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JEE Main & Advanced Sample Paper JEE Main Sample Paper-37

  • question_answer
    The fundamental frequency of a sonometer wire of length \[\ell \]is \[{{n}_{0}}\]. A bridge is now introduced at a distance of \[\Delta \ell (<<\ell )\]from the centre of the wire. The lengths of wire on the two sides of the bridge are now vibrated in their fundamental modes. Then, the beat frequency nearly is -

    A)  \[{{n}_{0}}\Delta \ell /\ell \]                 

    B)  \[8{{n}_{0}}\Delta \ell /\ell \]

    C)  \[2{{n}_{0}}\Delta \ell /\ell \]                            

    D)  \[{{n}_{0}}\Delta \ell /2\ell \]

    Correct Answer: B

    Solution :

     \[{{n}_{0}}=\frac{v}{2\ell },\,{{n}_{1}}=\frac{v}{2(\ell /2-\Delta \ell )},{{n}_{2}}=\frac{v}{2(\ell /2+\Delta \ell )}\] Beat frequency \[={{n}_{1}}-{{n}_{2}}\]                         \[\Rightarrow \] \[v\left[ \frac{1}{\ell -2\Delta \ell }-\frac{1}{\ell +2\Delta \ell } \right]=v\left[ \frac{(\ell +2\Delta \ell )-(\ell -2\Delta \ell )}{{{\ell }^{2}}-4\Delta {{\ell }^{2}}} \right]\]             \[=v\frac{4\Delta \ell }{{{\ell }^{2}}-4\Delta {{\ell }^{2}}}=\frac{8}{\ell }\frac{\Delta \ell v}{2\ell }=\frac{8\Delta \ell {{n}_{0}}}{\ell }\]


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