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Sound Test - 41

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Sound Test - 41
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Weekly Quiz Competition
  • Question 1
    1 / -0
    An empty vessel produces louder sound than a filled one because.
    Solution
    An empty vessel produces louder sound than a filled one because the air molecules in empty vessel have greater amplitude and hence greater intensity than liquid molecules in the filled vessel
  • Question 2
    1 / -0
    The tip of a pencil attached to a spring oscillates vertically, thereby etching an overlapping straight line on a white sheet of paper. The length of this line is 5 cm5 \ cm. What is the amplitude of the vibration
    Solution
    The line length gives the distance between two extremes = twice the amplitude of the vibration. Thus, amplitude of vibration $$= \dfrac{5}{2} = 2.5 \  cm$$

    The correct option is (C)
  • Question 3
    1 / -0
    The maximum displacement of a particle in a vibratory motion is known as 
    Solution
    As seen in above figure the maximum displacement of a particle from it's mean position in a vibratory motion is known as amplitude.

    The correct option is (c)

  • Question 4
    1 / -0
    A particle completes one vibration in 2 secs. How long does it take to vibrate 25 vibrations
    Solution
    For completing one vibration, time taken = 2 S2\ S
    Total time taken to complete 25 vibration =2×25=50sec = 2 \times 25 = 50 sec

    The correct option is (d)
  • Question 5
    1 / -0
    A particle performs 2020 vibrations in 5 s5\ s. The time period of vibration is:
    Solution
    Given,
    Number of vibrations =20=20
    Time taken to complete 2020 vibrations =5 s=5\ s
    Time taken to complete 11 vibration is known as time period TT.
    Time taken to complete 11 vibration, T=520T=\dfrac{5}{20}
    T=0.25 sT=0.25\ s
  • Question 6
    1 / -0
    A sound wave has a speed of 200 m/s in a medium and it travels a distance of 1 m. What is the time taken by the wave to cross the medium
    Solution
    We know that wave velocity = distance/Time taken
    200=1/t     t=1/200=0.005s200= 1/t  \implies t = 1/200 = 0.005 s

    The correct option is (d)
  • Question 7
    1 / -0
    A particle takes 0.01 s to complete 5 vibrations, what is its frequency
    Solution
    Given:
    Number of vibration = 55
    Time t=0.01 st=0.01\ s
    Frequency=Number of vibrationTime=50.01=500Hz\dfrac{\text{Number of vibration}}{\text{Time}}=\dfrac{5}{0.01} = 500 Hz

    The correct option is (d)
  • Question 8
    1 / -0
    The period of a source of a sound wave whose frequency is 400Hz400 Hz
    Solution
    Given the frequency of the wave is f=400 Hz f = 400\ Hz

    We know that time period is
    T=1f T = \dfrac{1}{f}

    T=1400=2.5×103=2.5 ms T = \dfrac{1}{400} \, = \, 2.5 \times 10^{-3} = 2.5\ ms

    So B is the correct answer.
  • Question 9
    1 / -0

    Directions For Questions

    Ultrasonic and Infrasonic Waves: The sound having frequency between 2020 Hz to 2020 kHz is called audible sound. Human ear cannot hear sound of frequencies lying outside this range. The limits of audibility of sound vary from individual. Generally the upper limit decreases with increase of age of individual; usually and people can not hear sound of frequency more than 1515 kHz.
    Sound of frequencies below 2020 Hz are called infrasonics, while frequencies beyond 2020 kHz are called ultrasonics. Ultrasonics can be heard by dogs, cat, bats, etc. Bats detect the path by means of ultrasonics. Ultrasonics can be used for diagnosis of turnover and other abnormalities within our body. This is known as ultrasonography. Ultrasonics are used to cure various diseases and to measure depth of oceans to detect submarines.

    ...view full instructions

    Ultrasonics can be detected by:
    Solution
    Ultrasonic sound can be detected by Bats. They use ultrasonic waves to navigate and find insect prey.

  • Question 10
    1 / -0
    A sound wave, whose time period is 3 s3\ s passes through a medium for 30 s30\ s. The number of waves that had passed through is:
    Solution
    Given,
    Time period, T=3 sT=3\ s
    Total time, t=30 st=30\ s
    Number of waves, nn

    We know,
    The time taken by the wave for one complete oscillation of the density or pressure of the medium is called the time period.
    T=tnT=\dfrac{t}{n}

         n=tT\implies n=\dfrac{t}{T}

    n=303n=\dfrac{30}{3}

    n=10n=10
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