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JEE Main & Advanced Physics Gravitation / गुरुत्वाकर्षण Question Bank Critical Thinking

  • question_answer
    A simple pendulum has a time period \[{{T}_{1}}\] when on the earth?s surface and \[{{T}_{2}}\] when taken to a height R above the earth?s surface, where R is the radius of the earth. The value of \[{{T}_{2}}/{{T}_{1}}\] is                                  [IIT-JEE 2001]

    A) 1         

    B)             \[\sqrt{2}\]

    C)             4         

    D)             2

    Correct Answer: D

    Solution :

                    If acceleration due to gravity is g at the surface of earth then at height R it value becomes \[g'=g{{\left( \frac{R}{R+h} \right)}^{2}}=\frac{g}{4}\]             \[{{T}_{1}}=2\pi \sqrt{\frac{l}{g}}\text{ and }\,{{T}_{2}}=2\pi \sqrt{\frac{l}{g/4}}\]\\[\frac{{{T}_{2}}}{{{T}_{1}}}=2\]


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