The magnitude of the potential energy per unit mass of the object at the surface of earth is E. Then the escape velocity of the object is

The correct option is √(2E).

Let us derive this step-by-step:

1. Potential Energy per Unit Mass:
The potential energy ($U$) of an object of mass $m$ at the surface of the Earth is given by the formula:
$U=-\frac{GMm}{R}$
where $G$ is the universal gravitational constant, $M$ is the mass of the Earth, and $R$ is the radius of the Earth.

The potential energy per unit mass is:
$\frac{U}{m}=-\frac{GM/}{R}$

The question defines $E$ as the magnitude of this potential energy per unit mass:
$E=|-\frac{GM}{R}|=\frac{GM}{R}$

2. Escape Velocity:
The escape velocity ($v_e$) of an object from the surface of the Earth is the minimum speed required for it to escape the Earth's gravitational field. It is given by the formula:
$v_e=\sqrt{\frac{2GM}{R}}$

3. Relating Escape Velocity to E:
We can rewrite the expression for escape velocity by substituting $E=\frac{GM}{R}$:
$v_e=\sqrt{2\left(\frac{GM}{R}\right)}=\sqrt{2E}$

Therefore, the escape velocity of the object is √(2E).