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The radii of two planets ‘A’ and ‘B’ are ‘R’ and ‘4R’ and their densities are $ρ$ and $ρ / 3$ respectively. The ratio of acceleration due to gravity at their surfaces $(g_{A} : g_{B})$ will be

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3 : 4

3 : 16

4 : 3

1 : 16

answer is D.

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Detailed Solution

$g = \frac{4}{3} πGρR$

$\Rightarrow \frac{g_{A}}{g_{B}} = \frac{R_{A}}{R_{B}} \times \frac{ρ_{A}}{ρ_{B}} = \frac{R}{4 R} \times \frac{ρ}{(\frac{ρ}{3})} = \frac{3}{4}$

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