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NEET AIPMT SOLVED PAPER SCREENING 2010

  • question_answer
    A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 s in earth's horizontal magnetic field of 24 \[\text{T}\text{.}\]When a horizontal field of \[\text{18 }\text{T}\]is produced opposite to the earth's field by placing a current carrying wire, the new time period of magnet will be

    A) 1s     

    B)                                        2s                           

    C)        3s                           

    D)        4s

    Correct Answer: B

    Solution :

    Time period in vibration magnetometer \[T=2\pi \sqrt{\frac{I}{M\times {{B}_{H}}}}\]                 \[\Rightarrow \]                               \[T\propto \frac{1}{\sqrt{B{{ & }_{H}}}}\]                                              \[\Rightarrow \]                               \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{({{B}_{H}})}_{2}}}{{{({{B}_{H}})}_{1}}}}\]                 \[\Rightarrow \]                               \[\frac{2}{{{T}_{2}}}=\sqrt{\frac{18}{24}}\]                                                 \[T=2.3s\approx 2s\]


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