AP 9th Class Physical Science Chapter 10 Solution – Sound
Andhra Pradesh SCERT 9th Class Physical Science Chapter 10 Sound Solution for AP 9th Class Physics/Chemistry Exam. Lots of Students of Andhra Pradesh Board will search on internet for Andhra Pradesh Class 9 Physical Science Textbook Solution or Study Material for AP 9th exam. Here you search will end! Here in this page we have provided for all question answer for AP SCERT Chapter 10 Sound.
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AP 9th Class Physics Chapter 10 Sound Solution
1.) Pick the correct answer
1.) When we say sound travels in a medium (AS1)
A) The medium travels
B) The particles of the medium travel
C) The source travels
D) The disturbance travels.
Ans: –D) The disturbance travels.
2.) A sound wave consists of (AS1)
A) Number of compression pulses only
B) number of rarefaction pulses only
C) number of compression and rarefaction pulses one after the other
D) vacuum only.
Ans: –c) number of compression and rarefaction pulses one after the other.
3.) Hertz stands for oscillations per (AS1)
D) mille second.
Ans: –a) second.
4.) When we increase the loudness of soundof a TV, the property of sound that changes is
Ans:- a) amplitude.
5.) The characteristic of the sound that describes how the brain interprets the frequency ofsound is called (AS1)
Ans: –a) pitch.
6.) In a stethoscope, sound of heart beats travel through stethoscopes tube (AS1)
A) By bending along the tube
B) In a straight line
C) Undergoing multiple reflections
D) all of the above.
Ans: –c) undergoing multiple reflections.
7.) The following terms a) amplitude b) wavelength c) frequency.
Amplitude: – The maximum possible displacement of a particle denote the amplitude of the particle.
Wavelength: -In a wave the distances between the two successive crests or troughs is known as wavelength.
Frequency: -Number of oscillation per minute time of a particle is known as frequency.
8.) Deduce the relation between wavelength, frequency and speed of sound.
Ans: – Let the frequency of a sound particle be n, speed is v, and wavelength is y.
As we all know that speed (v) = distance (y)/time (T).
Or, v=y/T; or, v= ny (as frequency=1/time).
So the relation between wavelength, frequency and speed of sound is v= n×y.
9.) How are multiple reflections of sound helpful to doctors and engineers?
Ans:- The multiple reflection create a huge impact in medical field as with the help of these property a doctor can listen the sound of the heartbeat and lungs.
In engineering field especially when the cinema hall build is made, that must have the design for multiple reflection to reach the sound to viewer.
10.) Name two quantities that vary periodically at a place in air as a sound wave travels throughit. (AS1)
Ans: – Refraction and series of compressionis affected by traveling of sound wave through them.
11.) Which has larger frequency – infrasonic sound or ultrasonic sound?
Ans: – Ultrasonic sound has larger frequency from infrasonic sound which has frequency less than 20 Hz.
12.) The grandparents and parents of two-year old girl are playing with her in a room. Asound source produces a 28-kHz sound. Who in the room is most likely to hear thesound?
Ans: –The range of sound frequency that can be heard by human being is 20Hz to 20 KHz. So the sound of 28 KHz could not be heard by them.
13.) Does the sound follow same laws of reflection as light does?
Ans: – The sound follow the same laws of reflection of light. And as results of that we see that incident wave of sound, reflected sound wave and normal all lies on a same point.
14.) Why is soft furnishing avoided in concert halls?
Ans: –Soft furnishing avoided in concert hall because of revibration. If it doesn’t follow then there will be multiple reflection of sound and quality of sound will be law and it will reached to someone late.
15.) Two sources A and B vibrate with the same amplitude. They produce sounds of frequencies 1 kHz and 30 kHz respectively. Which of the two waves will have larger power?
Ans: – As we all know that the power of sound is directly proportional to the frequency. As B has more frequency (30 KHz) has more energy so it will have larger power.
16.) What do you understand by a sound wave?
Ans: – Sound wave is longitudinal wave. In sound wave the particles propagate through air, water, solid etc medium from the source. The vibration create disturbance between the particles which transfer from one to another.
