Gravitation

JEE Main level, single correct, medium

90 of 223

Question 90 of 223

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 s2 \mathrm{~s}. The period of oscillation of the same pendulum on the planet would be:

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Answer: (D) 232 \sqrt{3} s

Acceleration due to gravity g=GMR2\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}

gpge=MpMe(ReRp)2=3(13)2=13\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∝1g⇒TpTe=gegp=3\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{g}}} \Rightarrow \frac{\mathrm{T}_{\mathrm{p}}}{\mathrm{T}_{\mathrm{e}}}=\sqrt{\frac{\mathrm{g}_{\mathrm{e}}}{\mathrm{g}_{\mathrm{p}}}}=\sqrt{3}

⇒Tp=23 s\Rightarrow \mathrm{T}_{\mathrm{p}}=2 \sqrt{3} \mathrm{~s}

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