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Sound Test - 20

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Sound Test - 20
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  • Question 1
    1 / -0
    Time period of a sound wave having the wavelength 0.2 m and frequency 10 Hz will be
    Solution
    Given -  $$    n=10Hz$$ ,
    we have , time period , $$T=1/n=1/10=0.1s$$
  • Question 2
    1 / -0
    Sound waves travel fastest in.
    Solution
    Answer is A.

    Sound travels fastest in solids and slowest in air. Water is in between. This is because the particles in solids are closer together than the particles in liquids or gases, and the particles in liquids are closer together than the particles in gases. The tighter particles are packed in a space, they collide more frequently. This allows sound, which is simply the combined collisions of particles, to travel fastest in solids. So, to reiterate, sound travels fastest in solids, then water, and slowest in air.
  • Question 3
    1 / -0
    The use of ultrasound waves to investigate the action of the heart is called.
    Solution
    Answer is D.

    The technique of echocardiography is based on the reflection property of ultrasonic waves.
    An echocardiogram is a type of ultrasound test that uses high-pitched sound waves that are sent through a device called a transducer. The device picks up echoes of the sound waves as they bounce off the different parts of our hearts. These echoes are turned into moving pictures of our hearts that can be seen on a video screen.
  • Question 4
    1 / -0
    A sound wave completes $$24$$ cycles in $$0.8\;s$$. The frequency of the wave is :
    Solution
    Frequency is the number of times the cycle repeats in a one-time unit.

    Given that, the sound wave completes $$24$$ cycles in $$0.8\,s$$.

    So, the number of cycles in $$0.1\,s =\dfrac{24}{0.8}=30\,Hz$$

    Hence the frequency is $$30\, Hz$$


  • Question 5
    1 / -0
    Light waves:
    Solution
    A mechanical wave is a wave that needs a material medium like air, water, steel, etc. for its propagation. It cannot travel through a vacuum. Light is not a mechanical wave. It does not require any material medium for its propagation.
  • Question 6
    1 / -0
    A tuning fork vibrates with 2 vibrations in 0.4 seconds. Its frequency (in Hz) is
    Solution
    Number of vibrations  $$n = 2$$
    Time taken  $$t = 0.4$$ s
    Thus frequency of sound  $$\nu = \dfrac{n}{t}$$
    $$\therefore$$  $$\nu = \dfrac{2}{0.4} = 5 \ Hz$$
  • Question 7
    1 / -0
    A tuning fork vibrates with $$2$$ vibrations in $$0.4\ s$$. Its frequency is:
    Solution
    Frequency is defined as the number of vibrations per second. 
    Given,
    A tuning fork vibrates with $$2$$ vibrations in $$0.4\ s$$.
    Number of vibrations $$=2$$
    Time $$=0.4\ s$$
    Frequency, $$f=\dfrac{number\ of\ vibrations}{time}$$
    $$f=\dfrac{2}{0.4}$$

    $$f=5\ Hz$$ 
  • Question 8
    1 / -0
    The _________ of sound depends on amplitude of vibration of the source.
    Solution
    The loudness of sound depends on the amplitude of vibration of the source. The loudness of sound is directly proportional to the square of the amplitude of vibration producing the sound.
  • Question 9
    1 / -0
    A sound wave completes $$24$$ cycles in $$0.8\;s$$. The time period of the wave would be:
    Solution
    A time period is the time taken to complete one cycle.

    Given that, the sound wave completes $$24$$ cycles in $$0.8\,s$$

    So, the time taken to complete one cycle $$=\dfrac{0.8}{24}=\dfrac{1}{30}\,s$$

    Hence the time period is $$\dfrac{1}{30}\,s$$
  • Question 10
    1 / -0
    A faint (or soft) sound can be changed to a loud sound by increasing its :
    Solution
    (First figure has more amplitude so more loudness) 
    Loudness of sound is directly proportional to square of Amplitude. 
    So increasing the amplitude intensifies the sound and can be heard over greater distances.

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