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JEE Main & Advanced Sample Paper JEE Main Sample Paper-40

  • question_answer
    56 tuning forks are so arranged in series that each fork gives 4 beats per sec with the previous one. The frequency of the last fork is 3 times that of the first. The frequency of the first fork is

    A)  110                            

    B) 56

    C) 60                               

    D) 52

    Correct Answer: A

    Solution :

    [a] \[{{n}_{last}}={{n}_{first}}+\left( n-1 \right)\times d\Rightarrow 3n=n+\left( 56-1 \right)\times 4\]or\[2n=55\times \Rightarrow n=110Hz\]


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