Surface Area of Original Sphere

The surface area of the original solid ball of radius $R = 10\text{ cm}$ is given by the formula for the surface area of a sphere: $S_1 = 4\pi R^2$.

$S_1 = 4\pi (10)^2 = 400\pi$

Surface Area of the Hollowed Ball

When a cylindrical hole of radius $r = 1\text{ cm}$ is drilled through the centre, two circular ends of area $\pi r^2$ each are removed from the outer surface, and a new inner cylindrical curved surface of height $2R$ (equal to the diameter of the ball) is created.

$S_2 = 4\pi R^2 - 2(\pi r^2) + 2\pi r (2R)$

Substitute the Values

Substitute $R = 10$ and $r = 1$ into the expression for the new surface area $S_2$: $S_2 = 4\pi(10)^2 - 2\pi(1)^2 + 2\pi(1)(2\times 10) = 400\pi - 2\pi + 40\pi = 438\pi$.

$S_2 = 400\pi - 2\pi + 40\pi = 438\pi$

Calculate the Ratio

Find the ratio of the total surface area of the hollowed ball to the surface area of the original ball: $\text{Ratio} = \frac{S_2}{S_1} = \frac{438\pi}{400\pi} = \frac{219}{200}$.

$\text{Ratio} = \frac{438\pi}{400\pi} = \frac{219}{200}$