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Question 13 Marks
(i) (a) What is an echo?
(b) State two conditions necessary for the formation of an echo.
(ii) What is sonar? State its principle. How is it used to find the depth of sea?
(iii) How do the following use echoes?
(a) army, (b) geologist, (c) medical sciences
Answer
(i) (a) Echo is the sound heard after reflection from a distant obstacle, after the original sound has died off.
(b) Conditions necessary for echo formation: (i) The minimum distance between the source of sound and reflecting object/obstacle should be 17 metres.
The size of the obstacle must be larger than the wavelength of sound.
(ii) Sonar stands for Sound Navigation and Ranging and the technique used to measure distances by using ultrasonic waves. To find the depth of an ocean, ultrasonic waves from the ship is sent towards the bottom of the ocean. These waves are then received by a suitable receiver, after reflection from the ocean floor. The distance of the bottom of ocean is calculated by measuring the time interval between the instant when waves are sent and the instant when reflected waves are received. Using the value of velocity of sound in water, the depth of ocean floor can be calculated.
(iii) (a) Army uses echoes to locate the gun positions of the enemy.
(b) Geologists use echoes for mineral prospecting.
(c) Echoes produced by ultrasonic waves are used for imaging internal human organs, such as prostate gland, womb, liver, gall bladder. Such imaging is called ultrasonography. Similarly, echo cardiography is used to take the image of heart.
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Question 23 Marks
(i) (a) A person is tuning his radio set to a particular station. What is the person trying to do to tune it?
(b) Name the phenomenon involved, in tuning the radio set.
(c) Define the phenomenon named by you in part (b).
(ii) (a) A wire of length 80 cm has a frequency of 256 Hz . Calculate the length of a similar wire under similar tension, which will have frequency 1024 Hz ,
(b) A certain sound has a frequency of 256 hertz and a wavelength of 1.3 m .
(i) Calculate the speed with which this sound travels.
(ii) What difference would be felt by a listener between the above sound and another sound traveling at the same speed, but of wavelength 2.6 m ?
(iii) Name the factor that determines: (a) Loudness of the sound heard. (b) Quality of the note. (c) Pitch of the note.
Answer
(i) (a) While tuning the radio, the person tries to match the frequency of his radio set with that of the radio station so that a resonance takes place and he hears the sound clearly.
(b) The phenomenon taking place in radio set is electromagnetic resonance.
(c) When the frequency of a given body corresponds to the frequency of the impressed force, the body vibrates with a greater amplitude and the phenomenon is called resonance.
(a) $\frac{f_1}{f_2}=\frac{l_2}{l_1}$
$\Rightarrow \frac{256 Hz}{1024 Hz}=\frac{l_2}{80 cm} \Rightarrow l_2=\frac{1}{4} \times 80 cm$
$\therefore l_2= 2 0 ~ c m$
(b) (i) Speed of sound $=$ frequency $\times$ wavelength
$=256 \times 1.3=332.8 ms^{-1}$
(ii) The sound of wavelength 2.6 m will have lesser pitch (bass) as compared to sound of wavelength 1.3 m .
(iii) (a) The square of amplitude of vibrating body determines its loudness.
(b) The waveform of vibrating body determines quality of the note.
(c) The frequency of a vibrating body determines its pitch.
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[3 Mark Question Answer] - Physics STD 10 Questions - Vidyadip