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Waves Test - 12

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Waves Test - 12
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  • Question 1
    1 / -0
    The phenomenon of interference is shown by 
    Solution

  • Question 2
    1 / -0
    In ordinary talk, the amplitude of vibration is approximately:
    Solution
    In ordinary talk, the amplitude of vibration is about $$10^{-8}$$ $$m$$.
  • Question 3
    1 / -0
    'Which is not true for a wave ?
    Solution
    (A) In a wave , frequency and wavelength are related by ,  $$v=n\lambda$$ .
    (B) During wave motion , a disturbance (energy) is transferred by medium particles in a wave .
    (C) Mechanical waves cannot travel through vacuum , because they are dependent on the properties of medium , so it is a wrong statement .
    (D) Wave velocity is given by ,   $$v=n(s^{-1})\lambda(m)=m/s$$
  • Question 4
    1 / -0
    The velocity of sound in a gas is 4 times that in air at the same temperature. When a tuning fork is sounded in a wave of frequency 480 and wavelength $$\lambda_1$$ is produced. The same fork is sounded in the gas and of $$\lambda_2$$ is the wavelength of the wave, then $$\lambda_2\,/\, \lambda_1$$ is :
    Solution
    We have ,  $$v=f\lambda$$ ,
    when tuning fork is sounded in air ,
                     $$v_{a}=f\lambda_{1}$$ ,  ........................eq1
    when tuning fork is sounded in gas ,
                     $$v_{g}=f\lambda_{2}$$ ,  .......................eq2 ,
    frequency will remain same as the source of sound is same ,
    dividing eq2 by eq1 ,
                     $$v_{g}/v_{a}=\lambda_{2}/\lambda_{1}$$  ,....................eq3
    now given ,  $$v_{g}=4v_{a}$$ ,
    putting  , $$v_{g}=4v_{a}$$ in eq3,
    we get ,     $$4v_{a}/v_{a}=\lambda_{2}/\lambda _{1}$$ ,
    or               $$4=\lambda_{2}/\lambda _{1}$$
  • Question 5
    1 / -0
    In the following properties of a wave, the one which is independent of the other is
    Solution
    We know that velocity of wave $$(v)=$$ frequency $$(f) \times $$ wavelength $$(\lambda)$$. 
    Thus, we can say that the amplitude (A) is independent of other means it is not related to the other quantities. 
  • Question 6
    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 7
    1 / -0
    The stethoscope used by doctors works on the principle of 
    Solution
    A stethoscope has an air filled tube with a diaphragm at its one end . When this diaphragm is placed on the chest of a patient , it vibrates due to faint sound coming from the chest , these vibrations are channelised through the air filled tube and amplified by multiple reflections by the inner walls of the tube , thus it is based on the principle of reflection .
  • Question 8
    1 / -0
    Two waves represented by $$y_1= a sin\omega t $$ and $$ y_2 = a sin(\omega t+ \phi) $$ and  $$ \phi =\dfrac{\pi}{2} $$ are superposed at any point at a particular instant. The resultant amplitude is
    Solution
    Since the angle between them is $$90$$, so $$cos\ 90^0=0$$
    So, $$R=\sqrt {a^2+a^2}=\sqrt 2a$$
  • Question 9
    1 / -0
    The SI unit of amplitude of oscillation is :
    Solution
    amplitude is nothing but displacement and si unit of displacement is metres 'm'
  • Question 10
    1 / -0
    A source of frequency $$500   Hz$$ emits waves of wavelength $$0.2  \  m$$. The time the wave takes to travel a distance of $$300\    m$$ is :
    Solution
    We have , speed of wave , $$v=n\lambda$$ ,
    where $$n=$$ frequency of wave ,
               $$\lambda=$$ wavelength of wave ,
    given $$n=500Hz  ,  \lambda=0.2m  ,  d=300m$$ ,
    hence  $$v=500\times0.2=100m/s$$ ,
    therefore , time taken to cover a distance of 300m ,
               $$t=d/v=300/100=3s$$   

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