m Example : A juice seller was serving his customers using glasses as shown in Fig. . . The inner diameter of the cylindrical glass was cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass.
If the height of a glass was cm, find the apparent capacity of the glass and its actual capacity. (Use = . .) Solution : Since the inner diameter of the glass = cm and height = cm, the apparent capacity of the glass = r h = . × .
× . × cm = . cm But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass. i.e., it is less by r = .
cm So, the actual capacity of the glass = apparent capacity of glass – volume of the hemisphere = ( . – . ) cm = . cm Fig.
. Example : A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is cm and the diameter of the base is cm. Determine the volume of the toy.
If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Take = . ) Solution : Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere (see Fig. .
). The radius BO of the hemisphere (as well as of the cone) = × cm = cm. So, volume of the toy = r h . ( ) .
( ) cm = . cm Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of the base of the right circular