If earth has a mass nine times and radius twice to the planet P. Then $\dfrac{v_e}{3}\sqrt{x}$ ms$^{-1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of P, where $v_e$ is escape velocity on earth. The value of $x$ is

A body is projected vertically from earth's surface of radius R with velocity equal to half the escape velocity. The maximum height reached by the body is

The density of a planet is three times that of earth and radius $2.5$ times that of earth. If $g_p$ and $g_e$ represents acceleration due to gravity on planet and earth respectively then ratio $g_p$ to $g_e$ is

Given below are two statements:
Statement I: Acceleration due to earth's gravity decreases as you go 'up' or 'down' from earth's surface.
Statement II: Acceleration due to earth's gravity is same at a height '$h$' and depths '$d$' from earth's surface, if $h = d$.
In the light of above statements, choose the most appropriate answer from the options given below.