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JEE Main & Advanced AIEEE Solved Paper-2003

  • question_answer
    A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes \[5T/3\], then the ratio of \[\frac{m}{M}\] is     AIEEE  Solved  Paper-2003

    A) \[\frac{3}{5}\]                                   

    B)       \[\frac{25}{9}\]                                

    C) \[\frac{16}{9}\]                

    D)       \[\frac{5}{3}\]

    Correct Answer: C

    Solution :

                    The time period for the SHM,                                 \[T=2\pi \sqrt{\frac{M}{k}}\]                      ... (i) and            \[T'=2\pi \sqrt{\frac{M+m}{k}}\] \[\Rightarrow \]   \[\frac{5T}{3}=2\pi \sqrt{\frac{M+m}{k}}\]           ... (ii) Dividing Eq. (i) by Eq. (ii), we have                     \[\frac{3}{5}=\sqrt{\frac{M}{M+m}}\] \[\Rightarrow \]   \[\frac{9}{25}=\frac{M}{M+m}\] \[\Rightarrow \]   \[9M+9m=25M\] \[\Rightarrow \]               \[16M=9m\] \[\therefore \]  \[\frac{m}{M}=\frac{16}{9}\]     


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