If the density of earth is doubled keeping its radius constant, then acceleration due to gravity (g) =
Answer & explanation
Correct answer: option 1
$g=\frac{GM}{R^2}=G \frac{4}{3} \pi \frac{R^3 \rho}{R^2}=\frac{4}{3} \pi GR \rho, g \propto \rho g$
If $\rho$ is doubled g is doubled
∴ $g^{\prime}-20 ms^{-2}$