Let r cm and R cm be the radius of the semi-circular sheet and base of the conical cup respectively.
Suppose the depth of the conical cup is H cm.
Given, 2r = 28 cm
⇒ r = 14 cm
When the semi-circular sheet of metal is bent into an open conical cup, then
Slant height of the cone, L = Radius of the semi-circular sheet = 14 cm
Circumference of base of cone = 2πR
∴ 2π R = 14π cm
⇒ 2R = 14 cm
⇒ R = 7 cm
Slant height of the cone, L = 14 cm
⇒ 49 cm2 + H 2 = (14 cm)2 = 196 cm2
⇒ H 2 = 196 cm2 – 49 cm2 = 147 cm2
Capacity or volume of the conical cup =