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NEET AIPMT SOLVED PAPER 2000

  • question_answer
                    Two strings A and B have lengths \[{{l}_{A}}\] and \[{{l}_{B}}\] and carry masses \[{{M}_{A}}\] and \[{{M}_{B}}\] at their lower ends, the upper ends being supported by rigid supports. If \[{{n}_{A}}\] and \[{{n}_{B}}\] are their frequencies of their vibrations and \[{{n}_{A}}=2{{n}_{B}},\] then:                                                                                                                                                                             

    A)                 \[{{l}_{A}}=4{{l}_{B}}\], regardless of masses     

    B)                 \[{{l}_{B}}=4{{l}_{A}}\], regardless of masses

    C)                 \[{{M}_{A}}=2{{M}_{B}},\,{{l}_{A}}=2{{l}_{B}}\]                

    D)                 \[{{M}_{B}}=2{{M}_{A}},\,{{l}_{B}}=2{{l}_{A}}\]

    Correct Answer: B

    Solution :

                    The frequency of vibrations of string is                 \[n=\frac{1}{2\pi }\sqrt{\frac{g}{l}}\] Given,   \[{{n}_{A}}=2{{n}_{B}}\] \[\therefore \]  \[\frac{1}{2\pi }\sqrt{\frac{g}{{{l}_{A}}}}=2\cdot \frac{1}{2\pi }\sqrt{\frac{g}{{{l}_{B}}}}\] or            \[\frac{1}{{{l}_{A}}}=\frac{4}{{{l}_{B}}}\] or            \[{{l}_{B}}=4{{l}_{A}}\]                 It is obvious from Eq. (i), the frequency of vibrations of strings does not depend on their mass.


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