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JEE Mains · Physics · STD 11 - Gravitation
  • JEE Mains 2019
  • Hard
  • MCQ

The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is \(2\,s\). The period of oscillation of the same pendulum on the planet would be

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Answer & Solution

Correct Answer

(D) \(2\sqrt 3 \,s\)

Step-by-step Solution

Detailed explanation

\(g = \frac{{GM}}{{{R^2}}}\)

\(\frac{{{g_p}}}{{{g_e}}} = \frac{{{M_p}}}{{{M_e}}}{\left( {\frac{{{R_e}}}{{{R_p}}}} \right)^2} = 3{\left( {\frac{1}{3}} \right)^2} = \frac{1}{3}\)

\(Also,\,T \propto \frac{1}{{\sqrt g }}\)

\( \Rightarrow \frac{{{T_p}}}{{{T_e}}} = \sqrt {\frac{{{g_e}}}{{{g_p}}}}  = \sqrt 3 \)

\( \Rightarrow {T_p} = 2\sqrt {3s} \)

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