NEET AIPMT SOLVED PAPER 2002

  • question_answer
                    Two bar magnets having same geometry with magnetic moments M and 2M, are firstly placed in such a way that their similar poles are on the same side; then its period of oscillation is \[{{T}_{1}}\]. Now the polarity of one of the magnets is reversed, then time period of oscillations will be:                                               

    A)                 \[{{T}_{1}}<{{T}_{2}}\]

    B)                 \[{{T}_{1}}>{{T}_{2}}\]

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

    D)                 \[{{T}_{2}}=\] \[\infty \]

    Correct Answer: A

    Solution :

                    The time period of bar magnet                 \[T=2\pi \sqrt{\frac{I}{MH}}\]                 where M= magnetic moment of magnet                 I = moment of inertia and                    H = horizontal component of magnetic field                 When same poles of magnets are placed on same side, then net magnetic moment                 \[{{M}_{1}}=M+2M=3M\]                 \[\therefore T{{ & }_{1}}=2\pi \,\sqrt{\frac{I}{{{M}_{1}}H}}\]                 \[=2\pi \,\sqrt{\frac{I}{3\,M\,H}}\]                                          ....(i)                 When opposite poles of magnets are placed on same side, then net magnetic moment                 \[{{M}_{2}}=2M-M=M\]                 \[\therefore {{T}_{2}}=2\pi \sqrt{\frac{I}{{{M}_{2}}H}}=2\pi \sqrt{\frac{I}{M\,H}}....(ii)\]                 From Eqs. (i) and (ii), we observe that                                 \[{{T}_{1}}<{{T}_{2}}\]


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