An object is kept at rest at a distance of 3R above the earth's surface where R is earth's radius. The minimum speed with which it must be projected so that it does not return to earth is : (Assume M = mass of earth, G = Universal gravitational constant)

The object is initially at rest at a height $$h = 3R$$ above Earth’s surface.
Therefore its initial distance from Earth’s centre is
$$r_0 = R + h = R + 3R = 4R$$.

To “never return to Earth” the object must just reach infinity with zero speed. Hence we apply conservation of mechanical energy between the initial point $$r_0$$ and infinity.

Initial mechanical energy
Kinetic: $$K_i = \frac12 m v^2$$ (where $$v$$ is the required minimum speed).
Gravitational potential: $$U_i = -\frac{GMm}{r_0} = -\frac{GMm}{4R}$$.

Final mechanical energy at infinity
Kinetic: $$K_f = 0$$ (minimum condition).
Potential: $$U_f = 0$$ (by convention).

Conservation of energy gives $$K_i + U_i = K_f + U_f$$ $$\frac12 m v^2 - \frac{GMm}{4R} = 0$$.

Solve for $$v$$:

$$\frac12 m v^2 = \frac{GMm}{4R}$$

$$v^2 = \frac{2GM}{4R} = \frac{GM}{2R}$$

$$v = \sqrt{\frac{GM}{2R}}$$.

The minimum speed required is $$\sqrt{\dfrac{GM}{2R}}$$, which corresponds to Option A.