Question : Three metallic spheres of radii 10 cm, 8 cm, and 6 cm, respectively are melted to form a single solid cone of radius 12 cm. Find the curved surface area of the cone, correct to two places of decimal. (Take $\pi=3.14$)
Option 1: $1664.50 \text{ cm}^2$
Option 2: $1669.86 \text{ cm}^2$
Option 3: $1876.79 \text{ cm}^2$
Option 4: $1864.41 \text{ cm}^2$
Correct Answer: $1864.41 \text{ cm}^2$
Solution : The volume of a sphere, where $r$ is the radius. $V = \frac{4}{3}\pi r^3$ The volume of a cone, where $r$ is the radius and $h$ is the height. $V = \frac{1}{3}\pi r^2 h$ The volume of the three spheres is equal to the volume of the cone since the spheres are melted to form the cone. $⇒\frac{4}{3}\pi (10^3 + 8^3 + 6^3) = \frac{1}{3}\pi r^2 h$ $⇒\frac{4}{3}\pi (10^3 + 8^3 + 6^3) = \frac{1}{3}\pi (12)^2 h$ $⇒ 4(10^3 + 8^3 + 6^3) = (12)^2 h$ $⇒ 6912= 144 h$ $⇒h = 48 \text{ cm}$ Now, the slant height, $l = \sqrt{r^2 + h^2}= \sqrt{(12)^2 + (48)^2} = 49.477 \text{ cm}$ The curved surface area of a cone, where $l$ is the slant height of the cone. $= \pi r l= \pi ×(12)× (49.477)=1864.41 \text{ cm}^2$ Hence, the correct answer is $1864.41 \text{ cm}^2$.
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