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A hoop and a solid cylinder of same mass and radius are made of permanent magnetic material with their magnetic moment parallel to their respective axes.But the magnetic moment of the hoopis twice of solid cylinder.They are placed in a uniform magnetic field in such a manner that there magnetic moment make a small angle with the field. If the oscillation periods of hoop and cylinder are T_{h} and T_{c} respectively, then:

 

 

  • Option 1)

     

    T_{h}=0.5T_{c}

  • Option 2)

     

    T_{h}=2T_{c}

  • Option 3)

     

    T_{h}=1.5T_{c}

  • Option 4)

     

    T_{h}=T_{c}

Answers (1)

best_answer

 

Time Period of Freely Suspended Magnet -

T= 2\pi \sqrt{\frac{I}{MB}}

- wherein

I - Moment of Inertia

T = 2\pi \sqrt{\frac{I}{MB}}

T_{h} = 2\pi \sqrt{\frac{mR^{2}}{2MB}}

T_{c} = 2\pi \sqrt{\frac{\frac{1}{2}mR^{2}}{MB}}

T_{h} = T_{c}

 


Option 1)

 

T_{h}=0.5T_{c}

Option 2)

 

T_{h}=2T_{c}

Option 3)

 

T_{h}=1.5T_{c}

Option 4)

 

T_{h}=T_{c}

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