If an object is having same weight at same distance above and below the surface of earth, find its distance from surface of earth.

The correct option is (√5-1)R/2.

Let $R$ be the radius of the Earth, and $h$ be the distance above and below the surface of the Earth where the weight of the object is the same.

The weight of an object of mass $m$ at a height $h$ above the Earth's surface is given by:
$W_{\text{above}}=m{g}_h=\frac{mg{R}^2}{{(R+h)}^2}$
where $g$ is the acceleration due to gravity at the Earth's surface.

The weight of the object at a depth $d=h$ below the Earth's surface is given by:
$W_{\text{below}}=m{g}_d=mg(1-\frac{h}{R})=\frac{mg(R-h)}{R}$

According to the problem statement, the weight of the object is the same at both locations:
$W_{\text{above}}=W_{\text{below}}$

Substituting the expressions for weight:
$\frac{mg{R}^2}{{(R+h)}^2}=\frac{mg(R-h)}{R}$

Canceling $mg$ from both sides:
$\frac{R^2}{{(R+h)}^2}=\frac{R-h}{R}$

Cross-multiplying to solve for $h$:
$R^3=(R-h){(R+h)}^2$

Expanding the right-hand side of the equation:
$R^3=(R-h)(R^2+2Rh+h^2)$
$R^3=R^3+2R^{2}h+R{h}^2-R^{2}h-2R{h}^2-h^3$
$R^3=R^3+R^{2}h-R{h}^2-h^3$

Subtracting $R^3$ from both sides yields:
$0=R^{2}h-R{h}^2-h^3$

Since the distance $h\ne 0$, we can divide the entire equation by $h$:
$0=R^2-Rh-h^2$
$h^2+Rh-R^2=0$

Applying the quadratic formula to solve for $h$:
$h=\frac{-R\pm \sqrt{R^2-4\left(1\right)(-R^2)}}{2}$
$h=\frac{-R\pm \sqrt{R^2+4R^2}}{2}$
$h=\frac{-R\pm \sqrt{5R^2}}{2}$
$h=\frac{-R\pm \sqrt{5}R}{2}$

Since the physical distance $h$ must be a positive quantity, we discard the negative root:
$h=\frac{(\sqrt{5}-1)R}{2}$