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BVP Medical BVP Medical Solved Paper-2005

  • question_answer
    Two identical straight wires are stretched so as to produce 6 beats per second when vibrating simultaneously. On changing the tension slightly in one of them, the beat frequency still remains unchanged. Denoting by \[{{T}_{1}}\] and \[{{T}_{2}}\], the higher and the lower initial tensions in the strings, it could be said that while making the above changes in tension :

    A) \[{{T}_{1}}\]was decreased        

    B) \[{{T}_{1}}\] was increased         

    C) \[{{T}_{2}}\]was increased          

    D) \[{{T}_{2}}\] was decreased

    Correct Answer: A

    Solution :

                    The relation for frequency and tension is given by \[f\propto \sqrt{T}\] As   \[{{T}_{1}}>{{T}_{2}}\] i.e. \[{{f}_{1}}>{{f}_{2}},\] So,  \[{{f}_{1}}-{{f}_{2}}=6Hz\] when we increase lower tension \[{{T}_{2}}\], then  \[{{f}_{2}}\] will be increased and \[{{f}_{1}}\] will decrease Hence,                 \[{{f}_{2}}-{{f}_{1}}=6Hz\]


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