17.) Define the wavelength of a sound wave. How is it related to the frequency and the wave speed?
Ans: – The maximum distance travelled by a particle in a particular one vibration is known as wavelength.
The wavelength (y), frequency (n), and speed (v) is related by v=n × y. Or, wavelength × frequency = wave speed.
18) Explain how echoes are used by bats to judge the distance of an obstacle in front of them?
Ans: – Bats find their way by echoes or the main navigation system of bat is echoes. In the night when bats roaming around they produce ultrasonic sound wave which after reflection on any object return to the bats by this way they predict the distance of obje1from them.
19.) With the help of a diagram describe how compression and rarefaction pulses are produced in air near a source of sound. (AS5)
Ans: – In the way of traveling a vibrating medium pushes the air in front of it which create a high pressure zone which is compression and in backwards the vibration create a low pressure density known as rarefaction.
20) How do echoes in a normal room affect the quality of the sounds that we hear? (AS7)
Ans: – The minimum distance to create an echo is 17.2 meters. In a normal room there length of source and object may be less than 17.2 m so echoes affect doesn’t seen. If condition of echoes present then heard a prolonged sound.
21) Explain the working and applications of SONAR.
Ans: – SONAR or sound navigation and ranging is the method of way to calculate distance of an object. Here an ultrasonic wave passes and after reflection by the object return, then this signal convert to electrical signal. The echo which produced by reflection measure distance. Example: -The submarine use this technique to measure depth of sea.
22.) Find the period of a source of a sound wave whose frequency is 400Hz.
Ans: – As we know the time period (T) and frequency (n) is related by the expression T = 1/n.
Here n= 400Hz, then time period T = 1/400 = 0.0025s.
23.) A sound wave travels at a speed of 340 m/s. If its wavelength is 2cm, what is the frequency of the wave? Will it be in the audible range?
Ans: –As we know speed (v) = wavelength (y) × frequency (n).
Here given v=340 m/s,y = 0.02m, then n= 340 / 0.02 = 17000Hz.
As the audible range frequency of human is 20 to 20 KHz so it’san audible sound.
24) Given that sound travels in air at 340 m/s, find the wavelength of the waves in air produced by a 20 kHz sound source. If the same source is put in a water tank, what would be the wavelength of the sound waves in water? Speed of sound in water = 1,480 m/s.
Ans: – As we know that speed (v) = wavelength (y) × frequency (n)
Here v=340 m/s, n=20 kHz, so wavelength (y) is (340 / 20000 = 0.017) 0.017m.
As we all know frequency remain unchanged so frequency in water will be same.
The speed of sound in water is 1480 m/s. So wavelength = (1480 / 20000) =0.074m.
25) A man is lying on the floor of a large, empty hemispherical hall, in such a way that his head is at the centre of the hall. He shouts “Hello!” and hears the echo of his voice after0.2 s. What is the radius of the hall? (AS7) (Speed of sound in air = 340 m/s)
Ans: –As the echo hears after 0.2s so the time of reflection of echoes is 0.1.
We know speed of sound = Distance / time;
Or, distance = speed × time = 340 × 0.1 = 34m.
So the radius of the hall is 34m.
26) “We know that sound is a form of energy. So, the large amount of energy produced due the sound pollution in cosmopolitan cities can be used to our day to day needs of energy. It also helps us to protect bio diversity in urban areas”. Do you agree with this statement? Explain. (AS7)
Ans: – From conservation of energy we know that energy not be created or destroyed it just change from one states to another. We know that sound is a from of energy and there is a huge source of sound of vehicle outside but to convert these in energy is not possible yet as there is no such technology available till now. So deep research is needed to do that.
27) How do you appreciate efforts of a musician to produce melodious sound using a musical instrument by simultaneously controlling frequency and amplitude of the sounds produced by it?
Ans:-The method that a musician take to produce melodious sound using a musical instruments by simultaneously controlling frequency and amplitude of the sound nothing but the different wave produce by different sound object. The difference in shapes of the sound wave create the difference between the melodious sounds.