If g is the acceleration due to gravity at the earth's surface and r is the radius of the earth, the escape velocity for the body to escape out of earth's gravitational field is
Correct option is B. $$\sqrt{2gr}$$
Let the mass of the body be $$m$$. When the body will escape earth's gravity into space it will have kinetic and potential energies as zero.
$$(KE+PE)_{surface}=(KE+PE)_{space}$$
$$\Rightarrow \frac{1}{2}mv_e^2-mgr=0$$ ( $$r$$ is the radius of earth)
$$\Rightarrow \frac{1}{2}mv_e^2=mgr$$
$$\Rightarrow v_e^2 = 2gr$$.$$\Rightarrow v_e=\sqrt{2gr}$$ is our required answer,