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Sound Test - 69

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Sound Test - 69
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  • Question 1
    1 / -0
    The wavelength of waves produced on the surface of water is 20 cm. If the wave velocity is 24 m s1s^{-1}, calculate the time in which one wave is produced ?
    Solution
    In the question, the wavelength (λ )\left( \lambda  \right) is given as 20 cm20\ cm, that is, 0.20 m 0.20\ m and the wave velocity, vv is given as 2424 m/sm/s.
    The frequency ν\nu , velocity vv and wavelength λ\lambda  relation is given as:
    ν=vλ \nu =\dfrac { v }{ \lambda  } .
    Number of waves produced in one second is called as frequency, i.e. ν=vλ \nu =\dfrac { v }{ \lambda  } =ν=240.20=\nu =\dfrac { 24 }{ 0.20 } =120= 120 waves per second.
    Time period of a wave is given as the inverse of frequency. 
    That is, T=1ν T=\dfrac { 1 }{ \nu  }  =1120=\dfrac { 1 }{ 120 } =8.33×103 seconds=8.33\times { 10 }^{ -3 }\ seconds
    Hence, the time period of the wave is given as 8.33×103 seconds8.33\times { 10 }^{ -3 }\ seconds.
  • Question 2
    1 / -0
    A man pronounces a, b, c, da,\space b, \space c, \space d and ee and hears last four syllables as an echo. Taking velocity of sound as 340 ms1340\space ms^{-1}, the distance of reflecting surface, if it takes one-fifth of a second to pronounce or hear one syllable is
    Solution
    Let the distance between the man and the reflecting surface be xx metres.
    Thus the distance travelled by sound to reach back to the man   d=2xd = 2 x
    The man cannot hear the syllable aa as at the instant when echo of syllable aa is about to reach to the man, he pronounces ee. Thus the time taken by man to pronounce all syllables is equal to the time taken by the sound to reach back to the man.  i.e   t=4×15=45 sect = 4 \times \dfrac{1}{5} = \dfrac{4}{5}   sec
    Thus  d=v td = v  t
    2 x=340 ×45          x=136 m2  x = 340   \times \dfrac{4}{5}             \implies x = 136   m
  • Question 3
    1 / -0
    A person is in front of a vertical mountain fires a bullet and hears its echo after a time 3 s. The person walks a distance 'd' towards mountain then fires another bullet and hears its echo after a time 2 s. If velocity of sound in air is 340 ms1^{-1}, the value of 'd' is
    Solution
    Given,
    Velocity of sound v=340v=340
    Case-1:  Echo is heard in 3 sec, let the distance between the mountain and the person is xx.

    Total distance travel by sound =2x=vt=2x=vt
    Distance from mountain in first case x=vt2x=\dfrac{v t}{2} =340×32=\dfrac{340\times 3}{2}

    Case-2: Echo is heard in 2 sec, let the distance between the mountain and the person is yy.
    Total distance travel by sound =2y=vt=2y=vt'
    Distance from mountain in second case y=340×22y=\dfrac{340\times 2}{2}

    Walking distance d=xyd=x-y

    d=340×32340×22=3402=170 m\therefore d=\dfrac{340 \times 3}{2}-\dfrac{340\times 2}{2}=\dfrac{340}{2}=170 \ m

    Option B

  • Question 4
    1 / -0
    Identify the longitudinal waves from the following
    Solution
    Ultrasonic waves are longitudinal waves of frequency greater than 20,000 Hz20,000\ Hz. This implies that a human being cannot hear beyond 20,000 Hz20,000\ Hz, i.e. the maximum frequency of audible sound is 20,000 Hz20,000\ Hz.
  • Question 5
    1 / -0
    The loudness of sound decreases with an:
    (1)(1) Increase in the distance between the ear and the source
    (2)(2) Decrease in the amplitude of the vibrating body
    (3)(3) Increase in frequency of sound
    (4)(4) Decrease in frequency
    Solution
    The loudness or softness of a sound is determined basically by its amplitude. A sound wave spreads out from its source. As it moves away from the source its amplitude as well as its loudness decreases.
    The loudness of sound decreases with an:
    (1) Increase in the distance between the ear and the source
    (2) Decrease in the amplitude of the vibrating body
  • Question 6
    1 / -0
    Which of the following is/are correct statement(s)?
    (1) Besides helping in finding the distance of an object, SONAR also indicates its direction and speed.
    (2) Megaphone is a device that works on the ~ principle of reflection of sound.
    (3) Ultrasonics are used to detect earthquakes.
    (4) Like light, sound waves also obey laws of reflection.
    Solution
    Sonar is a device that uses ultrasonic waves to measure the distance, direction, and speed of underwater objects.
    In a megaphone, a tube followed by a conical opening reflects sound successively to guide most of the sound waves from the source in the forward direction towards the audience.
    Ultrasonics are not used to detect earthquakes.
    Like light, sound waves also obey laws of reflection.

  • Question 7
    1 / -0
    When a tuning fork produces sound waves in air, which one of the following is the same in the material of tuning fork as well as in air ?
    Solution
    Sound waves are produced by vibrating objects.

     When the tuning fork is hit with a rubber hammer, the tines begin to vibrate. The back and forth vibration of the tines produce disturbances of surrounding air molecules. Since the vibration of air molecules is solely due to vibration of nearby tine of tuning fork,so it means the number of vibration of air molecules will be equal to number of vibration of tine of the tuning fork.So frequency ,which is number of vibrations per second, will be same for both the tuning fork and the air.
  • Question 8
    1 / -0
    Wavelength of the light frequency 100Hz100Hz is _________.
    Solution
    Given that the light frequency is 100Hz100\,Hz

    We know that the speed of the light is 3×108m/s3\times 10^{8}\,m/s

    We have the equation,

    V=nλV=n\lambda

    where,

    VV is the Wave velocity

    nn is the Frequency

    λ\lambda is the Wavelength

        \implies λ=Vn\lambda=\dfrac{V}{n}

    That is,

    λ=3×108100=3×106m\lambda=\dfrac{3\times 10^8}{100}=3 \times 10^6\,m

    Hence the wavelength is 3×106m3\times10^6\,m
  • Question 9
    1 / -0
    Which of the following is NOT an application of ultrasound waves?
    Solution
    Stethoscope is an acoustic medical device for listening to the internal sound of a living being. It works on reflection of sound.
  • Question 10
    1 / -0
    When a tuning fork is striked using a rubber prong, the compressions and rarefactions in the sound wave correspond to
    Solution
    Sound being a longitudinal wave show areas of compression and rarefaction : compressions are regions of high pressure due to particles being close together. rarefactions are regions of low pressure due to particles being spread further apart.
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