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 $\dfrac{2}{3}$ of the hemisphere. Calculate the height of the cone and the surface area of the buoy correct to 2 places of decimal.

The figure of the buoy made by surmounting a right cone on a hemisphere is shown below:

Radius of base of hemisphere = Radius of cone = 3.5 m.

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

Putting values,

$V = \dfrac{2}{3} \times \dfrac{22}{7} \times \Big(3.5)^3 \\[1em] = \dfrac{44}{21} \times 42.875 \\[1em] = \dfrac{44 \times 42.875}{21} \\[1em] = \dfrac{1886.5}{21} \\[1em] = 89.8 \text{ m}^3$

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

∴ Volume of cone = $\dfrac{2}{3} \times 89.8$

$= \dfrac{179.6}{3} \\[1em] = 59.87 \text{ m}^3.$

Volume of cone = $\dfrac{1}{3}πr^2h$.

$\therefore \dfrac{1}{3}πr^2h = 59.87 \\[1em] \Rightarrow \dfrac{1}{3} \times \dfrac{22}{7} \times (3.5)^2 \times h = 59.87 \\[1em] \Rightarrow \dfrac{22}{21} \times 12.25 \times h = 59.87 \\[1em] \Rightarrow h = \dfrac{59.87 \times 21}{22 \times 12.25} \\[1em] \Rightarrow h = \dfrac{1257.27}{269.5} \\[1em] \Rightarrow h = 4.67 m.$

Slant height of cone = l = $\sqrt{h^2 + r^2}$

Putting values we get,

$l = \sqrt{(4.67)^2 + (3.5)^2} \\[1em] l = \sqrt{21.81 + 12.25} \\[1em] l = \sqrt{34.06} \\[1em] l = 5.84 \text{ m}$

Surface area of the buoy = Curved Surface area of cone + Curved Surface area of hemisphere = πrl + 2πr².

∴ Surface area of buoy = $πr(l + 2r)$

$= \dfrac{22}{7} \times 3.5 \times (5.84 + 2 \times 3.5) \\[1em] = \dfrac{22}{7} \times 3.5 \times (5.84 + 7) \\[1em] = \dfrac{22}{7} \times 3.5 \times 12.84 \\[1em] = \dfrac{988.68}{7} \\[1em] = 141.17 \text{ m}^2.$

Hence, the height of cone = 4.67 m and surface area of buoy is 141.17 m².