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(D) The acceleration due to gravity on a planet is given by $g = \frac{GM}{R^2}$.Let $M_e$ and $R_e$ be the mass and radius of the Earth,and $M_p$ and

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$2\sqrt{3} \ s$

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(D) The acceleration due to gravity on a planet is given by $g = \frac{GM}{R^2}$.Let $M_e$ and $R_e$ be the mass and radius of the Earth,and $M_p$ and $R_p$ be the mass and radius of the planet.Given: $M_p = 3M_e$ and $D_p = 3D_e$,which implies $R_p = 3R_e$.The ratio of gravity on the planet to the Earth is $\frac{g_p}{g_e} = \frac{M_p}{M_e} \left( \frac{R_e}{R_p} \right)^2 = 3 \left( \frac{1}{3} \right)^2 = \frac{3}{9} = \frac{1}{3}$.The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$,so $T \propto \frac{1}{\sqrt{g}}$.Therefore,$\frac{T_p}{T_e} = \sqrt{\frac{g_e}{g_p}} = \sqrt{3}$.Given $T_e = 2 \ s$,we get $T_p = 2\sqrt{3} \ s$.

Column-$I$	Column-$II$
$(1)$ Maximum value of acceleration due to gravity $g$	$(a)$ At the center of the Earth
$(2)$ Minimum value of acceleration due to gravity $g$	$(b)$ At the poles
$(3)$ Zero value of acceleration due to gravity $g$	$(c)$ At the equator

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 $2 \ s$. The period of oscillation of the same pendulum on the planet would be

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Column-$I$ Column-$II$

$(1)$ Maximum value of acceleration due to gravity $g$ $(a)$ At the center of the Earth

$(2)$ Minimum value of acceleration due to gravity $g$ $(b)$ At the poles

$(3)$ Zero value of acceleration due to gravity $g$ $(c)$ At the equator

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