2. Solved Papers
  3. AMU Medical

AMU Medical AMU Solved Paper-2011

  • question_answer
    A person measures the time period of a simple pendulum inside a stationary lift and finds it to be T. If the lift starts accelerating upwards with an acceleration g/3, the time period of the pendulum will be

    A) \[\sqrt{3}T\]                                      

    B)  \[T/\sqrt{3}\]

    C)  \[\frac{\sqrt{3}T}{2}\]                                  

    D)  T/3

    Correct Answer: C

    Solution :

                     Time period of the simple pendulum in the lift                                 \[T=2\pi \sqrt{\frac{l}{ga}}\] Time period when lift starts accelerating upwards with acceleration g/3                                 \[T=2\pi \sqrt{\frac{l}{g+\frac{g}{3}}}\]                                 \[T=2\pi \times \sqrt{3}\sqrt{\frac{l}{4g}}\]                 or            \[T=2p\sqrt{\frac{l}{g}}\times \frac{\sqrt{3}}{2}\] or            \[T=\frac{\sqrt{3}}{2}T\]


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