 Answers in English & Hindi Medium

CBSE NCERT Solutions for Class 9 Science Chapter 12 Sound all intext question answers and Exercises question answers are given below to free download. Visit to Discussion Forum to ask your doubts and respond to the questions asked by other users. Download Offline Apps 2020-21 based on latest NCERT Books for session 2020-2021.

Class 9 Science Chapter 12 Solutions Related Pages

Class 9 Science Chapter 12 Important Questions

Distinguish between music and noise?

Music:
1. It has a pleasant effect on the ear.
2. It consists of a series of sound impulses which follow one another regularly.
3. The frequency of musical sound is high
4. There are no sudden changes in amplitude (loudness) of the wave constituting a musical sound.
Noise:
1. It has an unpleasant effect on the ear.
2. The sound impulses do not follow one another regularly.
3. The frequency of noise is low.
4. There are usually sudden changes in amplitude (loudness) of the wave forming a noise.

What is SONAR? How is it used to detect an underwater object and measure its distance?

It is an acronym which means Sound Navigation And Ranging.
Sonar is a device which measures the distance, direction and speed of objects lying under water ultrasonic waves.
The above method of finding the distance of an object is called echo-ranging as it is based on echo principle. The sonar technique is used to:
1. To locate the position of other ships or submarines. Ship to ship communication also uses ultrasonic waves. The advantage of using ultrasonic waves is that these waves cannot be heard without the aid of special instruments.
2. Determine depth of the sea , called echo depth ranging. When used for purpose , it is called fathometer as it directly fathoms the depth of the sea.
3. Locate underwater hills, valleys, icebergs, submarines and sunken ships.

Chandigarh radio station broadcast at 1200 kHz. At what metre Chandigarh station would be tuned in your transistor?

250 m

A hospital uses an ultrasonic scanner to locate tumour in a tissue. What is the wavelength of sound in a tissue in which the speed of sound is 1.7 km/s? The operating frequency of the scanner is 4.2 MHz .

0.0004 m

Questions for Practice on 9th Science Chapter 12

Question 1:
An engine is approaching a hill at constant speed when it is at a distance of 0.9 km, if blows a whistle, whose echo is heard by the driver after 5 s. if the speed of sound is 340 m/s, calculate the speed of the engine.
Answer 1:
20 m/s

Question 2:
A person standing between two vertical cliffs and 680 m away from the nearest cliff, shouted . he heard the first echo after 4 s and the second echo 3 s later, calculate (a) the speed of sound in air and (b) distance from the two cliffs.
Answer 2:
(a) 340 m/s
(b) 1870 m

Question 3:
A sonar emits pulses on the surface of water which are detected after reflection from its bottom at a depth of 1534 m. if the time interval between the emission and detection of the pulse is 2 s, find the speed of sound in water.
Answer 3:
1531 m/s

Question 4:
A bat emits unit ultrasonic sound of frequency 100 kHz in air. If this sound meets a water surface, what is the wavelength of (a) the reflected sound wave (b) the transmitted sound wave. Given speed of sound in air = 340 m/s and in water = 1486 m/s.
Answer 4:
(a)3.4 x 10-³ m
(b) 1.486 x 10-²

Important Questions on 9th Science Chapter 12

किसी माध्यम में ध्वनि द्वारा उत्पन्न विक्षोभ आपके कानों तक कैसे पहुँचता है?

किसी माध्यम में ध्वनि द्वारा उत्पन्न विक्षोभ अपने निकटवर्ती कणों में गति उत्पन्न कर देते हैं। ये कण इसी प्रकार की गति अन्य कणों में उत्पन्न करते हैं। माध्यम के कण स्वयं आगे नहीं बढ़ते, लेकिन विक्षोभ आगे बढ़ जाता है। ध्वनि की तरंगे, माध्यम के कणों की इसी गति द्वारा, हमारे कानो तक पहुँच जाती हैं।

What is sound and how is it produced?

Sound is a form of energy and it is produced due to vibrations of different types of object.
For example: A vibrating tuning fork, a bell, wires in a sitar and a guitar etc.

Why sound wave is called a longitudinal wave?

A sound wave is called a longitudinal wave because it travels in the form of compressions and rarefactions in the medium, where the particles of the medium vibrate in a direction which is parallel to the direction of propagation of the sound wave.

आपके विद्यालय की घंटी, ध्वनि कैसे उत्पन्न करती है?

जब विद्यालय की घंटी को हथौड़े से पीटा जाता है तो यह कंपन करने लगता है। कंपन करती हुई वस्तुएँ ध्वनि उत्पन्न करतीं हैं। अतः, घंटी ध्वनि उत्पन्न करने लगती है।

ध्वनि तरंगों को यांत्रिक तरंगें क्यों कहते है?

ध्वनि तरंगों को चलने के लिए द्रव्यात्मक माध्यम (जैसे: लोहा, ताँबा, वायु आदि) की आवश्यकता होती है। इसलिए इसे यांत्रिक तरगें कहते हैं।

मान लीजिए आप अपने मित्र के साथ चंद्रमा पर गए हुए है। क्या आप अपने मित्र द्वारा उत्पन्न ध्वनि को सुन पाएँगे?

ध्वनि तरंगों को चलने के लिए द्रव्यात्मक माध्यम (जैसे: लोहा, ताँबा, वायु आदि) की आवश्यकता होती है। चंद्रमा पर वायुमंडल नहीं है। इसलिए अपने मित्र द्वारा उत्पन्न ध्वनि को माध्यम के आभाव में हम नहीं सुन पाएँगे।

Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?

The timbre of sound is that characteristic which enables us to distinguish one sound from the other even when these are of the same pitch and loudness. Each person has its own timbre of sound and this characteristic helps us to identify a person from others even without looking at him (i.e., in a dark room).

The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?

Since the frequency of the source of sound is 100 Hz,
Number of vibrations of the source in 1 second = 100
Number of vibrations of the source in 1 minute (i.e., 60 second) =100×60 = 6000

अनुमान लगाइए कि निम्न में से किस ध्वनि का तारत्व अधिक है? (a) गिटार (b) कार का हॉर्न।

कार के हॉर्न का तारत्व अधिक है।

किसी ध्वनि तरंग की तरंगदैर्ध्य तथा आवृत्ति उसके वेग से किस प्रकार संबंधित है?

वेग = तरंगदैर्ध्य × आवृत्ति
⇒ v = λv

What is reverberation? How can it be reduced?

A sound created in a big hall will persist by repeated reflection from the walls until it is reduced to a value where it is no longer audible. The repeated reflection that results in this persistence of sound is called reverberation.
To reduce reverberation, the roof and walls of the auditorium are generally covered with sound-absorbent materials like compressed fibreboard, rough plaster or draperies. The seat materials are also selected on the basis of their sound absorbing properties.

How is ultrasound used for cleaning?

To clean any objects, it is placed in a cleaning solution and ultrasonic waves are sent into the solution. Due to the high frequency, the particles of dust, grease and dirt get detached and drop out. The objects thus get thoroughly cleaned.

ध्वनि की प्रबलता तथा तीव्रता में अंतर बताइए।

प्रबलता: ध्वनि के लिए कानों की संवेदनशीलता की माप को ध्वनि की प्रबलता कहते हैं।
तीव्रता: किसी एकांक क्षेत्रफल से एक सेकेंड में गुजरने वाली ध्वनि ऊर्जा को ध्वनि की तीव्रता कहते है।