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BHU PMT BHU PMT (Screening) Solved Paper-2005

  • question_answer
    There are 26 tuning forks arranged in the decreasing order of their frequencies. Each tuning fork gives 3 beats with the next. The first one is octave of the last. What is the frequency of 18th tuning fork?

    A)  100 Hz                                 

    B)  99 Hz

    C)  96 Hz                                   

    D)  103 Hz

    Correct Answer: B

    Solution :

                     Key Idea: Frequencies of 26 tuning forks will be in arithmetic progression. Frequency of two consecutive forks is 3. \[\therefore \] \[f={{f}_{1}}+(n-1)\]  (arithmetic progression) Given, \[{{f}_{1}}=2f,n=26,d=-3\]                 \[f=2f+(26-1)(-3)\]                 \[f=75\,Hz\] Frequency of 18th tuning fork is                 \[{{f}_{18}}={{f}_{1}}+(18-1)(-3)\]                 \[{{f}_{18}}=2\times 75++17\times (-3)\]                 \[{{f}_{18}}=150-51=99\,Hz\]


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