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JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Afternoon)

  • question_answer
    A string 2.0 m long and fixed at its ends is driven by a 240 Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is :- [JEE Main 9-4-2019 Afternoon]

    A) 320m/s, 120 Hz          

    B) 180m/s, 80 Hz

    C) 180m/s, 120 Hz

    D)   320m/s, 80 Hz

    Correct Answer: D

    Solution :

    \[3\left( \frac{\text{v}}{2\ell } \right)=240\]           \[3\frac{\text{v}}{2\times 2}=240\]           \[\text{v}=320m/s\] fundamental frequency\[=\frac{\text{v}}{2\ell }=\frac{320}{2\times 2}=80Hz.\]


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