2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. NLM, Friction, Circular Motion

JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Second Law of Motion

  • question_answer
    The time period of a simple pendulum measured inside a stationary lift is found to be T. If the lift starts accelerating upwards with an acceleration g/3, the time period is [EAMCET 1994; CMEET Bihar 1995; RPMT 2000]            

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

    B)                         \[T\sqrt{3}/2\]          

    C)                         \[T/\sqrt{3}\]               

    D)                         \[T/3\]

    Correct Answer: B

    Solution :

                              \[T=2\pi \sqrt{\frac{l}{g}}\] and \[T'=2\pi \sqrt{\frac{l}{4g/3}}\]             \[[As\ g'=g+a=g+\frac{g}{3}=\frac{4g}{3}\]]             \\[T'=\]\[\frac{\sqrt{3}}{2}T\]          


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