A semicircle sheet of metal of …

Given the diameter of semicircular sheet = 28 cm, radius of the semicircular sheet

= {tex}\left(\frac{1}{2} \times 28\right){/tex}cm = 14 cm
Length of the arc of semicircular sheet
= {tex}\frac{1}{2} \times 2 \pi r=\pi r{/tex}
= {tex}\left(\frac{22}{7} \times 14\right){/tex}cm = 44 cm
Let r cm be the radius of the cone, then
2nr = 44 {tex}\Rightarrow 2 \times \frac{22}{7} \times r=44 \Rightarrow{/tex} r = 7
Slant height of the cone = radius of semicircular sheet = 14 cm.
Let h cm be the height of the cone, then
p = h² + r² {tex}\Rightarrow{/tex} 14² = h² + 7²
{tex}\Rightarrow{/tex} h² = 196 - 49 = 147 {tex}\Rightarrow{/tex} h = {tex}\sqrt{147}=7 \sqrt{3}{/tex}
{tex}\therefore{/tex} The capacity of the conical cup = volume of cone = {tex}\frac{1}{3} \pi r^{2} h{/tex}
= {tex}\left(\frac{1}{3} \times \frac{22}{7} \times 7^{2} \times 7 \sqrt{3}\right) \mathrm{cm}^{3}=\frac{1078 \sqrt{3}}{3} \mathrm{cm}^{3}{/tex}
= 622.38 cm³ (approx)