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Sound Test - 43

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Sound Test - 43
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  • Question 1
    1 / -0
    The linear distance between the compression and the adjacent rarefaction is 0.8 m. The wavelength of the longitudinal wave is
    Solution
    Wave length λ\lambda is defined as the distance between two successive compressions or rarefactions , therefore the distance between two successive compression and rarefaction will be λ/2\lambda/2 ,
    now given ,  λ/2=0.8\lambda/2=0.8 ,
         or            λ=2×0.8=1.6m\lambda=2\times0.8=1.6m
  • Question 2
    1 / -0
    In the above figure, a boy is beating the drum using a stick and thus, drum produces a sound. In the other figure, another boy produces sound using a whistle. Choose the option which shows the correct waveforms in above figure.

    Solution
    Answer is B.

    A drum produces sound of lower frequency which is less shrill and has lower pitch, while a whistle produces sound of higher frequency which is shriller and is of higher pitch. The number of vibrations or oscillations per second is called frequency. In the figure B, the frequency of the first wave is lower and the frequency of the second wave is higher. This interprets the frequency of the drum and the whistle.
    Hence, the option B is correct.
  • Question 3
    1 / -0
    Waves inside a gas are
    Solution
    When a wave propagates through a gas, the displacement of the particle is parallel to wave propagation. Thus, it's a longitudinal wave. 
    Transverse waves cannot propagate in a gas or a liquid because there is no mechanism for driving motion perpendicular to the propagation of the wave.
    However transverse waves can travel through solids.
  • Question 4
    1 / -0
    Waves on water surface are
    Solution
    Water surface exhibits a combination of both longitudinal and transverse waves. When the wave travels through the water, the particles travel in a circular path, both parallel and perpendicular to the direction of wave propagation. 
  • Question 5
    1 / -0
    Take a metallic tumbler and a tablespoon. Strike the tablespoon gently at the brim of the tumbler (as shown in the figure above). Now suspend a small thermocole ball touching the rim of the tumbler and touch it to the vibrating tumbler. See how far the ball is displaced. Now, again strike the tablespoon at the brim of the tumbler but strike it hard this time. Again touch the thermocole to the vibrating tumbler and see how far the ball is displaced. Choose the correct option. 

    Solution
    In our observation, we see that the thermocole ball is displaced more when the tumbler is stricken hard, as the amplitude of vibration is more in this case.
    The loudness of sound is proportional to the square of the amplitude of the vibration producing the sound. For example, if the amplitude becomes twice, the loudness increases by a factor of 4. The loudness is expressed in a unit called decibel (dB). 
    The loudness of sound depends on its amplitude. When the amplitude of vibration is large, the sound produced is loud. When the amplitude is small, the sound produced is feeble.
    Hence, option C is correct.
  • Question 6
    1 / -0
    A person is listening to a tone of 500 Hz500 \ Hz sitting at a distance of 450 m450 \ m from the source of the sound. What is the time interval between the successive compression from the source?
    Solution
    Given -    n=500 Hz    n=500 \ Hz ,

    The time interval between two successive compressions or rarefactions is called , time period of wave , which is given by ,
          
    T=1n=1500=0.002 s=2 msT=\dfrac 1n= \dfrac{1}{500}= 0.002 \ s=2 \ ms

  • Question 7
    1 / -0
    A person is listening to a tone of 500500 Hz sitting at a distance of 450450 m from the source of the sound. What is the time interval between the successive compression from the source ? 
    Solution
    Given -    f=500Hz    f=500Hz ,
    The time interval between two successive compressions or rarefactions is the time period of the wave , which is given by ,
                T=1/f=1/500=0.002s=2 milli secondT=1/f=1/500=0.002s=2\ milli\ second

  • Question 8
    1 / -0
    Which of the following can be used to detect a flaw inside a metal block?
    Solution
    Ultrasonic waves are sound waves transmitted above the human-detectable frequency range, usually above 20,000 Hz20,000\ Hz. They are used by some animals and in medical or industrial-technological devices.
    Ultrasonic impact treatment (UIT) uses ultrasound to enhance the mechanical and physical properties of metals. It is a metallurgical processing technique in which ultrasonic energy is applied to a metal object.
  • Question 9
    1 / -0
    How far away should a deep sea diver be from an under-sea rock so that he can hear his own echo? (Speed of sound in water is 1500 m/s).
    Solution
    The sensation of sound persists in our brain for about 0.1 s. 

    To hear a distinct echo the minimum time interval between the original sound and the reflected sound = 0.1s0.1 s

    Speed of sound in water is given as 1500 m/s.1500 \ m/s.
    Here, the sound must go to the obstacle and reach back the ear of the listener on reflection after 0.1 s.

    Hence, the total distance covered by the sound from the point of generation to the reflecting surface and back = 1500×0.1=150m.1500 \times 0.1 = 150 m.

    Thus, for hearing the echo, the minimum distance of the rock from the deep sea diver must be half of this distance.
    =1502=75m=\dfrac{150}{2}=75m

    Answer: D

  • Question 10
    1 / -0
    Which of the following statements is not true?
    Solution
    The speed of sound depends on the properties of the medium through which it travels. The speed of sound in a medium depends on the temperature of the medium. If the temperature of the medium increases, the speed of sound increases. 

    speed=distancetimespeed=\dfrac{distance}{time}

        time=distancespeed\implies time=\dfrac{distance}{speed}

    An echo is heard sooner on a hot day than on a cold day.
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