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BHU PMT BHU PMT (Mains) Solved Paper-2010

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

    A)  \[\frac{6}{5}\]                                  

    B)  \[\frac{5}{6}\]

    C)  1                                            

    D)  4

    Correct Answer: A

    Solution :

                     According to the relation, \[y=k{{t}^{2}}\Rightarrow \frac{dy}{dt}=2kt\] \[\therefore \]  \[a=\frac{{{d}^{2}}y}{d{{t}^{2}}}=2k=2\times 1=2\]          \[[\because k=1]\] The point of suspension is moving upward with acceleration a, then effective acceleration due to gravity on pendulum \[g'=g+a=10+2=12\,m/{{s}^{2}}\]                 \[\therefore \]  \[{{T}_{1}}=2\pi \sqrt{\frac{l}{g}}\]                 And        \[{{T}_{2}}=2\pi \sqrt{\frac{l}{g+a}}\]                 \[\therefore \]  \[\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{g+a}{g}=\frac{12}{10}=\frac{6}{5}\]


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