At a certain depth $$d$$ below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height $$3R$$ above earth surface. Where $$R$$ is Radius of earth (Take $$R = 6400$$ km). The depth $$d$$ is equal to

We need to find the depth $$d$$ below the Earth's surface where the gravitational acceleration equals four times its value at height $$3R$$ above the surface.

To begin,

At a height $$h$$ above the Earth's surface, the acceleration due to gravity is:

$$ g_h = \frac{g}{\left(1 + \frac{h}{R}\right)^2} $$

This follows from Newton's law of gravitation: $$g_h = \frac{GM}{(R+h)^2} = \frac{g R^2}{(R+h)^2}$$, where $$g = GM/R^2$$ is the surface value.

Next,

$$ g_h = \frac{g}{\left(1 + \frac{3R}{R}\right)^2} = \frac{g}{(1+3)^2} = \frac{g}{16} $$

From this,

At depth $$d$$, assuming uniform density of the Earth:

$$ g_d = g\left(1 - \frac{d}{R}\right) $$

This is because only the mass in the sphere of radius $$(R-d)$$ contributes to gravity at that depth (by the shell theorem), and that mass is proportional to $$(R-d)^3$$ while the distance from the centre is $$(R-d)$$, giving $$g_d = g(R-d)/R$$.

Continuing,

$$ g\left(1 - \frac{d}{R}\right) = 4 \times \frac{g}{16} $$

$$ g\left(1 - \frac{d}{R}\right) = \frac{g}{4} $$

Dividing both sides by $$g$$:

$$ 1 - \frac{d}{R} = \frac{1}{4} $$

$$ \frac{d}{R} = 1 - \frac{1}{4} = \frac{3}{4} $$

Now,

$$ d = \frac{3}{4} \times R = \frac{3}{4} \times 6400 = 4800\;\text{km} $$

The depth is 4800 km.

The correct answer is Option 4: 4800 km.