Imagine a new planet having the same density as that of earth but it is 3 times bigger than the earth in size.
If the acceleration due to gravity on the surface of earth is g and that on the surface of the new planet is g', then

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$g^{'} = 9 g$

$g^{'} = 3 g$

$g^{'} = g / 9$

$g^{'} = 27 g$

answer is D.

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Detailed Solution

We know that

$\begin{array}{l} g = \frac{G M}{R^{2}} = \frac{G (\frac{4}{3} π R^{3}) ρ}{R^{2}} = \frac{4}{3} π G R ρ \\ \frac{g^{'}}{g} = \frac{R^{'}}{R} = \frac{3 R}{R} = 3 \\ g^{'} = 3 g \end{array}$

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