Correct Answer - Option 3 :

\(1 : \sqrt{rx}\)
The correct answer is option 3) i.e. \(1 : \sqrt{rx}\)

__Concept__:

Escape velocity is the minimum velocity with which a body is projected from the surface of the planet so as to reach infinity, by overcoming the pull by gravity.

Escape velocity at the surface of a planet is given by:

\(⇒ V_e=\sqrt{\frac{2GM}{R}}\)

Where,

G = gravitational constant (6.67 × 10-11 Nm2/kg2), M = mass of the planet and R = radius of the planet.

Acceleration due to gravity g is obtained from balancing the equation of force with the equation of gravitational force.

\(mg =\frac{GMm}{R^2}⇒ g =\frac{GM}{R^2}\)

Where M is the mass of the earth, m is the mass of an object, R is the radius of the earth, and G is the gravitational constant.

On comparing both the equations, we get

\(⇒ V_e= \sqrt{2gR}\)

__Explanation__:

Given that:

\(\frac{R_A}{R_B} = \frac{1}{r}\) and \(\frac{g_A}{g_B}=\frac{1}{x}\)

The ratio of escape speed,

\(\frac{V_A}{V_B} = \frac{\sqrt{2g_AR_A}}{\sqrt{2g_BR_B}}\)

\(\Rightarrow \frac{V_A}{V_B} = \sqrt{\frac{{g_A}}{{g_B}}} \times \sqrt{\frac{{R_A}}{{R_B}}}\)

\(\Rightarrow \frac{V_A}{V_B} = \sqrt{\frac{{1}}{{x}}} \times \sqrt{\frac{{1}}{{r}}}\)

\(\Rightarrow \frac{V_A}{V_B} = {\frac{{1}}{\sqrt{rx}}} \)