A semicircular sheet of paper of diameter 28 cm is bent to cover the exterior surface of an open conical ice-cream cup. The depth of the ice-cream cup is :

Slant height of the cup, l = Radius of sheet = 14 cm
Circumference of the base :
= Circumference of the paper sheet
= $$\left( {\frac{{22}}{7} \times 14} \right)$$  cm
= 44 cm
Let the radius of the base of the cone be r cm
$$\eqalign{ & \therefore 2\pi r = 44 \cr & \Rightarrow r = \frac{{44 \times 7}}{{2 \times 22}} \cr & \Rightarrow r = 7 \cr & {\text{Height, h:}} \cr & = \sqrt {{l^2} - {r^2}} \cr & = \sqrt {{{\left( {14} \right)}^2} - {{\left( 7 \right)}^2}} \cr & = \sqrt {147} {\text{ cm}} \cr & = 7\sqrt 3 {\text{ cm}} \cr & = 12.12{\text{ cm}} \cr} $$