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JEE Main & Advanced JEE Main Paper (Held On 8 April 2017)

  • question_answer
    Two wires \[{{W}_{1}}\] and \[{{W}_{2}}\] have the same radius r and respective densities \[{{\rho }_{1}}\] and \[{{\rho }_{2}}\] such that\[{{\rho }_{2}}=4{{\rho }_{1}}.\]. They are joined together at the point O, as shown in the figure. The combination is used as a sonometer wire and kept under tension T. The point O is midway between the two bridges. When a stationary waves is set up in the composite wire, the joint is found to be a node. The ratio of the number of antinodes formed in \[{{W}_{1}}\]to \[{{W}_{2}}\] is -                                 [JEE Online 08-04-2017]

    A)  4 : 1                                      

    B)  1 : 2

    C)  1 : 1                                      

    D)  1 : 3

    Correct Answer: B

    Solution :

    \[{{n}_{1}}={{n}_{2}}\] \[T\to \]same \[r\to \]same \[l\to \]same \[n=\frac{p}{2l}\sqrt{\frac{T}{\pi {{r}^{2}}d}}\] \[{{n}_{1}}={{n}_{2}}\] \[\frac{{{p}_{1}}}{\sqrt{{{d}_{1}}}}=\frac{{{p}_{2}}}{\sqrt{{{d}_{2}}}}\] \[\frac{{{p}_{1}}}{{{p}_{2}}}=\frac{1}{2}\]


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