Question
A solid metallic right circular cylinder of base diameter 16 m and height 2 cm is melted and recast into a right circular cone of height three times that of the cylinder. Find the curved surface area of the cone. [Use $$\pi $$ = 3.14]
Answer: Option C
Let the radius of the cone be r cm
Then,
$$\eqalign{
& \pi \times {\left( 8 \right)^2} \times 2 = \frac{1}{3} \times \pi \times {r^2} \times 6 \cr
& \Rightarrow r = 8 \cr} $$
Slant height,
$$\eqalign{
& l = \sqrt {{r^2} + {h^2}} \cr
& \,\,\, = \sqrt {{8^2} + {6^2}} \cr
& \,\,\, = \sqrt {100} \cr
& \,\,\, = 10\,cm \cr} $$
Curved surface area of cone :
$$\eqalign{
& = \pi rl \cr
& = \left( {3.14 \times 8 \times 10} \right){\text{ c}}{{\text{m}}^2} \cr
& = 251.2{\text{ c}}{{\text{m}}^2} \cr} $$
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Let the radius of the cone be r cm
Then,
$$\eqalign{
& \pi \times {\left( 8 \right)^2} \times 2 = \frac{1}{3} \times \pi \times {r^2} \times 6 \cr
& \Rightarrow r = 8 \cr} $$
Slant height,
$$\eqalign{
& l = \sqrt {{r^2} + {h^2}} \cr
& \,\,\, = \sqrt {{8^2} + {6^2}} \cr
& \,\,\, = \sqrt {100} \cr
& \,\,\, = 10\,cm \cr} $$
Curved surface area of cone :
$$\eqalign{
& = \pi rl \cr
& = \left( {3.14 \times 8 \times 10} \right){\text{ c}}{{\text{m}}^2} \cr
& = 251.2{\text{ c}}{{\text{m}}^2} \cr} $$
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