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JEE Main & Advanced JEE Main Paper Phase-I (Held on 09-1-2020 Evening)

  • question_answer
    A wire of length L and mass per unit length \[6.0\times {{10}^{3}}\text{ }kg\text{ }{{m}^{1}}\] is put under tension of 540 N. Two consecutive frequencies that it resonates at are: 420 Hz and 490 Hz. Then L in meters is [JEE MAIN Held on 09-01-2020 Evening]

    A) 1.1 m

    B) 5.1 m

    C) 8.1 m

    D) 2.1 m

    Correct Answer: D

    Solution :

    Fundamental frequency = 70 Hz. \[70=\frac{1}{21}\sqrt{\frac{T}{\mu }}\] \[I\approx 2.14m\]


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