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(D) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g}}$.From this formula,we can see that the time period is in

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$\sqrt{2}T$

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(D) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g}}$.From this formula,we can see that the time period is inversely proportional to the square root of the acceleration due to gravity: $T \propto \frac{1}{\sqrt{g}}$.Let $T_e$ and $g_e$ be the time period and acceleration due to gravity on Earth,and $T_p$ and $g_p$ be the time period and acceleration due to gravity on the other planet.Given that the acceleration due to gravity on the planet is half that of the Earth,we have $g_p = \frac{g_e}{2}$.Using the proportionality relation:$\frac{T_p}{T_e} = \sqrt{\frac{g_e}{g_p}} = \sqrt{\frac{g_e}{g_e/2}} = \sqrt{2}$.Therefore,the new time period $T_p = \sqrt{2}T_e = \sqrt{2}T$.

$A$ pendulum has a time period $T$. If it is taken to another planet having an acceleration due to gravity half that of the Earth and a mass $9$ times that of the Earth,then its time period on the other planet will be:

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