What is the inner radius of the circular opening of the pot so formed?

Given:

Inner radius of the hollow sphere (R) = 20 cm

Height of the pot (h) = 30 cm

Calculations:

Let 'r' be the inner radius of the circular opening of the pot.

Draw a vertical line from the center of the sphere to the center of the circular opening. This line segment will be perpendicular to the plane of the circular opening.

Let the center of the sphere be O. Let the center of the circular opening be C'.

The radius of the sphere (R) goes from O to any point on the surface of the sphere, including the edge of the circular opening.

Consider a right-angled triangle formed by:

The radius of the sphere (R) as the hypotenuse, from the center of the sphere to any point on the edge of the circular opening.

d = |R - h| = |20 - 30| = |-10| = 10 cm.

Now, substitute the values into the Pythagorean theorem:

R² = r² + d²

20² = r² + 10²

400 = r² + 100

r² = 400 - 100

r² = 300

r = \(\sqrt{300}\)

r = \(\sqrt{100 \times 3}\)

r = 10\(\sqrt{3}\) cm

∴ The inner radius of the circular opening of the pot is 10\(\sqrt{3}\) cm.