Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11} g$,where $g$ is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $\frac{2}{3}$ times that of the earth. If the escape speed on the surface of the earth is taken to be $11 \ km/s$,the escape speed on the surface of the planet in $km/s$ will be:

$A$ black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would Earth (mass $= 5.98 \times 10^{24} \, kg$) have to be compressed to be a black hole?

$A$ particle of mass $m$ is projected with a velocity $v = k V_{e}$ $(k < 1)$ from the surface of the earth. $(V_{e} = \text{escape velocity})$. The maximum height above the surface reached by the particle is:

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

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A :$ The escape velocities of planet $A$ and $B$ are same. But $A$ and $B$ are of unequal mass.
Reason $R :$ The product of their mass and radius must be same,$M_{1}R_{1} = M_{2}R_{2}$.
In the light of the above statements,choose the most appropriate answer from the options given below.

The condition for a uniform spherical mass $m$ of radius $r$ to be a black hole is [$G=$ gravitational constant and $c=$ speed of light]