The mass and diameter of a planet are twice those of earth. What will be the time period of the pendulum on this planet, which is a seconds pendulum on the earth?

Value of $g$ on the surface of earth, \[g=\frac{GM}{R^2}=\frac{4GM}{D^2}\] where $D$ is the diameter of the earth $D=2R.$

Value of $g$ on the surface of planet where $M’=2M$ and $D’=2D,$

\[g’=\frac{4GM’}{D’^2}=\frac{4G(2M)}{(2D)^2}=\frac{1}{2}\frac{4GM}{D^2}=\frac{g}{2}\]

The time period of a simple pendulum on earth is, \[T=2π\sqrt{\frac{l}{g}}\]

The time period of the pendulum on that planet is,

\[T’=2π\sqrt{\frac{l}{g’}}=2π\sqrt{\frac{l}{g/2}}\] \[=\sqrt{2}\left(2π\sqrt{\frac{l}{g}}\right)=\sqrt{2}T\]

For a seconds pendulum on earth, \[T’=\sqrt{2}×2=2.828\;\text{sec}\]

Therefore, the required time period will be $2.828$ sec.

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