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JEE Main & Advanced Sample Paper JEE Main Sample Paper-2

  • question_answer
    A tuning fork, when sounded with a column of air at temperature \[{{T}_{1}}\] gives 4 beats per second. When the temperature of air column is increased to \[{{T}_{2}}\] it gives 6 beats per second. If the original frequency of tuning fork is 4 Hz, then ratio of \[{{T}_{2}}/{{T}_{1}}\] becomes

    A)  \[\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{25}{16}\]                       

    B)  \[\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{16}{25}\]

    C)  \[\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{5}{4}\]                            

    D)  \[\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{4}{5}\]

    Correct Answer: A

    Solution :

    Beat frequency \[{{v}_{1}}-{{v}_{0}}=4\]                                               ?(i) \[{{v}_{2}}-{{v}_{0}}=6\]                                               ?(ii) \[\Rightarrow \,\,\,\,\,\,{{v}_{2}}-{{v}_{1}}=2\]                                  ?(iii) Also, \[v=\frac{1}{4\ell }\sqrt{\frac{\ell RT}{M}}\] \[\frac{{{v}_{1}}}{{{v}_{2}}}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\]                                   ?(iv) By solving above expression \[\frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{5}{4} \right)}^{2}}=\frac{15}{16}\]


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