JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Evening)

  • question_answer
    A hope and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moment parallel to their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner angle with the field. If the oscillation periods of hoop and cylinder are \[{{T}_{h}}\text{ }and\text{ }{{T}_{c}}\] respectively, then- [JEE Main Online Paper (Held On 10-Jan-2019 Evening]

    A) \[{{T}_{h}}=1.5\text{ }{{T}_{c}}\]                             

    B) \[{{T}_{h}}={{T}_{c}}\]

    C) \[{{T}_{h}}=2{{T}_{c}}\]       

    D)                  \[{{T}_{h}}=0.5{{T}_{c}}\]

    Correct Answer: B

    Solution :

    \[{{M}_{C}}=2M\text{ }\And {{M}_{H}}=M\] Using \[T=2\pi \sqrt{\frac{I}{MB}}\] \[\therefore \,\,\,\,\frac{{{T}_{H}}}{{{T}_{C}}}=\sqrt{\frac{{{I}_{H}}}{{{M}_{H}}}\,\,\frac{{{M}_{C}}}{{{I}_{C}}}}\] \[=\,\,\sqrt{\frac{m{{R}^{2}}}{\frac{1}{2}m{{R}^{2}}\,2M}}\,\,=\,\,\frac{1}{1}\] \[{{T}_{\ell }}=Tc\]


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