JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-II

  • question_answer
    5 beats/ second are heard when a turning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either \[0.95m\] or \[1m\]. The frequency of the fork will be:  [JEE Online 15-04-2018 (II)]

    A) \[195Hz\]                            

    B) \[251Hz\]            

    C)          \[150Hz\]            

    D)          \[300Hz\]

    Correct Answer: A

    Solution :

    \[{{\text{L}}_{\text{1}}}\text{=0}\text{.95m,}\,\,{{\text{L}}_{\text{2}}}\text{=1m}\] \[{{\text{L}}_{2}}>{{L}_{1}},{{n}_{1}}>N>{{n}_{2}}\] \[{{n}_{1}}-N=5andN-{{n}_{2}}=5\] On solving \[{{n}_{1}}-{{n}_{2}}=10\] \[{{n}_{2}}={{n}_{1}}-10\] By law of length of vibrating string \[{{n}_{1}}{{L}_{1}}={{n}_{2}}{{L}_{2}}\] On solving we get\[{{n}_{1}}=200Hz\] \[{{n}_{1}}-N=5\] \[N=195Hz\]


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