RAJASTHAN PMT Rajasthan - PMT Solved Paper-2008

  • question_answer
    A thin rectangular magnet suspended freely has a period of oscillation equal to T. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is T, the ratio T/T is

    A)  \[\frac{1}{2\sqrt{2}}\]                                  

    B)  \[\frac{1}{2}\]

    C)  2                            

    D)         \[\frac{1}{4}\]

    Correct Answer: B

    Solution :

    When magnet is divided into two equal parts, the magnetic dipole moment \[M=\]pole strength\[\times \frac{l}{2}=\frac{M}{2}\] (pole strength remains same) Also, the mass of magnet becomes half,\[ie,\]                 \[m=\frac{m}{2}\] Moment of inertia of magnet                 \[I=\frac{m{{l}^{2}}}{12}\] New moment of inertia                 \[I=\frac{1}{12}\left( \frac{m}{2} \right){{\left( \frac{l}{2} \right)}^{2}}=\frac{m{{l}^{2}}}{12\times 8}\] \[\therefore \]  \[I=\frac{I}{8}\] Now      \[T=2\pi \sqrt{\left( \frac{I}{MB} \right)}\]                 \[T=2\pi \sqrt{\left( \frac{I}{MB} \right)}=2\pi \sqrt{\left( \frac{I/8}{MB/2} \right)}\] \[\therefore \,\,\,\,\,\,\,\,\,T=\frac{T}{2}\,\Rightarrow \,\frac{T}{T}\,=\frac{1}{2}\]


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