The height 'h' at which the weight of a body will be the same as that at the same depth 'h' from the surface of the earth is (Radius of the earth is R and effect of the rotation of the earth is neglected):

Let the acceleration due to gravity on the surface of the Earth be denoted by $$g$$. We have two different positions to consider for the same body:

1. At a height $$h$$ above the surface. 2. At a depth $$h$$ below the surface.

According to Newton’s law of gravitation, the magnitude of $$g$$ varies with distance from the centre of the Earth. First we state the two standard formulae that connect $$g$$ with height and depth (rotation neglected):

• For a height $$h$$ above the surface, the distance from the centre becomes $$R+h$$.  The formula is $$g_h = g\left(\dfrac{R}{R+h}\right)^2$$.

• For a depth $$h$$ below the surface, only the mass enclosed within the radius $$R-h$$ contributes.  For a uniform Earth, $$g_d = g\left(1-\dfrac{h}{R}\right)$$.

The problem states that the weight (and hence the gravitational acceleration) is the same in both cases, so we equate the two expressions:

$$g\left(\dfrac{R}{R+h}\right)^2 \;=\; g\left(1-\dfrac{h}{R}\right).$$

Because the factor $$g$$ appears on both sides, it cancels out, leaving

$$\left(\dfrac{R}{R+h}\right)^2 = 1-\dfrac{h}{R}.$$

To simplify the algebra, we introduce the dimensionless variable

$$x = \dfrac{h}{R}.$$

Substituting $$h = xR$$ into the equation gives

$$\left(\dfrac{R}{R + xR}\right)^2 = 1 - x.$$

Inside the brackets, factor out $$R$$:

$$\left(\dfrac{R}{R(1 + x)}\right)^2 = 1 - x \;\;\Longrightarrow\;\; \left(\dfrac{1}{1 + x}\right)^2 = 1 - x.$$

Remove the fraction by multiplying both sides by $$(1 + x)^2$$:

$$(1 + x)^2\,(1 - x) = 1.$$

Now expand the left-hand side step by step. First, square the binomial:

$$(1 + x)^2 = 1 + 2x + x^2.$$

Next, multiply this result by $$(1 - x)$$:

$$(1 + 2x + x^2)(1 - x) = 1(1 - x) + 2x(1 - x) + x^2(1 - x).$$

Carrying out each product:

$$1 - x + 2x - 2x^2 + x^2 - x^3.$$

Combine like terms:

$$1 + x - x^2 - x^3.$$

Therefore the equation becomes

$$1 + x - x^2 - x^3 = 1.$$

Subtract $$1$$ from both sides to bring all terms to one side:

$$x - x^2 - x^3 = 0.$$

Factor out the common factor $$x$$:

$$x\bigl(1 - x - x^2\bigr) = 0.$$

One solution is $$x = 0$$, corresponding to $$h = 0$$, which is trivial. For a non-zero height we require

$$1 - x - x^2 = 0.$$

This is a quadratic in $$x$$. Rewriting it in standard form:

$$x^2 + x - 1 = 0.$$

We now apply the quadratic formula $$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a = 1,\; b = 1,\; c = -1$$:

$$x = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \dfrac{-1 \pm \sqrt{1 + 4}}{2} = \dfrac{-1 \pm \sqrt{5}}{2}.$$

The negative root gives a negative height, which is not meaningful here, so we choose the positive root:

$$x = \dfrac{-1 + \sqrt{5}}{2}.$$

Now revert to the original variable $$h$$ using $$x = \dfrac{h}{R}$$:

$$\dfrac{h}{R} = \dfrac{\sqrt{5} - 1}{2} \;\;\Longrightarrow\;\; h = \dfrac{\sqrt{5}\,R - R}{2}.$$

This expression exactly matches Option C.

Hence, the correct answer is Option C.