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BHU PMT BHU PMT (Mains) Solved Paper-2006

  • question_answer
    A simple pendulum has time period\[{{T}_{1}}\]. The point of suspension is now moved upward according to the relation\[y=k{{t}^{2}}(k=1\,m/{{s}^{2}})\]where y is the vertical displacement. The time period now becomes\[{{T}_{2}}\]. The ratio of\[\frac{T_{1}^{2}}{T_{2}^{2}}\]is: \[(g=10\,m/{{s}^{2}})\]

    A)  \[\frac{6}{5}\]  

    B)   \[\frac{5}{6}\]

    C)  1                                            

    D)  \[\frac{4}{5}\]

    Correct Answer: A

    Solution :

                     \[y=k{{t}^{2}}\] \[\frac{{{d}^{2}}y}{d{{t}^{2}}}=2k\] or            \[a=2m/{{s}^{2}}\]                    (as\[k=1m/{{s}^{2}}\])                 \[{{T}_{1}}=2\pi \sqrt{\frac{l}{g}}\] and        \[{{T}_{2}}=2\pi \sqrt{\frac{l}{g+{{a}_{y}}}}\] \[\therefore \]  \[\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{g+{{a}_{y}}}{g}\]                 \[=\frac{10+2}{10}=\frac{6}{5}\]


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