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(B) The acceleration due to gravity on Earth is $g = \frac{GM}{R^2}$.On the planet,the mass $M' = 2M$ and the radius $R' = 2R$ (since diameter is twic

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$2\sqrt{2} \ s$

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(B) The acceleration due to gravity on Earth is $g = \frac{GM}{R^2}$.On the planet,the mass $M' = 2M$ and the radius $R' = 2R$ (since diameter is twice that of Earth).The acceleration due to gravity on the planet is $g' = \frac{G(2M)}{(2R)^2} = \frac{2GM}{4R^2} = \frac{1}{2}g$.The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$,so $T \propto \frac{1}{\sqrt{g}}$.For a seconds pendulum on Earth,$T_1 = 2 \ s$.Therefore,$\frac{T_2}{T_1} = \sqrt{\frac{g}{g'}} = \sqrt{\frac{g}{g/2}} = \sqrt{2}$.$T_2 = T_1 	imes \sqrt{2} = 2\sqrt{2} \ s$.

The mass and diameter of a planet are twice those of Earth. What will be the period of oscillation of a pendulum on this planet if it is a seconds pendulum on Earth?

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