Question

The Classical Bead Problem A round hole is drilled through the center of a spherical solid of radius \(r .\) The resulting cylindrical hole has height 4 \(\mathrm{cm} .\) (a) What is the volume of the solid that remains? (b) What is unusual about the answer?

Short answer

The volume of the solid that remains is \(V= \frac{4}{3} \pi r^3 - \pi r^2 h\). The unusual thing about this problem is that the volume of the solid remaining doesn't depend on the radius of the original sphere, but only on the height of the cylindrical hole drilled.

Step-by-step solution

Determine the volume of the sphere

The formula for the volume of a sphere is given by \(V = \frac{4}{3} \pi r^3\). Here, r is the radius of the sphere. Plug the value of the radius into the formula to get the volume of the sphere.

Determine the radius of the cylindrical hole

The height of the cylindrical hole is same as the diameter of the original sphere, which is 2r. Since the hole is drilled through the center of the sphere, the diameter of the hole is also 2r. Therefore, the radius of the cylindrical hole is \(r=\frac{height}{2} =2cm\).

Determine the volume of the cylinder

The formula for the volume of a cylinder is given by \(V= \pi r^2 h\). Here h is height of the cylinder and r is radius of the base of the cylinder. Substitute the height and radius of the cylindrical hole to get the volume of the cylinder.

Calculate the volume of the remaining solid

The volume of the remaining solid can be calculated by subtracting the volume of the hole (cylinder) from the volume of the solid sphere.