Punjab Medical Punjab - MET Solved Paper-2004

  • question_answer
    On a planet a freely falling body takes\[2\,\,\sec \] when it is dropped from a height of \[8\,\,m\], the time period of simple pendulum of length \[1\,\,m\] on that planet is:

    A) \[3.14\sec \]                      

    B) \[16.28\sec \]

    C) \[1.57\sec \]                      

    D)  none of these

    Correct Answer: A

    Solution :

    Here: Time\[t=2\sec \], height\[h=8m\] Initial velocity\[v=0\] Length of pendulum\[=1m\] Relation for the height is given by,                 \[h=ut+\frac{1}{2}g{{t}^{2}}\]                 \[8=0\times 2+\frac{1}{2}g\times {{2}^{2}}=2\,\,g\]                 \[2g=8\]or\[g=4m/{{s}^{2}}\] Hence, the time period \[T\] is given by,                 \[=2\pi \sqrt{\frac{l}{g}}=2\pi \sqrt{\frac{1}{4}}=\pi \]


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