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JEE Main & Advanced Physics Magnetism Question Bank Magnetic Equipments

  • question_answer
    A magnet freely suspended in a vibration magnetometer makes 10 oscillations per minute at a place A and 20 oscillations per minute at a place B. If the horizontal component of earth?s magnetic field at A is \[36\times {{10}^{-6}}T,\] then its value at B is                        [EAMCET (Med.) 2001]

    A)            36 ´ 10?6 T                             

    B)            72 ´ 10?6 T

    C)            144 ´ 10?6 T                           

    D)            288 ´ 10?6 T

    Correct Answer: C

    Solution :

               \[\frac{{{T}_{A}}}{{{T}_{B}}}=\sqrt{\frac{{{({{B}_{H}})}_{B}}}{{{({{B}_{H}})}_{A}}}}\Rightarrow \frac{60/10}{60/20}=\sqrt{\frac{{{({{B}_{H}})}_{B}}}{36\times {{10}^{-6}}}}\] \[\Rightarrow {{({{B}_{H}})}_{B}}=144\times {{10}^{-6}}T\]


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