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MGIMS WARDHA MGIMS WARDHA Solved Paper-2005

  • question_answer
    The period of oscillation of a simple pendulum   of constant length at surface of the earth is T. Its time period inside a mine will be :

    A) cannot be compared.

    B) equal to T

    C) less than T

    D) more than T

    Correct Answer: D

    Solution :

    Time period of a simple pendulum at the surface of earth, \[T=2\pi \sqrt{\frac{l}{g}}\] \[T\propto \frac{1}{\sqrt{g}}\] In a mine (below the surface of earth) value of 'g' will decrease, so time period 'T' will increase.


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