JEE Main & Advanced Sample Paper JEE Main - Mock Test - 45

  • question_answer
    Sound of wavelength \[100\text{ }cm\]travels in air. At a given point the difference in maximum and minimum pressure is \[0.2\,N{{m}^{-2}}.\]If the bulk modulus of air is \[1.5\times 10h5N{{m}^{-2}},\] find the amplitude of vibration of the particles of the medium.

    A) \[1.2\times {{10}^{-7}}m\] 

    B)                    \[2.4\times {{10}^{-7}}m\]

    C) \[2.0\times {{10}^{-7}}m\]       

    D)        \[1.0\times {{10}^{-7}}m\]

    Correct Answer: D

    Solution :

    [d] \[{{P}_{\max }}-{{P}_{\min }}=2\Delta {{P}_{0}}\] where \[\Delta {{P}_{0}}\] is pressure amplitude. \[\therefore \,\,\,\,\,2\Delta {{P}_{0}}=0.2\] \[\Rightarrow \,\,\,\Delta {{P}_{0}}=0.1N{{m}^{-2}}\Rightarrow BAk=0.1\] \[\Rightarrow \,\,\,A=\frac{0.1\times \lambda }{2\pi B}=\frac{0.1\times 1.0}{2\times 3.14\times 1.5\times {{10}^{5}}}=1.0\times {{10}^{-7}}m\]


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