Question

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 m 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 decimal places.

Answer 1

Radius of the hemisphere part (r) = 3.5 m = 7/2 m

Volume of hemisphere

\(\begin{array}{*{35}{l}} =\text{ }2/3\text{ }\pi {{r}^{3}}  \\ =\text{ }2/3\text{ x }22/7\text{ x }7/2x\text{ }7/2\text{ x }7/2  \\ =\text{ }539/6\text{ }{{m}^{3}}  \\ \end{array}\)

Volume of conical part = 2/3 x 539/6 m³ [2/3 ^rd of the hemisphere]

Let height of the cone

\(\begin{array}{*{35}{l}} =\text{ }h  \\ 1/3\text{ }\pi {{r}^{2~}}h\text{ }=\text{ }\left( 2\text{ x }539 \right)/\text{ }\left( 3\text{ x }6 \right)  \\ 1/3\text{ x }22/7\text{ x }7/2\text{ x }7/2\text{ x }h\text{ }=\text{ }\left( 2\text{ x }539 \right)/\text{ }\left( 3\text{ x }6 \right)  \\ h\text{ }=\text{ }\left( 2\text{ x }539\text{ x }2\text{ x }2\text{ x }7\text{ x }3 \right)/\text{ }\left( 3\text{ x }6\text{ x }22\text{ x }7\text{ x }7 \right)  \\ h\text{ }=\text{ }14/3\text{ }m\text{ }=\text{ }4.67\text{ }m  \\ \end{array}\)

Height of the cone = 4.67 m

\(\begin{array}{*{35}{l}} Surface\text{ }area\text{ }of\text{ }buoy\text{ }=\text{ }2\text{ }\pi {{r}^{2}}~+\text{ }\pi rl  \\ l\text{ }=\text{ }{{\left( {{r}^{2~}}+\text{ }{{h}^{2}} \right)}^{1/2}}  \\ l\text{ }=\text{ }{{\left( {{\left( 7/2 \right)}^{2~}}+\text{ }{{\left( 14/3 \right)}^{2}} \right)}^{1/2}}  \\ l\text{ }=\text{ }{{\left( 1225/36 \right)}^{1/2}}~=\text{ }35/6\text{ }m  \\ Surface\text{ }area\text{ }=\text{ }2\pi {{r}^{2}}~+\text{ }\pi rl  \\ =\text{ }\left( 2\text{ x }22/7\text{ x }7/2\text{ x }7/2 \right)\text{ }+\text{ }\left( 22/7\text{ x }7/2\text{ x }35/6 \right)  \\ =\text{ }77\text{ }+\text{ }385/6\text{ }=\text{ }847/6  \\ =\text{ }141.17\text{ }{{m}^{2}}  \\ \end{array}\)