What is the escape velocity for the surface of a black hole?

The escape velocity from a spherical planet $A$ is $10 \ km/s$. The escape velocity from another planet $B$ whose density and radius are $10\%$ of those of planet $A$,is . . . . . . $m/s$.

Planet $A$ has mass $M$ and radius $R$. Planet $B$ has half the mass and half the radius of Planet $A$. If the escape velocities from the Planets $A$ and $B$ are $v_{A}$ and $v_{B}$ respectively,then $\frac{v_{A}}{v_{B}}=\frac{n}{4}$. The value of $n$ is

The escape velocity for the earth is $v_e$. The escape velocity for a planet whose radius is four times and density is nine times that of the earth,is (in $,v_e$)

The escape velocity from the moon's surface is less than that on the earth's surface,because:

$v_e$ and $v_p$ denote the escape velocity from the Earth and another planet having twice the radius and the same mean density as the Earth. Then: