2. Solved Papers
  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    A simple pendulum has a time period\[{{T}_{1}}\]when on the earths surface and\[{{T}_{2}}\]when taken to a height 2R above the earths surface where R is the radius of the earth. The value of\[({{T}_{1}}/{{T}_{2}})\]is:

    A)  \[\frac{1}{9}\]                  

    B)  \[\frac{1}{3}\]

    C)  \[\sqrt{3}\]                       

    D)         \[9\]

    E)  3

    Correct Answer: E

    Solution :

    The periodic time of a simple pendulum is given by, \[T=2\pi \sqrt{\frac{l}{g}}\] Where taken to height 2R. \[g=g{{\left( 1+\frac{h}{{{R}_{e}}} \right)}^{2}}=g{{\left( 1+\frac{2R}{R} \right)}^{-2}}=g{{(3)}^{-2}}\] \[\therefore \]  \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{1}{{{3}^{2}}}}\] \[\Rightarrow \]               \[{{T}_{2}}=3{{T}_{1}}\] \[\Rightarrow \]               \[\frac{{{T}_{2}}}{{{T}_{1}}}=3\]


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