The height of the point vertically above the earth's surface, at which acceleration due to gravity becomes 1% of its value at the surface is (Radius of the earth = R)

The correct answer is 9R.

Step-by-step explanation:

Let $g$ be the acceleration due to gravity at the surface of the Earth, and $g_h$ be the acceleration due to gravity at a height $h$ above the Earth's surface.

The variation of acceleration due to gravity with height $h$ is given by the formula:
$g_h=g{\left(\frac{R}{R+h}\right)}^2$
where $R$ is the radius of the Earth.

We are given that the acceleration due to gravity at height $h$ is 1% of its value at the surface:
$g_h=1\%\text{of}g=\frac{g}{100}$

Substitute this into the formula:
$\frac{g}{100}=g{\left(\frac{R}{R+h}\right)}^2$

Divide both sides by $g$:
$\frac{1}{100}={\left(\frac{R}{R+h}\right)}^2$

Take the square root on both sides:
$\frac{1}{10}=\frac{R}{R+h}$

Cross-multiply to solve for $h$:
$R+h=10R$

Subtract $R$ from both sides:
$h=10R-R$
$h=9R$

Therefore, the height of the point vertically above the earth's surface is 9R.