A semicircular sheet of metal of radius 14 cm is bent to form an open conical cup of the largest size. Find the capacity of the cup. (Take $\pi = 3\dfrac{1}{7}$)

Semicircular sheet of metal of radius 14 cm is shown in the figure below:

On bending the semicircular sheet in the form of cone,

Slant height of cone = Radius of sheet (r) = 14 cm

Also, the circumference of base of cone = Circumference of semi-circular sheet.

Let radius of cone be R cm and height be h cm.

⇒ 2πR = πr

⇒ r = 2R

⇒ R = $\dfrac{r}{2} = \dfrac{14}{2}$ = 7 cm.

By formula,

⇒ (Slant height)² = Height² + Radius²

⇒ l² = h² + R²

⇒ 14² = h² + 7²

⇒ 196 = h² + 49

⇒ h² = 196 - 49

⇒ h² = 147

⇒ h = $\sqrt{147}$ = 12.124 cm

Volume of cone = $\dfrac{1}{3}πR^2h$

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 7^2 \times 12.124 \\[1em] = \dfrac{22 \times 7}{3} \times 12.124 \\[1em] = \dfrac{1867.096}{3} \\[1em] = 622.37 \text{ cm}^3.$

Hence, capacity of cup = 622.37 cm³.