A semi circular sheet of paper …

Diameter of semicircular sheet is 28 cm. It is bent to from an open conical cup. The radius of sheet becomes the slant height of the cup. The circumference of the sheet becomes the circumference of the base of the cone.
{tex}\therefore \;l = {/tex} Slant height of conical cup = 14 cm.
Let r cm be the radius and h cm be the height of the conical cup circumference of conical cup of the semicircular sheet
{tex}\therefore \;2\pi r = \pi \times 14 \Rightarrow r = 7cm{/tex} 
Now, {tex}{l^2} = {r^2} + {h^2} \Rightarrow h = \sqrt {{l^2} - {r^2}}{/tex}
{tex}= \sqrt {{{(14)}^2} - {{(7)}^2}} = \sqrt {196 - 49}{/tex}{tex}= \sqrt {147} = 12.12cm{/tex}
{tex}\therefore{/tex} Capacity of the cup
{tex} = \frac{1}{3}\pi {r^2}h = \frac{1}{3} \times \frac{{22}}{7} \times 7 \times 7 \times 12.12{/tex}
= 622.16 cm³