A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 metres and its volume is two-third of the hemisphere. Calculate the height of the cone and the surface area of the buoy, correct to two places of decimal.

From figure,

Radius of hemisphere (r) = Radius of cone (R) = 3.5 m.

Volume of hemisphere = $\dfrac{2}{3}πr^3$

= $\dfrac{2}{3} \times \dfrac{22}{7} \times (3.5)^3$

= $\dfrac{2}{3} \times 22 \times 0.5 \times 3.5 \times 3.5$

= $\dfrac{269.5}{3}$ m³.

Let height of the cone = h.

By formula,

Volume of conical part = $\dfrac{1}{3}πR^2h$ ……..(1)

Given,

Volume of cone = $\dfrac{2}{3}$ Volume of hemisphere

= $\dfrac{2}{3} \times \dfrac{269.5}{3} = \dfrac{539}{9}$ m³ ………..(2)

From (1) and (2), we get :

$\Rightarrow \dfrac{1}{3}πR^2h = \dfrac{539}{9} \\[1em] \Rightarrow \dfrac{1}{3} \times \dfrac{22}{7} \times (3.5)^2 \times h = \dfrac{539}{9} \\[1em] \Rightarrow h = \dfrac{539 \times 7 \times 3}{9 \times (3.5)^2 \times 22} \\[1em] \Rightarrow h = \dfrac{11319}{2425.5} \\[1em] \Rightarrow h = 4.67 \text{ m}.$

By formula,

⇒ l² = R² + h²

⇒ l² = (3.5)² + (4.67)²

⇒ l² = 12.25 + 21.81

⇒ l² = 34.06

⇒ l = $\sqrt{34.06}$

⇒ l = 5.83 cm.

Surface area of buoy = Surface area of hemisphere + Surface area of cone = 2πr² + πRl

$= 2 \times \dfrac{22}{7} \times (3.5)^2 + \dfrac{22}{7} \times 3.5 \times 5.83 \\[1em] = 2 \times 22 \times 0.5 \times 3.5 + 22 \times 0.5 \times 5.83 \\[1em] = 77 + 64.13 \\[1em] = 141.13 \text{ m}^2.$

Hence, height = 4.67 and surface area of buoy = 141.13 m².