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Uttarakhand PMT Uttarakhand PMT Solved Paper-2011

  • question_answer
    A simple pendulum of length; has a brass bob attached at its lower end. Its period is T. If a   steel bob of same size, having density \[x\] times that of brass, replaces the brass bob and its length is changed, so that period becomes, 2T then new length is

    A)  \[2l\]              

    B)  \[4l\]

    C)  \[4lx\]              

    D)  \[\frac{41}{x}\]

    Correct Answer: B

    Solution :

     Time period of simple pendulum \[T=2\pi \sqrt{\frac{l}{g}}\] Time period does not depend on density of bob. \[T\propto \sqrt{l}\]   (\[l=\]length of pendulum) \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{l}_{1}}}{{{l}_{2}}}}\] \[\frac{T}{2T}=\sqrt{\frac{l}{{{l}_{2}}}}\] Hence,  \[{{l}_{2}}=4l\] Hence, new length is\[4l.\]


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