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(D) The time of fall $t$ for an object falling from height $h$ is given by $t = \sqrt{\frac{2h}{g}}$. Thus,$t \propto \frac{1}{\sqrt{g}}$.The time per

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$t' = 2 T'$

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(D) The time of fall $t$ for an object falling from height $h$ is given by $t = \sqrt{\frac{2h}{g}}$. Thus,$t \propto \frac{1}{\sqrt{g}}$.The time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$. Thus,$T \propto \frac{1}{\sqrt{g}}$.Taking the ratio,we get $\frac{t}{T} = \frac{k_1 / \sqrt{g}}{k_2 / \sqrt{g}} = 	ext{constant}$.Since the ratio $\frac{t}{T}$ is independent of the acceleration due to gravity $g$,the relationship remains the same on any planet.Given that on Earth $t = 2T$,it follows that on the other planet $t' = 2T'$.

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