What happens when a cone is immersed in a cylinder?

Question

A cylindrical container of base radius 8 cm is filled with water to a depth of 10 cm. (The height of the container is more than 10 cm). A solid right circular cone of base radius 6 cm and height h cm is immersed into the water. The water in the container covers the cone completely, with its vertex exactly below the water surface.
1. Find the value of h.
2. Find the total surface area of the cone.

Answer 1

Volume of water = πR²h = π (8cm)² x 10 cm = 640 π

Volume of cone = 1/3 xπ r²H =1/3 x π x (6 cm)² x H

∴ volume of water in cylinder when the cone is inside = πR²H- Volume of cone

= π x (8cm)² x H - 1/3 x π x (6 cm)² x H = 52πH

⇒ 640π =52πH

⇒ H = 640/52 = 12.31 cm

 

The total surface area of the cone = πr² + πrL

where the slant height L = √(H² + r²)

 

I leave this last part as an exercise for you to do.

 

Answer 2

1) To solve this, look at the problem as a sum of volumes.

Volume of water = π*(radius of cylinder)^2*(height of water) = 640*π cm^3 ≈ 2010 cm^3.

Volume of water with cone = π*(radius of cylinder)^2*(height of cone) = 201*h cm^3 (h is unknown).

Volume of cone = Volume of water with cone - Volume of water: 201*h - 2010.

We know that the volume of a cone is π/3*(radius of cone)^2*(height of cone) = 37.7*h.

Thus:

37.7*h = 201*h - 2010

=> 163.3*h = 2010

=> h = 12.3 cm.

 

2) Now that we have the height of the cone, we can simply use the surface area formula:

Surface area of cone = π*r^2 + π*r*l. Note that l is the length of the side of the cone, which is simply (using pythagorean's theorem) l = sqrt(r^2 + h^2). Thus the Surface area is π*r^2 + π*r*sqrt(r^2+h^2) = 370.9 cm^2.

 

UPDATE:

I accidentally reversed the volumes earlier, giving an incorrect height. I updated to the correct answer now.