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JEE Main & Advanced JEE Main Paper (Held On 9 April 2014)

  • question_answer
    A transverse wave is represented by\[y=\frac{10}{\pi }\sin \left( \frac{2\pi }{T}t-\frac{2\pi }{\lambda }x \right)\] For what value of the wavelength the wave velocity is twice the maximum particle velocity?   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) 40 cm                   

    B) 20 cm

    C) 10 cm                   

    D) 60 cm

    Correct Answer: A

    Solution :

    Given, amplitude a = 10 cm wave velocity = 2 × maximum particle velocity i.e.,\[\frac{\omega \lambda }{2\pi }=2\frac{a\omega }{\pi }\] or,\[\lambda =4a=4\times 10=40cm\]


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