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NEET AIPMT SOLVED PAPER SCREENING 2009

  • question_answer
    Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to \[k,{{k}_{0}},{{k}_{1}}\]When both the strings vibrate simultaneously the number of beats is

    A) 5                                             

    B) 7                             

    C)        8                             

    D)        3

    Correct Answer: B

    Solution :

    Key Idea The number of beats will be the difference of frequencies of the two strings. Frequency of first string \[c{{m}^{-2}}{{s}^{-1}}\] \[\varepsilon \]\[\varepsilon \] Similarly, frequency of second string \[\varepsilon \]\[\varepsilon \] Number of beats\[\hat{i}+2\hat{j}+\hat{k}\] = 7 beats


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