The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be :
${{\sqrt 3 } \over 2}$ s
${3 \over 2}$ s
${2 \over {\sqrt 3 }}$ s
$2\sqrt 3 $ s
Solution
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