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JEE Main & Advanced Physics Gravitation / गुरुत्वाकर्षण Question Bank Acceleration Due to Gravity

  • question_answer
    R is the radius of the earth and \[\omega \] is its angular velocity and \[{{g}_{p}}\] is the value of g at the poles. The effective value of g at the latitude \[\lambda =60{}^\circ \] will be equal to      [MP PMT 1999]

    A)             \[{{g}_{p}}-\frac{1}{4}R{{\omega }^{2}}\]

    B)               \[{{g}_{p}}-\frac{3}{4}R{{\omega }^{2}}\]

    C)             \[{{g}_{p}}-R{{\omega }^{2}}\]

    D)               \[{{g}_{p}}+\frac{1}{4}R{{\omega }^{2}}\]

    Correct Answer: A

    Solution :

         \[g={{g}_{p}}-R{{\omega }^{2}}{{\cos }^{2}}\lambda \]=\[{{g}_{p}}-{{\omega }^{2}}R{{\cos }^{2}}60{}^\circ \] =\[{{g}_{p}}-\frac{1}{4}R{{\omega }^{2}}\]


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