KVPY Sample Paper KVPY Stream-SX Model Paper-21

  • question_answer
    A simple pendulum has a period T on the earth. What is the period T of this same pendulum on the moon, where the acceleration due w gravity is 1/6 that of the earth?

    A) \[\frac{T}{\sqrt{6}}\]

    B) \[\frac{T}{6}\]

    C) \[\sqrt{6}T\]

    D) 36T

    Correct Answer: C

    Solution :

    The length of the pendulum hasn?t changed and the only other factor that determines the period of a pendulum is the acceleration due to gravity
    \[T=2\pi \sqrt{\frac{\ell }{g}}\]
    \[T=2\pi \sqrt{\frac{\ell }{\frac{1}{6g}}}\]
    \[T=\sqrt{6}\left( 2\pi \sqrt{\frac{\ell }{g}} \right)\]

You need to login to perform this action.
You will be redirected in 3 sec spinner